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Current Electricity MCQs for NEET — Physics Questions with Answers

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The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100 W and 200 V lamp when not in use?

14,000 $Ω$

400 $Ω$

40 $Ω$

4 $Ω$

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Explanation

When the filament is hot, $R = \frac{V^{2}}{P} = \frac{200 \times 200}{100} = 400 Ω$

Hence, when cold, the resistance is 40 $Ω$ .

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Two electric bulbs A and B are rated 60 W and 100 W, respectively. If they are connected in parallel to the same source, then,

both the bulbs draw the same current

bulb A draws more current than bulb B

bulb B draws more current than bulb A

currents drawn in the bulbs are in the ratio of their resistances

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Explanation

Given $P_{A} = 60 W, P_{B} = 100 W$ . We know that current flowing through the bulb is I = P/V. We also know that as both the bulbs are connected in parallel, potential difference (V) across both the bulbs is the same. Thus, $I α P$ . Since the power of bulb B is greater than that of bulb A, bulb B draws more current than bulb A.

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A factory is served by a 220 V supply line. In a circuit protected by a fuse marked 10 A, the maximum number of 100 W lamps in parallel that can be turned on is :

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Explanation

The current required by each bulb is :

$I = \frac{P}{V} = \frac{100}{220} A$

If n bulbs are joined in parallel, then,

$n I = I_{f u s e} o r n \times \frac{100}{220} = 10 o r n = 22$ .

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A 100 W bulb designed to operate on 100 V is to be connected across a 500 V source. Find the resistance to be put in series so that bulb consumes 100 W only.

100 $Ω$

500 $Ω$

400 $Ω$

300 $Ω$

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The resistance of an ideal voltmeter is

Zero

Very low

Very large

Infinite

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Explanation

The resistance of an ideal voltmeter is considered as infinite.

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The net resistance of a voltmeter should be large to ensure that :

It does not get overheated

It does not draw excessive current

It can measure large potential difference

It does not appreciably change the potential difference to be measured

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Explanation

The resistance of voltmeter is too high, so that it draws negligible current from the circuit, hence potential drop in the external circuit is also negligible.

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A voltmeter of resistance 1000 Ω gives full-scale deflection when a current of 100 mA flow through it. The shunt resistance required across it to enable it to be used as an ammeter reading 1 A at full-scale deflection is :

10000 Ω

9000 Ω

222 Ω

111 Ω

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Explanation

By using $\frac{i}{i_{g}} = 1 + \frac{G}{S}$

⇒ $\frac{i}{100 \times 10^{- 3}} = 1 + \frac{1000}{S}$

⇒ $S = \frac{1000}{9} = 111 Ω$

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A galvanometer has 30 divisions and a sensitivity 16 $μ A / div.$ It can be converted into a voltmeter to read 3 V by connecting (approximately):

Resistance nearly 6 k Ω in series

6 k Ω in parallel

500 Ω in series

It cannot be converted

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In a circuit, 5 percent of total current passes through a galvanometer. If the resistance of the galvanometer is G then the value of the shunt is :

19 G

20 G

$\frac{G}{20}$

$\frac{G}{19}$

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Explanation

$\frac{i_{g}}{i} = \frac{S}{G + S}$

⇒ $\frac{5}{100} = \frac{S}{G + S}$

⇒ $S = \frac{G}{19}$

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A voltmeter has a range 0-V with a series resistance R. With a series resistance 2R, the range is 0-V'. The correct relation between V and V' is :

$V^{'} = 2 V$

$V^{'} > 2 V$

$V^{'} > > 2 V$

$V' < 2 V$

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Explanation

For conversion of galvanometer (of resistances) into voltmeter, a resistance R is connected in series.

∴ $i_{g} = \frac{V_{}}{R + G}$ and $i_{g} = \frac{{V^{'}}_{}}{2 R + G}$

⇒ $\frac{V}{R + G} = \frac{{V_{}}^{'}}{2 R + G}$ ⇒ $\frac{V^{'}}{V_{}} = \frac{2 R + G}{R + G} = \frac{2 (R + G) - G}{(R + G)}$

$= 2 - \frac{G}{(R + G)}$ ⇒ ${V^{'_{}}}_{} = 2 V - \frac{V_{} G}{(R + G)}$ ⇒ ${V^{'}}_{} < 2 V_{}$

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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.