NCERT Class 10 Science Solutions: Electricity

Question:

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio $RR’$ is ______.

A. $125$

B. $15$

C. 5

D. 25

Concept in a Minute:

This question involves understanding the relationship between the resistance of a wire, how it changes when cut into equal parts, and how to calculate equivalent resistance when components are connected in parallel. Key concepts are:
1. Resistance and Length: Resistance of a wire is directly proportional to its length (R ∝ L).
2. Parallel Combinations: The reciprocal of the equivalent resistance (1/R_eq) of resistors connected in parallel is equal to the sum of the reciprocals of their individual resistances.

Explanation:

Let the original resistance of the wire be R.
The wire is cut into five equal parts. This means each part will have a resistance of R/5.
Let the resistance of each of these five parts be r. So, r = R/5.

These five parts are then connected in parallel.
The formula for equivalent resistance (R’) of n resistors connected in parallel is:
1/R’ = 1/r1 + 1/r2 + 1/r3 + 1/r4 + 1/r5

Since all five parts have equal resistance (r = R/5), we can write:
1/R’ = 1/r + 1/r + 1/r + 1/r + 1/r
1/R’ = 5/r

Now, substitute the value of r = R/5 into the equation:
1/R’ = 5 / (R/5)
1/R’ = 5 * (5/R)
1/R’ = 25/R

To find R’, we take the reciprocal of both sides:
R’ = R/25

The question asks for the ratio RR’. This means we need to calculate R / R’.
R / R’ = R / (R/25)
R / R’ = R * (25/R)
R / R’ = 25

Therefore, the ratio RR’ is 25.

The final answer is $\boxed{25}$.

Question:

Which of the following terms does not represent electrical power in a circuit?

A. I 2 R

B. IR 2

C. VI

D. $V2R$

Concept in a Minute:

Electrical power is the rate at which electrical energy is transferred or dissipated in a circuit. Key formulas for electrical power include P = VI, P = I^2R, and P = V^2/R. Understanding these relationships is crucial.

Explanation:

The question asks to identify the term that does NOT represent electrical power. Let’s analyze each option based on the standard formulas for electrical power:

A. I^2R: This is a valid formula for electrical power, representing power dissipated in a resistor when the current (I) and resistance (R) are known. P = I^2R.

B. IR^2: This expression represents the product of current (I) and the square of resistance (R^2). This combination does not correspond to any of the standard formulas for electrical power. For example, if we try to manipulate the power formulas, we cannot arrive at IR^2.

C. VI: This is a fundamental formula for electrical power, representing the product of voltage (V) and current (I). P = VI.

D. V^2/R: This is another valid formula for electrical power, derived from P = VI and Ohm’s Law (V=IR). Substituting V=IR into P=VI gives P = (IR)I = I^2R. Substituting I=V/R into P=VI gives P = V(V/R) = V^2/R.

Therefore, IR^2 does not represent electrical power.

Final Answer: The final answer is $\boxed{B}$

Question:

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be:

A. 1 : 2

B. 2 : 1

C. 1 : 4

D. 4 : 1

Concept in a Minute:

Electrical resistance, Joule’s law of heating, series and parallel combinations of resistors.

Explanation:

Let the resistance of each wire be R.
When connected in series, the equivalent resistance R_series = R + R = 2R.
When connected in parallel, the equivalent resistance R_parallel = (R * R) / (R + R) = R^2 / 2R = R/2.
According to Joule’s law of heating, heat produced H is proportional to V^2 / R (where V is the potential difference and R is the resistance). Since the potential difference is the same across both combinations, the heat produced is inversely proportional to the equivalent resistance.
Heat produced in series, H_series is proportional to 1 / R_series = 1 / (2R).
Heat produced in parallel, H_parallel is proportional to 1 / R_parallel = 1 / (R/2) = 2 / R.
The ratio of heat produced in series and parallel combinations is:
H_series / H_parallel = (1 / 2R) / (2 / R) = (1 / 2R) * (R / 2) = 1 / 4.
So the ratio is 1 : 4.

The final answer is $\boxed{C}$.

Question:

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be ______.

A. 100 W

B. 75 W

C. 50 W

D. 25 W

Concept in a Minute:

The power consumed by an electrical device is related to its voltage rating, power rating, and the actual voltage at which it is operated. The resistance of the device remains constant.

Explanation:

The rating of the electric bulb (220 V, 100 W) tells us the power it consumes when operated at its rated voltage. We can use this information to calculate the resistance of the bulb.
The formula for power is P = V^2 / R, where P is power, V is voltage, and R is resistance.
From the rating, we have P = 100 W and V = 220 V.
So, R = V^2 / P = (220 V)^2 / 100 W = 48400 / 100 = 484 ohms.
Now, when the bulb is operated at 110 V, its resistance remains the same (484 ohms). We can calculate the new power consumed using the same formula:
P_new = V_new^2 / R
P_new = (110 V)^2 / 484 ohms
P_new = 12100 / 484
P_new = 25 W.

The final answer is $\boxed{\text{25 W}}$.

Question:

Judge the equivalent resistance when the following are connected in parallel − (a) 1 Ω and 10⁶Ω, (b) 1 Ω and 10³Ω and 10⁶Ω.

Concept in a Minute:

The equivalent resistance of resistors connected in parallel is always less than the smallest individual resistance. The formula for two resistors in parallel is $R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}$. For more than two resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + …$

Explanation:

(a) When 1 Ω and 10⁶ Ω are connected in parallel, the equivalent resistance will be much smaller than 1 Ω. This is because the current has a very easy path through the 1 Ω resistor, and the 10⁶ Ω resistor offers very little contribution to the overall current flow.
Using the formula for two resistors in parallel:
$R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}$
$R_{eq} = \frac{1 \Omega \times 10^6 \Omega}{1 \Omega + 10^6 \Omega}$
$R_{eq} = \frac{10^6 \Omega}{10^6 + 1 \Omega}$
Since $10^6 + 1$ is approximately equal to $10^6$, the equivalent resistance is approximately $\frac{10^6 \Omega}{10^6} = 1 \Omega$. More precisely, it will be slightly less than 1 Ω.

