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Current Electricity MCQs for NEET — Physics Questions with Answers

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n identical cells each of e.m.f. E and internal resistance r are connected in series. An external resistance R is connected in series to this combination. The current through R is

$\frac{n E}{R + n r}$

$\frac{n E}{n R + r}$

$\frac{E}{R + n r}$

$\frac{n E}{R + r}$

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Explanation

Total e.m.f. = nE, Total resistance R + nr ⇒ $i = \frac{n E}{R + n r}$

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Two identical cells send the same current in 2 Ω resistance, whether connected in series or in parallel. The internal resistance of the cell should be

1 Ω

2 Ω

$\frac{1}{2} Ω$

2.5 Ω

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Explanation

In series , $i_{1} = \frac{2 E}{2 + 2 r}$

In parallel, $I_{2} = \frac{E}{2 + \frac{r}{2}} = \frac{2 E}{4 + r}$

Since $i_{1} = i_{2}$ ⇒ $\frac{2 E}{4 + r} = \frac{2 E}{2 + 2 r}$ ⇒ $r = 2 Ω$

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Two non-ideal identical batteries are connected in parallel. Consider the following statements :

(i) The equivalent e.m.f. is smaller than either of the two e.m.f.s

(ii) The equivalent internal resistance is smaller than either of the two internal resistances

Both (i) and (ii) are correct

(i) is correct but (ii) is wrong

(ii) is correct but (i) is wrong

Both (i) and (ii) are wrong

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Explanation

Because $E_{e q} = E$ and $r_{e q} = \frac{r}{2}$

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The number of dry cells, each of e.m.f. 1.5 volt and internal resistance 0.5 ohm that must be joined in series with a resistance of 20 ohm so as to send a current of 0.6 ampere through the circuit is

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Explanation

In series $i = \frac{n E}{n r + R}$

⇒ $0 .6 = \frac{n \times 1 .5}{n \times 0 .5 \times 20}$ ⇒ n = 10

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For driving a current of 2 A for 6 minutes in a circuit, 1000 J of work is to be done. The e.m.f. of the source in the circuit is

1.38 V

1.68 V

2.04 V

3.10 V

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Explanation

$P = \frac{W}{t} = V i$

⇒ $V = \frac{W}{i t} = \frac{1000}{2 \times 6 \times 60} = 1 .38 V$

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Four identical cells each having an electromotive force (e.m.f.) of 12V, are connected in parallel. The resultant electromotive force (e.m.f.) of the combination is :

48 V

12 V

4 V

3 V

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Explanation

In parallel combination $E_{e q} = E = 12 V$

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The internal resistance of a cell of e.m.f. 12V is $5 \times 10^{- 2} Ω$ . It is connected across an unknown resistance. The voltage across the cell, when a current of 60 A is drawn from it, is :

15 V

12 V

9 V

6 V

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Explanation

$V = E - i r$ = $12 - 60 \times 5 \times 10^{- 2}$ = 9V.

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A battery is charged at a potential of 15 V for 8 hours when the current flowing is 10 A. The battery on discharge supplies a current of 5 A for 15 hours. The mean terminal voltage during discharge is 14 V. The "Watt-hour" efficiency of the battery is :

82.5%

80 %

90%

87.5%

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Explanation

Watt-hour efficiency $= \frac{Discharging energy}{Charging energy}$

$= \frac{14 \times 5 \times 15}{15 \times 8 \times 10} = 0 .875 = 87 .5 %$

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A capacitor is connected to a cell of emf E having some internal resistance r. The potential difference across the

Cell is < E

Cell is E

Capacitor is > E

Capacitor is < E

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Explanation

In the given case cell is in open circuit (i = 0) so voltage across the cell is equal to its e.m.f.

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The maximum power drawn out of the cell from a source is given by (where r is internal resistance)

$E^{2} / 2 r$

$E^{2} / 4 r$

$E^{2} / r$

$E^{2} / 3 r$

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Explanation

$i = \frac{E}{r + R}$ ⇒ $P = i^{2} R$ ⇒ $P = \frac{E^{2} R}{{(r + R)}^{2}}$

Power is maximum when r = R ⇒ $P_{\max} = E^{2} / 4 r$

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Frequently Asked Questions

Yes. You can attempt every Current Electricity question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.