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Current Electricity MCQs for NEET — Physics Questions with Answers

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Consider the following two statements :

Kirchhoff's junction law follows from the conservation of charge.

Kirchhoff's loop law follows from the conservation of energy.

Which of the following is correct?

(a) Both (A) and (B) are wrong

(b) (A) is correct and (B) is wrong

(A) is wrong and (B) is correct

Both (A) and (B) are correct

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Explanation

Kirchhoff's first law follows from the conservation of charge.

Kirchhoff's second law follow from the conservation of energy.

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A student measures the terminal potential difference (V) of a cell (of emf $ε$ and internal resistance r) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I, then respectively, equal :

$ε a n d ‐ r$

$‐ r a n d ε$

$r a n d ‐ ε$

$‐ ε a n d r$

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Explanation

According to Ohm's law

$\frac{d V}{d I} = - r$

and $V = ε i f I = 0 [A s V + I r = ε]$

$∴ S l o p e o f t h e g r a p h = - r a n d i n t e r c e p t = ε$

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A wire of a certain material is streched slowly by ten per cent. Its new resistance and specific resistance become respectively

1.2 times, 1.1 times

1.21 times,same

both remain the same

1.1 times, 1.1 times

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Explanation

Key Idea: In streching, specific resistance remains unchanged. After streching, specific resistance $(ρ)$ will remain same. original resistance of the wire,

$R = \frac{p l}{A}$

or $R \propto \frac{l}{A} o r R \propto \frac{l^{2}}{V} (a s V = A l)$

and $R^{'} \propto \frac{{(l + 10 % l)}^{2}}{V}$

Therefore, $\frac{R^{'}}{R} = \frac{{(l + \frac{10}{100} l)}^{2}}{l^{2}}$

or $\frac{R^{'}}{R} = \frac{(\frac{{(11 l)}^{2}}{10})}{l^{2}}$ = $\frac{121}{100}$

or $R^{'} = 1.21 R$

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An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature $20^{°}$ C? The temperature of boiling water is $100^{°}$ C

6.3 min

8.4 min

12.6 min

4.2 min

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Explanation

Heat evovled due to Joule's effect is used up in boiling water.

VIt=ms $∆ t$

or t= $\frac{m s V t}{V I}$

putting under given values

I=4 A, V=220 volt, m= 1 kg

$∆ t = {(100 - 20)}^{°} C, s = 4200 J / k g^{°} C ∴ t = \frac{1 \times 4200 \times 80}{220 \times 4} = 6.3 m i n$

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A cell can be balanced against 110cm and 100 cm of potentiometer wire, respectively with and without being short-circuited through a resistance of $10 Ω$ . Its internal resistance is

1.0 $Ω$

0.5 $Ω$

2.0 $Ω$

zero

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Explanation

This problem is based on the application of potentiometer in which we find the internal resistance of a cell. In potentiometer experiment in which we find internal resistance of a cell, let E be the emf of the cell and V the terminal potential difference, then $\frac{E}{V} = \frac{l_{1}}{l_{2}}$

where $l_{1} a n d l_{2}$ are lengths of potentiometer wire with and without short circuited through a resistance.

since, $\frac{E}{V} = \frac{R + r}{R} [∴ E = I (R + r) a n d V = I R] ∴ \frac{R + r}{R} = \frac{I_{2}}{I_{2}} o r 1 + \frac{r}{R} = \frac{110}{100} o r \frac{r}{R} = \frac{110}{100} - 1 o r r = \frac{1}{10} \times 10 = 1 Ω$

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Two similar headlight lamps are connected in parallel to each other. Together, they consume 48 W from a 6 V battery. What is the resistance of each filament?

6 $Ω$

4 $Ω$

3.0 $Ω$

1.5 $Ω$

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Explanation

The power consumed by each lamp is 24 W.

Hence, using R = $(V^{2} / P)$ , we find R = (36/24) = 1.5 $Ω$ .

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Two electric bulbs, rated for the same voltage, have powers of 200 W and 100 W, respectively. If their resistances are $r_{1} a n d r_{2}$ , respectively then :

$r_{1} = 2 r_{2}$

$r_{2} = 2 r_{1}$

$r_{2} = 4 r_{1}$

$r_{1} = 4 r_{2}$

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Explanation

$P = \frac{V^{2}}{R}, \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}} o r r_{2} = 2 r_{1}$

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If the current in an electric bulb decreases by 0.5%, the power in the bulb decreases by approximately :

0.5%

0.25%

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Explanation

$P = i^{2} R, \frac{d P}{P} = 2 \frac{d I}{I} = 2 \times 0.5 % = 1 %$

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An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance R that must be put in series with the bulb, so that the bulb draws 500 W, is :

18 $Ω$

20 $Ω$

40 $Ω$

700 $Ω$

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A wire, when connected to a 220 V mains supply, has power dissipation $P_{1}$ . Now, the wire is cut into two equal pieces, which are connected in parallel to the same supply. Power dissipation in this case is $P_{2}$ . Then $P_{2} : P_{1}$ is :

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Explanation

$P = V^{2} / R$ .

R is reduced by a factor of 4.

So, P is increased by a factor of 4.

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