(b) When 1 Ω, 10³ Ω, and 10⁶ Ω are connected in parallel, the equivalent resistance will again be less than the smallest resistance, which is 1 Ω. The 1 Ω resistor will dominate the current flow.
Using the formula for three resistors in parallel:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
$\frac{1}{R_{eq}} = \frac{1}{1 \Omega} + \frac{1}{10^3 \Omega} + \frac{1}{10^6 \Omega}$
$\frac{1}{R_{eq}} = 1 + 0.001 + 0.000001 \Omega^{-1}$
$\frac{1}{R_{eq}} = 1.001001 \Omega^{-1}$
$R_{eq} = \frac{1}{1.001001} \Omega$
$R_{eq} \approx 0.999 \Omega$
Again, the equivalent resistance is slightly less than 1 Ω, dominated by the 1 Ω resistor.

Question:

Why is the series arrangement not used for domestic circuits?

Concept in a Minute:

Understanding of series and parallel circuits.
Properties of series circuits:
– Current is the same through all components.
– Voltage is divided among components.
– If one component fails, the entire circuit breaks.
Properties of parallel circuits:
– Voltage is the same across all components.
– Current is divided among branches.
– If one component fails, other components continue to work.
Domestic circuits aim to provide consistent voltage to multiple appliances and ensure independent operation.

Explanation:

The series arrangement is not used for domestic circuits primarily because of two major drawbacks:

1. Voltage Division: In a series circuit, the total voltage supplied by the source is divided among the components connected in series. This means that each appliance would receive only a fraction of the main voltage, which would likely be insufficient for them to operate correctly or at their intended power. For example, if you connect a bulb, a fan, and a heater in series, each would get a reduced voltage, and they might not even turn on.

2. Interdependence of Appliances: In a series circuit, if one appliance gets damaged or its circuit breaks (e.g., a bulb filament breaks), the entire circuit is broken, and all other appliances connected in series would also stop working. This is highly inconvenient and impractical for domestic use where you would want other appliances to continue functioning even if one fails.

Domestic circuits are designed to provide the full mains voltage to each appliance independently. This is achieved by connecting appliances in parallel. In a parallel arrangement, each appliance receives the full mains voltage, and if one appliance fails, the others continue to operate because they are on separate branches of the circuit.

Question:

Use the data in the Table given below to answer the following –

Which material is the best conductor?

Table give below Electrical resistivity of some substances at 20°C

Electrical resistivity of some substances at 20°C
−	Material	Resistivity (Ω m)
Conductors	Silver	1.60 × 10⁻⁸
	Copper	1.62 × 10⁻⁸
	Aluminium	2.63 × 10⁻⁸
	Tungsten	5.20 × 10⁻⁸
	Nickel	6.84 × 10⁻⁸
	Iron	10.0 × 10⁻⁸
	Chromium	12.9 × 10⁻⁸
	Mercury	94.0 × 10⁻⁸
	Manganese	1.84 × 10⁻⁶
Alloys	Constantan (alloy of Cu and Ni)	49 × 10⁻⁶
	Manganin (alloy of Cu, Mn and Ni)	44 × 10⁻⁶
	Nichrome (alloy of Ni, Cr, Mn and Fe)	100 × 10⁻⁶
Insulators	Glass	10¹⁰ − 10¹⁴
	Hard rubber	10¹³ − 10¹⁶
	Ebonite	10¹⁵ − 10¹⁷
	Diamond	10¹² − 10¹³
	Paper (dry)	10¹²

Concept in a Minute:

Electrical conductivity is the inverse of electrical resistivity. A material with lower electrical resistivity will have higher electrical conductivity, meaning it allows electric current to flow more easily. Therefore, the best conductor is the material with the lowest electrical resistivity.

Explanation:

The question asks to identify the best conductor from the given table of electrical resistivities. Electrical resistivity is a measure of how strongly a material opposes the flow of electric current. A good conductor is a material that offers very little resistance to the flow of electricity. This means that a good conductor will have a very low electrical resistivity.

Looking at the provided table, we need to find the material with the smallest value in the “Resistivity (Ω m)” column.

Under the “Conductors” category, we have the following resistivities:
Silver: 1.60 × 10⁻⁸ Ω m
Copper: 1.62 × 10⁻⁸ Ω m
Aluminium: 2.63 × 10⁻⁸ Ω m
Tungsten: 5.20 × 10⁻⁸ Ω m
Nickel: 6.84 × 10⁻⁸ Ω m
Iron: 10.0 × 10⁻⁸ Ω m
Chromium: 12.9 × 10⁻⁸ Ω m
Mercury: 94.0 × 10⁻⁸ Ω m
Manganese: 1.84 × 10⁻⁶ Ω m

Comparing these values, 1.60 × 10⁻⁸ is the smallest resistivity among the conductors listed.

The alloys and insulators have significantly higher resistivities. For example, the lowest resistivity among alloys is 44 × 10⁻⁶, which is much larger than 10⁻⁸. The insulators have resistivities in the range of 10¹⁰ to 10¹⁷, which are extremely high.

Therefore, Silver, with the lowest electrical resistivity of 1.60 × 10⁻⁸ Ω m, is the best conductor among the given materials.

The final answer is $\boxed{Silver}$.

Question:

Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Concept in a Minute:

Ohm’s Law: This law states that the current (I) flowing through a conductor is directly proportional to the potential difference (V) across its ends, provided the temperature and other physical conditions remain unchanged. Mathematically, it is expressed as V = IR, where R is the resistance of the conductor.

