Electric Current MCQs with Solutions

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Electric current is the flow of electric charges in a closed circuit. The SI unit of current is Ampere (A). 1 Ampere is the current produced due to the flow of 1 coulomb of charges in one second. Depending upon the direction of flow of current, it is divided into two types - Alternating current and Direct current.

Electric current is measured with the help of a device called Ammeter. An ammeter is always connected in series in a circuit. It has negligible resistance, thus it draws no current from the circuit to which it is connected.

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MCQs on Electric Current

Ques 1. While connecting 6 cells in a battery in series, in a tape recorder, by mistake one cell is connected with reverse polarity. If the effective resistance of load is 24 Ω and internal resistance of each cell is 1 Ω and emf is 1.5 V, the current delivered by the battery is:

0.1 A
0.2 A
0.3 A
0.4 A

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Ans. (b). 0.2 A

Explanation: Applying KVL to the loop,

4E = I [R + 6r]

4 × 1.5 = I [30]

I = 6/30

I = 1/5 = 0.2 A

Ques 2. A 5°C rise in temperature is observed in a conductor by passing a current, when the current is doubled, the rise in temperature will be nearly

10°C
16°C
20°C
12°C

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Ans. P = I²R

If I is doubled, the power will be 4 times

and, ΔT ∝ P

If power is increased 4 times, then ΔT will also increase by 4 times

so, 4 × 5°C = 20°C

Ques 3. Assertion: When current through a bulb decreases by 0.5%, the glow of the bulb decreases by 1%.

Reason: Glow (Power) is directly proportional to square of current.

Both assertion and reason are true and reason is the correct explanation of the assertion
Both assertion and reason are true but reason is not the correct explanation for assertion
Assertion is correct but reason is incorrect
Both assertion and reason are incorrect

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Ans. (a). Both assertion and reason are true and reason is the correct explanation of the assertion

Explanation: P = I²R

P_f / P_i= I_f² / I_i² (R = constant) ………(i)

I_f = I_i - I_i (0.5)/100 = 0.995 I_i

I_f / I_i = 0.995

From (i)

P_f / P_i = (0.995)² = 0.99

Power decreased = (P_i - P_f / P_i) × 100 = 1%

Ques 4. Three resistors each of 10Ω are connected in series to a battery of potential difference 150 V. The current flowing through it is:

45 A
15 A
20 A
5 A

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Ans. (d). 5 A

Explanation: V = IR

V = I × R_eq

I = V / R_eq

The three resistances are in series. So, the equivalent resistance will be;

R_eq = R₁ + R₂ + R₃

R_eq = 10 + 10 + 10

R_eq = 30 Ω

I = 150/30

I = 5 A

Ques 5. The resistance of an iron wire is 10 Ω and its temperature coefficient of resistance is 5 × 10^-3/°C. A current of 30 mA is flowing in it at 20°C. Keeping potential difference across its ends constant, if its temperature is increased to 120°C then the current flowing in the wire will be (in mA)

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Ans. (a). 20

Explanation: Temperature coefficient of resistance, α = 5 × 10^-3 /°C

Initial Temperature, T_o = 20°

Initial Resistance at T_o , R_o = 10 Ω

Initial Current at T_o , I_o = 30 mA

Increased temperature, T₁ = 120°

Increased resistance,

R₁ = R_o [1 + α (T₁ - T_o)]

R₁ = 10 [1 + 5 × 10^-3 (120 - 20)]

R₁ = 10 [1 + 5 × 10^-1]

R₁ = 10 [1 + 0.5]

R₁ = 10 [1.5]

R₁ = 15 Ω

Now, potential difference across the wire is constant,

V = I_oR_o = I₁R₁

I₁ = I_oR_o / R₁

I₁ = 30 × 10 / 15

I₁ = 300/15

I₁ = 20 mA

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Ques 6. A steady current is passing through a linear conductor of non-uniform cross-section. The net quantity of charge crossing any cross-section per second is:

Independent of area of cross-section
Directly proportional to the length of the conductor
Directly proportional to the area of the cross section
Inversely proportional to the length of the conductor

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Ans. (a). Independent of area of cross-section

Explanation: The quantity of charge passing through the cross-section is the current flowing in it. It is given in the question that the current is steady. Therefore, it is independent of the area of cross-section.

Ques 7. For a chosen non-zero value of voltage, there can be more than one value of current in:

Copper wire
Thermistor
Nichrome wire
Manganin wire

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Ans. (b). Thermistor

Explanation: Thermistor are made up of substance that have a high value of temperature coefficient of resistance (α), thus a slight change in temperature can vary its resistance to a high amount. This increases the possibility of more than one value of current.

Ques 8. A 24 V potential difference is applied across a parallel combination of four 6Ω resistor. The current in each resistor is:

4 A
1 A
36 A
16 A

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Ans. (a). 4 A

Explanation: As the resistors are connected in parallel, the voltage across each resistor is the same i.e. 24 V.

So, current in each resistor is,

I = V/R

I = 24/6 = 4 A

Ques 9. What steady current can be drawn from an accumulator of capacity 80 Ah?

4 A for 20 hours
40 A for 6 hours
20 A for 5 hours
2 A for 100 hours

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Ans. (a). 4 A for 20 hours

Explanation: The capacity of the accumulator is 80 Ah. So, either a current of 80 A can be drawn for 1 hour or less current can be drawn for a longer period.

I = dQ/dt = Q/t

Q = 4 A × 20 hrs = 80 Ah

Ques 10. Name the instrument used to control the current in an electric circuit.

Rheostat
Voltmeter
Ammeter
Galvanometer

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Ans. (a). Rheostat

Explanation: A rheostat can be used to control current in an electric circuit. It is a variable resistor that can control electrical resistance of a circuit without influencing the flow of current. Rheostats are used in applications that require high voltage and current.

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Electric Current MCQs with Solutions

MCQs on Electric Current

CBSE CLASS XII Related Questions

1.
A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

2.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

3.
A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:

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Electric Current MCQs with Solutions

MCQs on Electric Current

CBSE CLASS XII Related Questions

1.A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

2.Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

3.A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:

4.A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle \(i\) such that it is refracted out grazing the face AD. Find the value of angle \(i\).

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

2.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

3.
A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is: