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Current Electricity MCQs for NEET — Physics Questions with Answers

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A student has 10 resistors of resistance ‘r’. The minimum resistance made by him from given resistors is :

10 r

$\frac{r}{10}$

$\frac{r}{100}$

$\frac{r}{5}$

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Explanation

To obtain minimum resistance, all resistors must be connected in parallel.

Hence equivalent resistance of combination $= \frac{r}{10}$

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Two wires of same metal have the same length but their cross-sections are in the ratio 3 : 1. They are joined in series. The resistance of the thicker wire is 10 Ω. The total resistance of the combination will be

40 Ω

$\frac{40}{3} Ω$

$\frac{5}{2} Ω$

100 Ω

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Explanation

For same material and same length

$\frac{R_{2}}{R_{1}} = \frac{A_{1}}{A_{2}} = \frac{3}{2}$ ⇒ $R_{2} = 3 R_{1}$

Resistance of thick wire $R_{1} = 10 Ω$

∴ Resistance of thin wire $R_{2} = 30 Ω$

Total resistance in series = 10 + 30 = 40 Ω

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A wire of resistance R is cut into ‘n’ equal parts. These parts are then connected in parallel. The equivalent resistance of the combination will be :

$\frac{R}{n}$

$\frac{n}{R}$

$\frac{R}{n^{2}}$

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Explanation

The resistance of each part will be = $\frac{R}{n}$

Such n parts are joined in parallel so, $R_{e q} = \frac{R}{n^{2}}$ .

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n equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum to the minimum resistance

$\frac{1}{n^{2}}$

n²

$\frac{1}{n}$

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Explanation

$R_{\max} = n R$ and $R_{\min} = R / n$

⇒ $\frac{R_{\max}}{R_{\min}} = n^{2}$

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Two wires of equal diameters, of resistivities ρ₁and ρ₂ and lengths l₁ and l₂, respectively, are joined in series. The equivalent resistivity of the combination is :

$\frac{ρ_{1} l_{1} + ρ_{2} l_{2}}{l_{1} + l_{2}}$

$\frac{ρ_{1} l_{2} + ρ_{2} l_{1}}{l_{1} - l_{2}}$

$\frac{ρ_{1} l_{2} + ρ_{2} l_{1}}{l_{1} + l_{2}}$

$\frac{ρ_{1} l_{1} - ρ_{2} l_{2}}{l_{1} - l_{2}}$

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Explanation

$R_{1} = \frac{ρ_{1} l_{1}}{A} and R_{2} = \frac{ρ_{2} l_{2}}{A}$ .

In series $R_{e q} = R_{1} + R_{2}$

$\frac{ρ_{e q .} (l_{1} + l_{2})}{A} = \frac{ρ_{1} l_{1}}{A} + \frac{ρ_{2} l_{2}}{A}$ $\Rightarrow ρ_{e q} = \frac{ρ_{1} l_{1} + ρ_{2} l_{2}}{l_{1} + l_{2}}$ .

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Four resistances of 100 Ω each are connected in the form of square. Then, the effective resistance along the diagonal points is :

200 Ω

400 Ω

100 Ω

150 Ω

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The resistors of resistances 2 Ω, 4 Ω and 8 Ω are connected in parallel, then the equivalent resistance of the combination will be :

$\frac{8}{7} Ω$

$\frac{7}{8} Ω$

$\frac{7}{4} Ω$

$\frac{4}{9} Ω$

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Explanation

$\frac{1}{R_{e q}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{4 + 2 + 1}{8}$

⇒ $R_{e q} = \frac{8}{7} Ω$ .

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The effective resistance of two resistors in parallel is $\frac{12}{7} Ω$ . If one of the resistors is disconnected the resistance becomes 4 Ω. The resistance of the other resistor is :

4 Ω

3 Ω

$\frac{12}{7} Ω$

$\frac{7}{12} Ω$

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Explanation

$\frac{7}{12} = \frac{1}{4} + \frac{1}{R} \Rightarrow R = 3 Ω$

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Three resistors are connected to form the sides of a triangle ABC, the resistance of the sides AB, BC and CA are 40 ohms, 60 ohms and 100 ohms respectively. The effective resistance between the points A and B in ohms will be

200

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Two wires of the same dimensions but resistivities ρ₁ and ρ₂ are connected in series. The equivalent resistivity of the combination is

ρ₁ + ρ₂

$\frac{ρ_{1} + ρ_{2}}{2}$

$\sqrt{ρ_{1} ρ_{2}}$

$2 (ρ_{1} + ρ_{2})$

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Explanation

$R_{e q} = R_{1} + R_{2}$

⇒ $\frac{ρ_{e f f . 2 l}}{A} = \frac{ρ_{1} l}{A} + \frac{ρ_{2} l}{A}$

$\Rightarrow ρ_{e f f .} = \frac{ρ_{1} + ρ_{2}}{2}$ .

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