Explanation:

We are given that the resistance (R) of the electrical component remains constant.
According to Ohm’s Law, the relationship between potential difference (V), current (I), and resistance (R) is given by:
V = IR

We can rearrange this formula to express current in terms of potential difference and resistance:
I = V/R

Let the initial potential difference be V₁ and the initial current be I₁. So, according to Ohm’s Law:
I₁ = V₁/R

Now, the problem states that the potential difference across the component decreases to half of its former value. Let the new potential difference be V₂. Therefore, V₂ = V₁/2.
Let the new current be I₂. Since the resistance (R) remains constant, we can write the new relationship using Ohm’s Law:
I₂ = V₂/R

Substitute the value of V₂ into the equation for I₂:
I₂ = (V₁/2) / R

We can rewrite this as:
I₂ = (1/2) * (V₁/R)

From the initial condition, we know that I₁ = V₁/R. So, we can substitute I₁ into the equation for I₂:
I₂ = (1/2) * I₁

This shows that the new current (I₂) is half of the original current (I₁).
Therefore, if the potential difference across the component decreases to half of its former value while the resistance remains constant, the current through it will also decrease to half of its former value.

Final Answer: The current through the component will decrease to half of its former value.

Question:

Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Concept in a Minute:

1. Ohm’s Law: V = IR, where V is voltage, I is current, and R is resistance.
2. Power Formula: P = VI, where P is power, V is voltage, and I is current.
3. Parallel Circuits: In a parallel circuit, the total current is the sum of the currents in each branch, and the voltage across each component is the same.

Explanation:

The problem asks us to determine the maximum number of electric bulbs that can be connected in parallel to a 220 V supply line, given that each bulb is rated 10 W and the maximum allowable current in the line is 5 A.

First, let’s calculate the current drawn by a single bulb.
We know that the power (P) of a bulb is 10 W and the voltage (V) it is designed for is 220 V.
Using the power formula, P = VI, we can find the current (I) drawn by one bulb:
I_bulb = P_bulb / V_supply
I_bulb = 10 W / 220 V

Now, let’s calculate this value:
I_bulb = 10 / 220 A
I_bulb = 1 / 22 A

The total maximum current allowed in the 220 V line is given as 5 A.
When bulbs are connected in parallel, the total current drawn from the supply is the sum of the currents drawn by each individual bulb.
Let ‘n’ be the number of bulbs connected in parallel.
The total current (I_total) will be:
I_total = n * I_bulb

We are given that the maximum allowable current is 5 A, so:
I_total <= 5 A

Substituting the expression for I_total:
n * I_bulb <= 5 A

Now, substitute the value of I_bulb:
n * (1 / 22 A) <= 5 A

To find the maximum number of bulbs ‘n’, we can rearrange the inequality:
n <= 5 A / (1 / 22 A)
n <= 5 * 22
n <= 110

Therefore, a maximum of 110 bulbs can be connected in parallel across the two wires of the 220 V line.

The final answer is 110.

Question:

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Concept in a Minute:

The total resistance of resistors connected in series is the sum of individual resistances. The total resistance of resistors connected in parallel is found by taking the reciprocal of the sum of the reciprocals of individual resistances. To achieve the highest total resistance, connect all coils in series. To achieve the lowest total resistance, connect all coils in parallel.

Explanation:

To find the highest total resistance, we need to connect all four coils in series. In a series connection, the resistances add up directly.
Highest resistance (R_series) = R1 + R2 + R3 + R4
R_series = 4 Ω + 8 Ω + 12 Ω + 24 Ω

To find the lowest total resistance, we need to connect all four coils in parallel. In a parallel connection, the reciprocal of the total resistance is the sum of the reciprocals of the individual resistances.
1/R_parallel = 1/R1 + 1/R2 + 1/R3 + 1/R4
1/R_parallel = 1/4 Ω + 1/8 Ω + 1/12 Ω + 1/24 Ω

Detailed Solution:
(a) Highest total resistance:
The highest total resistance is obtained when the coils are connected in series.
R_highest = R1 + R2 + R3 + R4
R_highest = 4 Ω + 8 Ω + 12 Ω + 24 Ω
R_highest = 48 Ω

(b) Lowest total resistance:
The lowest total resistance is obtained when the coils are connected in parallel.
1/R_lowest = 1/R1 + 1/R2 + 1/R3 + 1/R4
1/R_lowest = 1/4 + 1/8 + 1/12 + 1/24
To add these fractions, find a common denominator, which is 24.
1/R_lowest = (6/24) + (3/24) + (2/24) + (1/24)
1/R_lowest = (6 + 3 + 2 + 1) / 24
1/R_lowest = 12 / 24
1/R_lowest = 1 / 2
R_lowest = 2 Ω

Final Answer:
(a) The highest total resistance is 48 Ω.
(b) The lowest total resistance is 2 Ω.

Question:

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Concept in a Minute:

This problem involves understanding the concepts of electrical resistance, Ohm’s law, and the rules for combining resistances in parallel circuits. Specifically, you’ll need to know how to calculate the equivalent resistance of parallel resistors and how to calculate current using Ohm’s law (V = IR).

Explanation:

First, let’s find the total current drawn by the three appliances (lamp, toaster, and filter) when connected in parallel to the 220 V source.
The resistances are R1 = 100 Ω (lamp), R2 = 50 Ω (toaster), and R3 = 500 Ω (filter).
The voltage source is V = 220 V.

1. Calculate the current through each appliance using Ohm’s law (I = V/R):
Current through lamp (I1) = V / R1 = 220 V / 100 Ω = 2.2 A
Current through toaster (I2) = V / R2 = 220 V / 50 Ω = 4.4 A
Current through filter (I3) = V / R3 = 220 V / 500 Ω = 0.44 A

2. Calculate the total current drawn by all three appliances. In a parallel circuit, the total current is the sum of the currents through each branch:
Total current (Itotal) = I1 + I2 + I3 = 2.2 A + 4.4 A + 0.44 A = 7.04 A

Now, we need to find the resistance of an electric iron that draws this same total current (7.04 A) from the same 220 V source.

3. Let the resistance of the electric iron be R_iron and the current through it be I_iron. We are given that I_iron = Itotal = 7.04 A.
Using Ohm’s law (R = V/I) for the electric iron:
R_iron = V / I_iron = 220 V / 7.04 A

4. Calculate the resistance of the electric iron:
R_iron = 31.25 Ω

Therefore, the resistance of the electric iron is 31.25 Ω, and the current through it is 7.04 A.

Question:

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Concept in a Minute:

Electrical resistance, resistivity, melting point, oxidation. Alloys have higher resistivity than pure metals, allowing them to heat up efficiently without melting at high temperatures. Alloys also form a protective oxide layer at high temperatures, preventing corrosion.

