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Current Electricity MCQs for NEET — Physics Questions with Answers

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An unknown resistance R₁ is connected in series with a resistance of 10 Ω. This combinations is connected to one gap of a metre bridge while a resistance R₂ is connected in the other gap. The balance point is at 50 cm. Now, when the 10 Ω resistance is removed the balance point shifts to 40 cm. The value of R₁ is (in ohm)

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Explanation

For first balancing condition $\frac{10 + R_{1}}{R_{2}} = \frac{50}{50}$

⇒ $R_{2} = 10 + R_{1}$ . For second balancing condition $\frac{R_{1}}{R_{2}} = \frac{40}{60}$

⇒ $\frac{R_{1}}{10 + R_{1}} = \frac{2}{3}$ ⇒ $R_{1} = 20 Ω$

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A wire has a resistance of 6 Ω. It is cut into two parts and both half values are connected in parallel. The new resistance is :

12 Ω

1.5 Ω

3 Ω

6 Ω

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Explanation

Given $R = 6 Ω$ .

When resistor is cut into two equal parts and connected in parallel, then

$R_{e q} = \frac{R / 2}{2} = \frac{R}{4} = \frac{6}{4} = 1 .5 Ω$

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An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3, then the ratio of the currents passing through the wire will be

1/3

8/9

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Explanation

$\frac{i_{1}}{i_{2}} = \frac{R_{2}}{R_{1}} = \frac{l_{2}}{l_{1}} \times {(\frac{r_{1}}{r_{2}})}^{2} = \frac{3}{4} {(\frac{2}{3})}^{2} = \frac{1}{3}$

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When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, the resistance between any two of diametrically opposite points will be :

$\frac{R}{4}$

$\frac{R}{8}$

$\frac{R}{2}$

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By using only two resistance coils-singly, in series, or in parallel one should be able to obtain resistances of 3, 4, 12, and 16 ohms. The separate resistances of the coil are :

3 and 4

4 and 12

12 and 16

16 and 3

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Explanation

When resistances $4 Ω$ and $12 Ω$ are connected in series $= 4 + 12 = 16 Ω$

When these resistances are connected in parallel,

$\frac{1}{R_{P}} = \frac{1}{4} + \frac{1}{12} \Rightarrow R_{P} = \frac{4 \times 12}{4 + 12} = \frac{4 \times 12}{16} = 3 Ω$

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The e.m.f. of a cell is E volts and internal resistance is r ohm. The resistance in external circuit is also r ohm. The p.d. across the cell will be

E/2

E/4

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Explanation

Since both the resistors are same, therefore potential difference $= V + V = E$

⇒ $V = \frac{E}{2}$

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Kirchhoff's first law i.e. $Σ i = 0$ at a junction is based on the law of conservation of :

Charge

Energy

Momentum

Angular momentum

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Explanation

Kirchhoff's first law is based on the law of conservation of charge.

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The terminal potential difference of a cell when short-circuited is (E = E.M.F. of the cell)

E/2

Zero

E/3

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Explanation

In short circuiting R = 0, so V = 0

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The potential difference in open circuit for a cell is 2.2 volts. When a 4-ohm resistor is connected between its two electrodes the potential difference becomes 2 volts. The internal resistance of the cell will be :

1 ohm

0.2 ohm

2.5 ohm

0.4 ohm

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Explanation

$(4 + r) i = 2 .2$ ......(1)

and $4 i = 2$ ⇒ $i = \frac{1}{2}$ .....(2)

Putting the value of i in (1), we get r = 0.4 ohm.

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A cell whose e.m.f. is 2 V and internal resistance is 0.1 Ω, is connected with a resistance of 3.9 Ω. The voltage across the cell terminal will be :

0.50 V

1.90 V

1.95 V

2.00 V

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Explanation

The voltage across cell terminal will be given by

$= \frac{E}{R + r} \times R = \frac{2}{(3 .9 + 0 .1)} \times 3 .9 = 1 .95 V$

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