Blog Log in Get Started Free

Current Electricity MCQs for NEET — Physics Questions with Answers

Q: Are these Current Electricity questions free to attempt?

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.

Q: Do I need an account to practise Current Electricity MCQs?

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

Q: Are these previous year NEET questions?

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

Practice free Current Electricity (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

Two resistances of 400 Ω and 800 Ω are connected in series with a 6-volt battery of negligible internal resistance. A voltmeter of resistance 10,000 Ω is used to measure the potential difference across 400 Ω. The error in the measurement of potential difference in volts approximately is :

0.01

0.02

0.03

0.05

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

The length of a wire of a potentiometer is 100 cm, and the emf of its standard cell is E volt. It is employed to measure the e.m.f of a battery whose internal resistance is 0.5 Ω. If the balance point is obtained at l = 30 cm from the positive end, the e.m.f. of the battery is :

$\frac{30 E}{100}$

$\frac{30 E}{100 .5}$

$\frac{30 E}{(100 - 0 .5)}$

$\frac{30 (E - 0 .5 i)}{100}$ , where i is the current in the potentiometer

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

From the principle of potentiometer $V \propto l$

⇒ $\frac{V}{E} = \frac{l}{L}$ ; where V = emf of battery, E = emf of standard cell, L = Length of potentiometer wire

$V = \frac{E l}{L} = \frac{30 E}{100}$ .

● ● ● ● ●

The current flowing in a coil of resistance 90 Ω is to be reduced by 90%. What value of resistance should be connected in parallel with it

9 Ω

90 Ω

1000 Ω

10 Ω

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$i_{g} = \frac{i}{10}$ ⇒ Required shunt $S = \frac{G}{(n - 1)} = \frac{90}{(10 - 1)} = 10 Ω$

● ● ● ● ●

A galvanometer of 50 ohm resistance has 25 divisions. A current of 4 × 10^–4 ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of :

2500 Ω as a shunt

2450 Ω as a shunt

2550 Ω in series

2450 Ω in series

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Full deflection current $i_{g} = 25 \times 4 \times 10^{- 4} = 100 \times 10^{- 4} A$

Using $R = \frac{V}{I_{g}} - G = \frac{25}{100 \times 10^{- 4}} - 50 = 2450 Ω$ in series.

● ● ● ● ●

In a metre bridge experiment, the null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y

50 cm

80 cm

40 cm

70 cm

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

In balancing condition, $\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} = \frac{l_{1}}{100 - l_{1}}$

⇒ $\frac{X}{Y} = \frac{20}{80} = \frac{1}{4}$ .....(i)

and $\frac{4 X}{Y} = \frac{l}{100 - l}$ .....(ii)

From equation (i) and (ii) :

⇒ $\frac{4}{4} = \frac{l}{100 - l}$ ⇒ $l = 50 c m$

● ● ● ● ●

In a potentiometer experiment, the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2 Ω, the balancing length becomes 120 cm. The internal resistance of the cell is :

4 Ω

2 Ω

1 Ω

0.5 Ω

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$r = R (\frac{l_{1}}{l_{2}} - 1) = 2 (\frac{240}{120} - 1) = 2 Ω$

● ● ● ● ●

Potentiometer wire of length 1 m is connected in series with 490 Ω resistance and 2V battery. If 0.2 mV/cm is the potential gradient, then the resistance of the potentiometer wire is :

4.9 Ω

7.9 Ω

5.9 Ω

6.9 Ω

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Potential gradient $x = \frac{e}{(R + R_{h} + r)} . \frac{R}{L}$

⇒ ⇒ R = 4.9 Ω.

● ● ● ● ●

Two uniform wires A and B are of the same metal and have equal masses. The radius of wire A is twice that of wire B. The total resistance of A and B when connected in parallel is :

4 Ω when the resistance of wire A is 4.25 Ω

5 Ω when the resistance of wire A is 4.25 Ω

4 Ω when the resistance of wire B is 4.25 Ω

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$\frac{R_{A}}{R_{B}} = {(\frac{r_{B}}{r_{A}})}^{4}$ ⇒ $\frac{R_{A}}{R_{B}} = {(\frac{1}{2})}^{4} = \frac{1}{16}$ ⇒ $R_{B} = 16 R_{A}$

When R_A and R_B are connected in parallel then equivalent resistance $R_{e q} = \frac{R_{A} R_{B}}{(R_{A} + R_{B})} = \frac{16}{17} R_{A}$

If $R_{A} = 4 .25 Ω$ then $R_{e q} = 4 Ω$ i.e. option is correct.

● ● ● ● ●

You are given several identical resistances each of value R = 10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5 Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this job

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Suppose n resistors are used for the required job. Suppose equivalent resistance of the combination is R' and according to energy conservation, it's current rating is $i' .$

Energy consumed by the combination = n × (Energy consumed by each resistance)

⇒ $i'^{2} R' = n \times i^{2} R$

⇒ $n = {(\frac{i'}{i})}^{2} \times (\frac{R'}{R}) = {(\frac{4}{1})}^{2} \times (\frac{5}{10}) = 8$

● ● ● ● ●

There are three resistance coils of equal resistance. The maximum number of resistances you can obtain by connecting them in any manner you choose, being free to use any number of the coils in any way is :

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Maximum number of resistance $= 2^{n - 1} = 2^{3 - 1} = 4$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

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.