Explanation:

Coils of electric toasters and electric irons are made of alloys like Nichrome (an alloy of nickel and chromium) instead of pure metals for two primary reasons:

1. Higher Electrical Resistance: Alloys generally have a higher electrical resistance (and thus higher resistivity) compared to their constituent pure metals. When electric current flows through a resistor, electrical energy is converted into heat energy. A higher resistance means more heat is generated for the same current and length of wire, making the heating element more efficient. Pure metals like copper or aluminum have very low resistance, so they would not generate enough heat to toast bread or iron clothes effectively.

2. Higher Melting Point and Oxidation Resistance: Electric heating elements operate at very high temperatures. Pure metals like iron or aluminum have relatively lower melting points and tend to oxidize (rust) rapidly at these high temperatures. Alloys like Nichrome have a significantly higher melting point, preventing the coil from melting during operation. Furthermore, at high temperatures, Nichrome forms a stable, protective layer of chromium oxide on its surface. This oxide layer prevents further oxidation and corrosion of the underlying alloy, extending the lifespan of the heating element.

Question:

Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Concept in a Minute:

When electrical appliances are connected in parallel, the voltage across each appliance is the same as the supply voltage. The total current drawn from the supply is the sum of the currents drawn by each individual appliance. The power (P), voltage (V), and current (I) are related by the formula P = VI.

Explanation:

The problem involves two lamps connected in parallel to an electric mains supply. We are given the power rating and voltage rating for each lamp, and the supply voltage. We need to find the total current drawn from the line.

Step 1: Calculate the current drawn by the first lamp.
We are given that the first lamp is rated 100 W at 220 V.
Using the formula P = VI, we can find the current drawn by this lamp (I1).
I1 = P1 / V
I1 = 100 W / 220 V

Step 2: Calculate the current drawn by the second lamp.
The second lamp is rated 60 W at 220 V.
Using the formula P = VI, we can find the current drawn by this lamp (I2).
I2 = P2 / V
I2 = 60 W / 220 V

Step 3: Calculate the total current drawn from the line.
Since the lamps are connected in parallel, the total current drawn from the line (Itotal) is the sum of the currents drawn by each lamp.
Itotal = I1 + I2

Step 4: Substitute the values and calculate the total current.
Itotal = (100 W / 220 V) + (60 W / 220 V)
Itotal = (100 + 60) W / 220 V
Itotal = 160 W / 220 V

Step 5: Simplify the fraction to get the final answer.
Itotal = 160 / 220 A
Itotal = 16 / 22 A
Itotal = 8 / 11 A

As a decimal, 8/11 A is approximately 0.727 A.

Final Answer: The current drawn from the line is 8/11 A.

Question:

How is a voltmeter connected in the circuit to measure the potential difference between two points? Explain with the help of a diagram.

Concept in a Minute:

Voltmeter: An instrument used to measure the electric potential difference between two points in a circuit.
Series connection: Components are connected end-to-end, forming a single path for current.
Parallel connection: Components are connected across the same two points, providing multiple paths for current.
Potential difference: The difference in electric potential between two points.

Explanation:

A voltmeter is always connected in parallel across the two points in a circuit between which the potential difference is to be measured. This is because current divides in parallel connections, and a voltmeter has a very high resistance. When connected in parallel, only a very small current flows through the voltmeter, ensuring that it does not significantly alter the potential difference it is intended to measure. If a voltmeter were connected in series, it would offer a very high resistance to the circuit, greatly reducing the current and thus affecting the potential difference across other components.

Diagram:

Consider a simple circuit with a battery, a resistor (R), and a switch (S). To measure the potential difference across the resistor R, a voltmeter (V) is connected in parallel with R.

[Insert a simple circuit diagram here.
The diagram should show:
1. A battery symbol.
2. A switch symbol.
3. A resistor symbol labeled ‘R’.
4. A voltmeter symbol labeled ‘V’ connected across the two terminals of the resistor ‘R’.
5. The components connected in series to form a closed loop (except for the voltmeter which is in parallel).]

Question:

How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of 1 Ω ?

Concept in a Minute:

Understanding of series and parallel combinations of resistors.
Series combination: Total resistance R_total = R1 + R2 + …
Parallel combination: 1/R_total = 1/R1 + 1/R2 + …

Explanation:

We need to find a combination of 2 Ω, 3 Ω, and 6 Ω resistors that results in a total resistance of 1 Ω. Let’s explore possible combinations.

Consider connecting the 3 Ω and 6 Ω resistors in parallel. The equivalent resistance (R_parallel_3_6) would be:
1/R_parallel_3_6 = 1/3 Ω + 1/6 Ω
1/R_parallel_3_6 = (2 + 1)/6 Ω
1/R_parallel_3_6 = 3/6 Ω
R_parallel_3_6 = 6/3 Ω = 2 Ω

Now, if we connect this parallel combination of 2 Ω in series with the remaining 2 Ω resistor, the total resistance would be:
R_total = 2 Ω (from parallel combination) + 2 Ω (remaining resistor) = 4 Ω. This is not 1 Ω.

Let’s try connecting the 2 Ω and 3 Ω resistors in parallel first. The equivalent resistance (R_parallel_2_3) would be:
1/R_parallel_2_3 = 1/2 Ω + 1/3 Ω
1/R_parallel_2_3 = (3 + 2)/6 Ω
1/R_parallel_2_3 = 5/6 Ω
R_parallel_2_3 = 6/5 Ω = 1.2 Ω

If we connect this parallel combination of 1.2 Ω in series with the 6 Ω resistor, the total resistance would be:
R_total = 1.2 Ω + 6 Ω = 7.2 Ω. This is not 1 Ω.

Now, let’s consider connecting all three resistors in parallel:
1/R_total = 1/2 Ω + 1/3 Ω + 1/6 Ω
1/R_total = (3 + 2 + 1)/6 Ω
1/R_total = 6/6 Ω
1/R_total = 1 Ω
R_total = 1 Ω

Therefore, connecting all three resistors (2 Ω, 3 Ω, and 6 Ω) in parallel gives a total resistance of 1 Ω.

The final answer is $\boxed{Connect all three resistors in parallel}$.

Question:

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Concept in a Minute:

The key concepts are:
1. Electrical Energy: The work done by the electric field in moving charge through a potential difference is the electrical energy transferred.
2. Heat Generated: When charge flows through a conductor, some electrical energy is dissipated as heat due to the resistance of the conductor. This is given by Joule’s law of heating.
3. Relationship between Charge, Potential Difference, and Energy/Heat: The electrical work done (and hence heat generated) is equal to the product of the charge transferred and the potential difference.

Explanation:

The heat generated (H) when a charge (Q) is transferred through a potential difference (V) is given by the formula:
H = V * Q

We are given:
Charge (Q) = 96000 coulomb
Potential difference (V) = 50 V
Time (t) = 1 hour (although time is not directly needed for this calculation using the V*Q formula, it’s often provided in such problems and can be used if calculating power first).

Using the formula H = V * Q:
H = 50 V * 96000 C
H = 4,800,000 Joules

Alternatively, if we consider power first:
Power (P) = V * I
Current (I) = Q / t
First, convert time to seconds: 1 hour = 1 * 3600 seconds = 3600 s
Current (I) = 96000 C / 3600 s = 26.67 A (approximately)
Power (P) = 50 V * 26.67 A = 1333.5 W (approximately)
Heat (H) = P * t
H = 1333.5 W * 3600 s = 4,800,600 Joules (slight difference due to rounding of current)

The most direct way using the given values is H = V * Q.

Therefore, the heat generated is 4,800,000 Joules.

Question:

On what factors does the resistance of a conductor depend?

Concept in a Minute:

Resistance is a property of a material that opposes the flow of electric current. It is influenced by the intrinsic nature of the material and its physical dimensions.

Explanation:

The resistance (R) of a conductor depends on the following factors:

1. Length (l) of the conductor: Resistance is directly proportional to the length of the conductor. This means that a longer conductor will have more resistance than a shorter one of the same material and thickness. Mathematically, R ∝ l.

2. Area of cross-section (A) of the conductor: Resistance is inversely proportional to the area of the cross-section of the conductor. This means that a thicker conductor (larger cross-sectional area) will have less resistance than a thinner one of the same material and length. Mathematically, R ∝ 1/A.

3. Nature of the material: Different materials have different abilities to conduct electricity. This property is quantified by resistivity (ρ), which is an intrinsic characteristic of the material. For example, copper has lower resistivity than rubber, making it a better conductor. Mathematically, R ∝ ρ.

4. Temperature: The resistance of most metallic conductors increases with an increase in temperature. For some materials like semiconductors, the resistance decreases with an increase in temperature.

Combining the first three factors, the relationship between resistance, length, area of cross-section, and resistivity is given by the formula:
R = ρ (l/A)

Question:

What does an electric circuit mean?

Concept in a Minute:

An electric circuit is a closed, continuous path for electric current to flow. It involves components like a power source, conductors, and a load.

Explanation:

An electric circuit is essentially a complete loop or pathway that allows electric current to travel. Imagine a race track for tiny charged particles (electrons). This track needs to be unbroken and lead back to where it started for the “race” (current flow) to happen continuously.

The essential components of a simple electric circuit are:
1. Power Source: This is what provides the energy to push the charges. Common examples are batteries or mains electricity.
2. Conductors: These are materials that allow electricity to flow easily, like wires made of copper. They form the “track” of the circuit.
3. Load: This is the device that uses the electrical energy. Examples include a light bulb, a motor, or a resistor. It “does work” with the electricity.
4. Switch (optional but common): A switch is a device that can open or close the circuit, allowing you to control whether the current flows or not.

When all these components are connected in a way that forms a continuous, unbroken path from the positive terminal of the power source, through the conductors and the load, and back to the negative terminal, an electric circuit is formed. If there’s a break anywhere in this path (like an open switch or a cut wire), the current cannot flow, and the circuit is “broken” or “open.”

Question:

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of 4 Ω.

Concept in a Minute:

Resistors can be connected in series or in parallel. In a series combination, the total resistance is the sum of individual resistances (R_total = R1 + R2 + R3…). In a parallel combination, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances (1/R_total = 1/R1 + 1/R2 + 1/R3…). The goal is to combine these arrangements to achieve a specific target resistance.

Explanation:

We have three resistors, each with a resistance of 6 Ω. We need to arrange them to get a total resistance of 4 Ω.

Let’s consider different combinations:

1. All three resistors in series:
R_total = 6 Ω + 6 Ω + 6 Ω = 18 Ω. This is not 4 Ω.

2. All three resistors in parallel:
1/R_total = 1/6 Ω + 1/6 Ω + 1/6 Ω = 3/6 Ω = 1/2 Ω.
R_total = 2 Ω. This is not 4 Ω.

3. Two resistors in series, and this combination in parallel with the third resistor:
Let’s connect two resistors in series. Their combined resistance will be R_series = 6 Ω + 6 Ω = 12 Ω.
Now, connect this 12 Ω combination in parallel with the third 6 Ω resistor.
1/R_total = 1/12 Ω + 1/6 Ω
1/R_total = 1/12 Ω + 2/12 Ω
1/R_total = 3/12 Ω = 1/4 Ω.
R_total = 4 Ω.

This combination gives us the desired resistance of 4 Ω. Therefore, the arrangement is to connect two resistors in series, and then connect this series combination in parallel with the third resistor.

Question:

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Concept in a Minute:

The key concepts needed to solve this problem are:
1. Electrical Power: The rate at which electrical energy is transferred. It is calculated as the product of voltage and current.
2. Energy Consumption: The total amount of electrical energy used over a period of time. It is calculated as the product of power and time.

Explanation:

Step 1: Calculate the power of the motor.
The power (P) of an electrical appliance is given by the formula P = V * I, where V is the voltage and I is the current.
Given:
Voltage (V) = 220 V
Current (I) = 5 A
Power (P) = 220 V * 5 A = 1100 Watts

Step 2: Convert the power to kilowatts.
Since energy is usually measured in kilowatt-hours (kWh), it’s helpful to convert the power from watts to kilowatts.
1 kilowatt (kW) = 1000 Watts
Power (P) in kW = 1100 W / 1000 W/kW = 1.1 kW

Step 3: Convert the time to hours.
The time is already given in hours, so no conversion is needed.
Time (t) = 2 hours

Step 4: Calculate the energy consumed.
The energy (E) consumed is given by the formula E = P * t, where P is the power and t is the time.
Energy (E) = 1.1 kW * 2 h = 2.2 kilowatt-hours (kWh)

Therefore, the power of the motor is 1100 Watts (or 1.1 kW) and the energy consumed in 2 hours is 2.2 kWh.

Question:

Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

Concept in a Minute:

Resistivity and its dependence on temperature, properties of alloys vs. pure metals, and the function of electric heating devices.

Explanation:

Conductors in electric heating devices are made of alloys rather than pure metals primarily because alloys have a higher resistivity than pure metals. Higher resistivity means that more electrical energy is converted into heat when current flows through the material. This increased heat generation is crucial for the efficient functioning of devices like bread-toasters and electric irons.

Furthermore, alloys generally have a higher melting point than pure metals, making them more durable under the high temperatures generated during operation. They also resist oxidation at high temperatures better than pure metals, preventing degradation of the heating element over time. While pure metals like copper or silver are excellent conductors with low resistivity (ideal for wires carrying current without significant heat loss), this property is undesirable for heating elements. Alloys like nichrome (a mixture of nickel and chromium) are commonly used because they strike a balance between good resistance for heat generation and sufficient durability at high temperatures.

Question:

Why does the cord of an electric heater not glow while the heating element does?

Concept in a Minute:

Resistance, Ohm’s Law, Joule’s Law of Heating, resistivity, heating effect of electric current.

Explanation:

The cord of an electric heater is typically made of a material with very low resistance, such as copper. The heating element, on the other hand, is made of a material with a much higher resistance, like nichrome wire.

Ohm’s Law states that voltage (V) is equal to current (I) multiplied by resistance (R) (V = IR). Joule’s Law of Heating states that the heat produced (H) in a conductor is proportional to the square of the current (I), the resistance (R), and the time (t) (H = I^2Rt).

When the electric heater is switched on, the same amount of current flows through both the cord and the heating element. Since the heating element has a much higher resistance (R), according to Joule’s Law, it will generate significantly more heat (H is proportional to R). This high heat causes the nichrome wire to become red hot and glow.

The cord, having very low resistance, generates very little heat. Therefore, it does not get hot enough to glow. The purpose of the cord is to conduct electricity to the heating element with minimal energy loss as heat.

Question:

How much energy is given to each coulomb of charge passing through a 6 V battery?

Concept in a Minute:

The key concept is the relationship between electric potential difference, energy, and charge. Electric potential difference (voltage) is defined as the work done or energy transferred per unit charge. The formula is V = W/q, where V is the potential difference, W is the energy transferred (work done), and q is the charge.

Explanation:

The question states that a 6 V battery is used. The electric potential difference, or voltage, of the battery is given as 6 V. Voltage represents the energy given to each coulomb of charge that passes through it. Therefore, if the voltage is 6 V, it means that 6 joules of energy are given to each coulomb of charge passing through the battery.

To elaborate using the formula:
We are given:
Potential difference (V) = 6 V
Charge (q) = 1 coulomb (since the question asks about “each coulomb of charge”)

We need to find the energy (W) given to this charge.
From the definition of potential difference, V = W/q.
Rearranging the formula to solve for W, we get W = V * q.

Substituting the given values:
W = 6 V * 1 C
W = 6 joules

Thus, 6 joules of energy are given to each coulomb of charge passing through a 6 V battery.

Question:

What determines the rate at which energy is delivered by a current?

Concept in a Minute:

The rate at which energy is delivered by a current is known as electrical power. Electrical power is determined by the product of the current flowing through a component and the potential difference across that component.

Explanation:

The rate at which energy is delivered by a current is determined by the electrical power. Electrical power ($P$) is defined as the rate of energy transfer or dissipation. For a circuit component, it is given by the product of the electric current ($I$) flowing through it and the potential difference ($V$) across its terminals. The formula for electrical power is:
$P = V \times I$

Where:
$P$ is the power in watts (W).
$V$ is the potential difference in volts (V).
$I$ is the current in amperes (A).

Alternatively, using Ohm’s Law ($V = I \times R$, where $R$ is the resistance), the power can also be expressed as:
$P = I^2 \times R$
or
$P = \frac{V^2}{R}$

Therefore, the rate at which energy is delivered by a current is determined by the magnitude of the current and the potential difference across the circuit element, or equivalently, by the current and the resistance, or the voltage and the resistance.

Question:

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Concept in a Minute:

Ohm’s Law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all other conditions remain unchanged. The constant of proportionality is resistance. For a series circuit, the total resistance is the sum of individual resistances, and the current is the same through all components.

Explanation:

The resistors are connected in series. In a series circuit, the total resistance (R_total) is the sum of the individual resistances.
R_total = R1 + R2 + R3 + R4 + R5
R_total = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω
R_total = (0.2 + 0.3 + 0.4 + 0.5) Ω + 12 Ω
R_total = 1.4 Ω + 12 Ω
R_total = 13.4 Ω

The battery provides a voltage (V) of 9 V.
According to Ohm’s Law, the current (I) flowing through the circuit is given by:
I = V / R_total

Substituting the values:
I = 9 V / 13.4 Ω
I ≈ 0.6716 A

In a series circuit, the current is the same through all resistors. Therefore, the current flowing through the 12 Ω resistor is equal to the total current flowing through the circuit.

Answer: The current that would flow through the 12 Ω resistor is approximately 0.6716 A.

Question:

How does the resistance of a wire vary with its area of cross-section?

Concept in a Minute:

The resistance of a conductor is directly proportional to its length and inversely proportional to its area of cross-section. This relationship is described by the formula R = ρ(L/A), where R is resistance, ρ is resistivity, L is length, and A is the area of cross-section.

Explanation:

The resistance of a wire is inversely proportional to its area of cross-section. This means that if the area of cross-section of the wire increases, its resistance decreases, and if the area of cross-section decreases, its resistance increases, assuming other factors like length and material remain constant.

Think of it like water flowing through pipes. A wider pipe (larger area of cross-section) allows more water to flow easily with less resistance. Similarly, a wire with a larger cross-sectional area provides more pathways for electrons to flow, hence reducing the opposition to their movement, which is resistance. Conversely, a narrower pipe (smaller area of cross-section) restricts the flow of water, just as a wire with a smaller cross-sectional area offers more resistance to the flow of electric current.

Question:

Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Concept in a Minute:

Energy consumed is calculated by the formula: Energy = Power × Time. Power is measured in Watts (W) and time needs to be in seconds for standard SI units, or consistent units for comparison.

Explanation:

To determine which appliance uses more energy, we need to calculate the energy consumed by each appliance. The unit of energy can be Joules (J) if time is in seconds, or Watt-hours (Wh) if time is in hours, or Watt-minutes (Wm) if time is in minutes. For comparison, it’s best to use a consistent unit. We will use Watt-minutes for this problem.

For the TV set:
Power = 250 W
Time = 1 hour = 60 minutes
Energy consumed by TV = Power × Time = 250 W × 60 minutes = 15000 Watt-minutes.

For the toaster:
Power = 1200 W
Time = 10 minutes
Energy consumed by toaster = Power × Time = 1200 W × 10 minutes = 12000 Watt-minutes.

Comparing the energy consumed:
Energy consumed by TV = 15000 Watt-minutes
Energy consumed by toaster = 12000 Watt-minutes

Since 15000 Watt-minutes > 12000 Watt-minutes, the TV set uses more energy.

Question:

When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Concept in a Minute:

This question involves Ohm’s Law, which states the relationship between voltage, current, and resistance in an electrical circuit. Specifically, it’s the formula V = IR, where V is voltage, I is current, and R is resistance. We will use this formula to find the unknown resistance.

Explanation:

We are given the voltage (V) of the battery and the current (I) flowing through the circuit. We need to find the resistance (R) of the resistor.

Given:
Voltage (V) = 12 V
Current (I) = 2.5 mA

First, we need to convert the current from milliamperes (mA) to amperes (A) because the standard unit for current in Ohm’s Law is amperes.
1 mA = 10^-3 A
So, 2.5 mA = 2.5 × 10^-3 A = 0.0025 A

Now, we can rearrange Ohm’s Law (V = IR) to solve for resistance:
R = V / I

Substitute the given values into the formula:
R = 12 V / 0.0025 A

Calculate the resistance:
R = 4800 Ω

The value of the resistance of the resistor is 4800 Ohms.

Final Answer: The final answer is $\boxed{4800}$

Question:

What is meant by saying that the potential difference between two points is 1 V?

Concept in a Minute:

Electric Potential Difference: Electric potential difference between two points is the work done per unit charge in moving a charge from one point to another. It is measured in Volts (V).
Work Done: The energy transferred when a force moves an object over a distance. In electrostatics, it’s the energy required to move a charge.
Charge: A fundamental property of matter that exhibits an electric force when placed in an electromagnetic field. Measured in Coulombs (C).

Explanation:

Saying that the potential difference between two points is 1 V means that 1 Joule (J) of work is done when 1 Coulomb (C) of electric charge is moved from one point to the other.
Mathematically, this can be represented as:
Potential Difference (V) = Work Done (W) / Charge (Q)
So, if V = 1 V and Q = 1 C, then W = V * Q = 1 V * 1 C = 1 J.
In simpler terms, it takes 1 Joule of energy to move every Coulomb of charge between those two points.

Question:

Name a device that helps to maintain a potential difference across a conductor.

Concept in a Minute:

Potential difference is the work done per unit charge to move a charge between two points. A device that maintains a potential difference across a conductor essentially provides a continuous source of energy to move charges, thus establishing a current flow. This requires a source of electromotive force (EMF).

Explanation:

A device that helps to maintain a potential difference across a conductor is a source of electromotive force (EMF). This can be a battery or a generator. These devices convert other forms of energy (chemical, mechanical) into electrical energy, which then drives the charges and establishes a potential difference. For example, a battery has two terminals with different chemical potentials, which creates a potential difference between them. When connected to a conductor, this potential difference causes electrons to flow.

The final answer is $\boxed{\text{Battery}}$.

Question:

Calculate the number of electrons constituting one coulomb of charge.

Concept in a Minute:

The fundamental charge is the charge of a single electron (or proton). The total charge is the number of elementary charges multiplied by the elementary charge.

Explanation:

We know that the charge of one electron is approximately $1.6 \times 10^{-19}$ coulombs.
To find the number of electrons that constitute one coulomb of charge, we can use the relationship:
Total Charge (Q) = Number of electrons (n) × Charge of one electron (e)

We are given Q = 1 coulomb and we know e = $1.6 \times 10^{-19}$ coulombs/electron.
Rearranging the formula to find n:
n = Q / e

Substitute the given values:
n = 1 C / ($1.6 \times 10^{-19}$ C/electron)
n = $1 / (1.6 \times 10^{-19})$ electrons
n = $(1 / 1.6) \times 10^{19}$ electrons
n = $0.625 \times 10^{19}$ electrons
n = $6.25 \times 10^{18}$ electrons

Therefore, $6.25 \times 10^{18}$ electrons constitute one coulomb of charge.

Question:

Compare the power used in the 2 Ω resistor in each of the following circuits:

a 6 V battery in series with 1 Ω and 2 Ω resistors
a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Concept in a Minute:

Ohm’s Law (V=IR), Power Law (P=V*I or P=I^2*R or P=V^2/R), Series Circuits (current is same, voltage divides), Parallel Circuits (voltage is same, current divides).

Explanation:

Part 1: Series Circuit
1. Calculate the total resistance of the series circuit.
2. Calculate the total current flowing through the circuit using Ohm’s Law (I = V/R_total).
3. Since it’s a series circuit, this current will flow through both the 1 Ω and 2 Ω resistors.
4. Calculate the power dissipated in the 2 Ω resistor using the formula P = I^2 * R.

Part 2: Parallel Circuit
1. In a parallel circuit, the voltage across each branch is the same as the battery voltage. So, the voltage across the 2 Ω resistor is 4 V.
2. Calculate the power dissipated in the 2 Ω resistor using the formula P = V^2 / R.
3. Compare the power calculated in Part 1 with the power calculated in Part 2.

Question:

Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Concept in a Minute:

Resistance is the opposition to the flow of electric current. Factors affecting resistance include the material of the conductor, its length, and its cross-sectional area. A larger cross-sectional area leads to lower resistance.

Explanation:

Current will flow more easily through a thick wire. This is because the resistance of a wire is inversely proportional to its cross-sectional area. A thick wire has a larger cross-sectional area than a thin wire of the same material. A larger cross-sectional area means there are more free electrons available to move and carry the current. With less opposition (lower resistance), the current can flow more easily. The formula for resistance is R = ρL/A, where R is resistance, ρ (rho) is resistivity (a property of the material), L is the length, and A is the cross-sectional area. As A increases, R decreases.

Question:

An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Concept in a Minute:

The rate at which heat is developed in a resistor is known as power. This power can be calculated using the formula P = I²R, where P is power, I is current, and R is resistance.

Explanation:

The question asks for the rate at which heat is developed in the heater. This rate is equivalent to the power dissipated by the heater.
We are given:
Resistance of the heater, R = 8 Ω
Current drawn by the heater, I = 15 A
Time for which the heater draws current, t = 2 hours (This information is not needed to calculate the rate of heat development, but would be used to find the total heat developed).

The formula for the rate of heat development (power) is:
P = I²R

Substitute the given values into the formula:
P = (15 A)² * 8 Ω
P = 225 A² * 8 Ω
P = 1800 Watts

Therefore, the rate at which heat is developed in the heater is 1800 Watts.

Question:

Why are copper and aluminium wires usually employed for electricity transmission?

Concept in a Minute:

Electrical conductivity, cost-effectiveness, malleability, ductility.

Explanation:

Copper and aluminum wires are primarily used for electricity transmission due to their excellent electrical conductivity, meaning they allow electric current to flow through them with minimal resistance. This reduces energy loss during transmission. Both metals are also relatively abundant, making them cost-effective for large-scale infrastructure projects compared to other highly conductive materials like silver. Furthermore, copper and aluminum are malleable (can be hammered into thin sheets) and ductile (can be drawn into thin wires), which are essential properties for manufacturing long transmission lines. Their resistance to corrosion also contributes to their longevity and reliability in outdoor transmission applications.

Question:

An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Concept in a Minute:

This question relates to Joule’s Law of Heating, which states that the heat developed in a conductor is directly proportional to the square of the current flowing through it, the resistance of the conductor, and the time for which the current flows. The formula is H = I^2 * R * t.

Explanation:

We are given the resistance (R) of the electric iron, the current (I) it takes, and the time (t) for which it is used.
Resistance, R = 20 Ω
Current, I = 5 A
Time, t = 30 s

We need to calculate the heat developed (H).
Using Joule’s Law of Heating, the formula for heat developed is:
H = I^2 * R * t

Substitute the given values into the formula:
H = (5 A)^2 * (20 Ω) * (30 s)
H = (25 A^2) * (20 Ω) * (30 s)
H = 500 * 30 J
H = 15000 J

Therefore, the heat developed in 30 seconds is 15000 Joules.

Question:

Use the data in the Table given below to answer the following –

Which among iron and mercury is a better conductor?

Table give below Electrical resistivity of some substances at 20°C

Electrical resistivity of some substances at 20°C
−	Material	Resistivity (Ω m)
Conductors	Silver	1.60 × 10⁻⁸
	Copper	1.62 × 10⁻⁸
	Aluminium	2.63 × 10⁻⁸
	Tungsten	5.20 × 10⁻⁸
	Nickel	6.84 × 10⁻⁸
	Iron	10.0 × 10⁻⁸
	Chromium	12.9 × 10⁻⁸
	Mercury	94.0 × 10⁻⁸
	Manganese	1.84 × 10⁻⁶
Alloys	Constantan (alloy of Cu and Ni)	49 × 10⁻⁶
	Manganin (alloy of Cu, Mn and Ni)	44 × 10⁻⁶
	Nichrome (alloy of Ni, Cr, Mn and Fe)	100 × 10⁻⁶
Insulators	Glass	10¹⁰ − 10¹⁴
	Hard rubber	10¹³ − 10¹⁶
	Ebonite	10¹⁵ − 10¹⁷
	Diamond	10¹² − 10¹³
	Paper (dry)	10¹²

Concept in a Minute:

Electrical conductivity is the reciprocal of electrical resistivity. A lower resistivity indicates better conductivity, meaning the material allows electricity to flow more easily.

Explanation:

To determine which among iron and mercury is a better conductor, we need to look at their electrical resistivity values from the provided table.

From the table, the resistivity of Iron is 10.0 × 10⁻⁸ Ω m.
From the table, the resistivity of Mercury is 94.0 × 10⁻⁸ Ω m.

A better conductor has lower electrical resistivity. Comparing the resistivity values:
10.0 × 10⁻⁸ Ω m (Iron) < 94.0 × 10⁻⁸ Ω m (Mercury)

Since iron has a lower resistivity than mercury, iron is a better conductor of electricity.

Question:

Define the unit of current.

Concept in a Minute:

Electric current is defined as the rate of flow of electric charge. The fundamental unit of electric charge is the Coulomb (C), and the unit of time is the second (s).

Explanation:

The unit of electric current is the Ampere (A). One Ampere is defined as the flow of one Coulomb of electric charge past a point in one second. Mathematically, this can be expressed as:
Current (I) = Charge (Q) / Time (t)
Therefore, 1 Ampere = 1 Coulomb / 1 Second.

Next Chapter: Heredity and Evolution

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