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Current Electricity MCQs for NEET — Physics Questions with Answers

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Two wires of resistance R₁ and R₂ have temperature coefficient of resistance $α_{1} and α_{2}$ , respectively. These are joined in series. The effective temperature coefficient of resistance is :

$\frac{α_{1} + α_{2}}{2}$

$\sqrt{α_{1} α_{2}}$

$\frac{α_{1} R_{1} + α_{2} R_{2}}{R_{1} + R_{2}}$

$\frac{\sqrt{R_{1} R_{2} α_{1} α_{2}}}{\sqrt{R_{1}^{2} + R_{2}^{2}}}$

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Explanation

$R_{t_{1}} = R_{1} (1 + α_{1} t)$ and $R_{t_{2}} = R_{2} (1 + α_{2} t)$

Also $R_{e q .} = R_{t_{1}} + R_{t_{2}} \Rightarrow R_{e q} = R_{1} + R_{2} + (R_{1} α_{1} + R_{2} α_{2}) t$

⇒ $R_{e q} = (R_{1} + R_{2}) \{1 + (\frac{R_{1} α_{1} + R_{2} α_{2}}{R_{1} + R_{2}}) . t\}$

So $α_{e f f} = \frac{R_{1} α_{1} + R_{2} α_{2}}{R_{1} + R_{2}}$

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When connected across the terminals of a cell, a voltmeter measures 5V and a connected ammeter measures 10 A of current. A resistance of 2 ohms is connected across the terminals of the cell. The current flowing through this resistance will be :

2.5 A

2.0 A

5.0 A

7.5 A

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Explanation

Emf E = 5V , Internal resistance $r = \frac{5}{10} = 0 .5 Ω$

Current through the resistance $i = \frac{5}{(2 + 0 .5)} = 2 A$

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Two resistances R₁ and R₂ are made of different materials. The temperature coefficient of the material of R₁ is α and of the material of R₂ is –β. The resistance of the series combination of R₁ and R₂ will not change with temperature, if R₁/ R₂ equals :

$\frac{α}{β}$

$\frac{α + β}{α - β}$

$\frac{α^{2} + β^{2}}{α β}$

$\frac{β}{α}$

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Explanation

$R_{1} + R_{2} = R_{1} (1 + α t) + R_{2} (1 - β t)$

⇒ $R_{1} + R_{2} = R_{1} + R_{2} + R_{1} α t - R_{2} β t$

⇒ $\frac{R_{1}}{R_{2}} = \frac{β}{α}$

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An ionization chamber with parallel conducting plates as anode and cathode has $5 \times 10^{7}$ electrons and the same number of singly-charged positive ions per cm³. The electrons are moving at 0.4 m/s. The current density from anode to cathode is $4 μ A / m^{2}$ . The velocity of positive ions moving towards cathode is :

0.4 m/s

16 m/s

Zero

0.1 m/s

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Explanation

Current density of drifting electrons j = nev

$n = 5 \times 10^{7} c m^{- 3} = 5 \times 10^{7} \times 10^{6} m^{- 3}$ .

$v = 0 .4 m s^{- 1}, e = 1 .6 \times 10^{- 19} C$ ⇒ $j = 3 .2 \times 10^{- 6} A m^{- 2}$

Current density of ions = (4 – 3.2) × 10^–6 = $0 .8 \times 10^{- 6} \frac{A}{m^{2}}$

This gives v for ions = 0.1 ms^–1.

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A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to current, the temperature of the wire is raised by ΔT in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross–section but of length 2 L. The temperature of the wire is raised by the same amount ΔT in the same time t. the value of N is

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Explanation

Let R and m be the resistance and mass of the first wire, then the second wire has resistance 2R and mass 2m. Let E = emf of each cell, S = specific heat capacity of the material of the wire. For the first wire, current $i_{1} = \frac{3 E}{R}$ and $i_{1}^{2} R t = m S Δ T$

For the second wire, $i_{2} = \frac{N E}{2 R}$ and $i_{2}^{2} (2 R) t = 2 m S Δ T$ . Thus, $i_{1} = i_{2}$ or $N = 6$ .

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The current in a conductor varies with time t as $I = 2 t + 3 t^{2}$ where I is in ampere and t in seconds. The electric charge flowing through a section of the conductor during t = 2 sec to t = 3 sec is :

10 C

24 C

33 C

44 C

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Explanation

dQ = Idt ⇒ $Q = \int_{t = 2}^{t = 3} I d t = [2 \int_{2}^{3} t d t + 3 \int_{2}^{3} t^{2} d t]$

= ${[t^{2}]}_{2}^{3} + {[t^{3}]}_{2}^{3}$ = (9 – 4) + (27 – 8) = 5 + 19 = 24C.

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Length of a hollow tube is 5m, it’s outer diameter is 10 cm and thickness of it’s wall is 5 mm. If the resistivity of the material of the tube is 1.7 × 10^–8 Ω×m then the resistance of the tube will be :

5.6 × 10^–5 Ω

2 × 10^–5 Ω

4 × 10^–5 Ω

None of these

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The resistance of the series combination of two resistance is S. When they are joined in parallel the total resistance is P. If S = nP, then the minimum possible value of n is :

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Explanation

If two resistances are $R_{1}$ and $R_{2}$ then

$S = R_{1} + R_{2}$ and $P = \frac{R_{1} R_{2}}{(R_{1} + R_{2})}$

From given condition S = nP i.e. $(R_{1} + R_{2}) = n (\frac{R_{1} R_{2}}{R_{1} + R_{2}})$

⇒ ${(R_{1} + R_{2})}^{2} = n R_{1} R_{2}$ ⇒ ${(R_{1} - R_{2})}^{2} + 4 R_{1} R_{2} = n R_{1} R_{2}$

So $n = 4 + \frac{{(R_{1} - R_{2})}^{2}}{R_{1} R_{2}} .$

Hence minimum value of n is 4.

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The V-I graph for a conductor makes an angle θ with V-axis. Here V denotes the voltage and I denotes current. The resistance of the conductor is given by :

$\sin θ$

$\cos θ$

$\tan θ$

$\cot θ$

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Explanation

At an instant approach the student will choose tanθ will be the right answer. But it is to be seen here the curve makes the angle θ with the V-axis. So it makes an angle (90 – θ) with the i-axis.

So resistance = slope = tan (90 – θ) = cotθ.

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The resistance of a wire is R ohm. If it is melted and stretched to n times its original lenght, its new resistance will be

$\frac{R}{n}$

$n^{2} R$

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

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Explanation

(c)Thinking process volume of material remains same in streching .

As volume remains same,

$A_{1} l_{1} = A_{2} l_{2}$

Now,given $l_{2} = n l_{1}$

So, New area $A_{2} = \frac{A_{1} l_{1}}{l_{2}} = \frac{A_{1}}{n}$

Resistance of wire after stretching

$R_{2} = ρ \frac{l_{2}}{A_{2}}$

= $ρ . \frac{n l_{2}}{A_{2} \ln}$

$= (ρ \frac{l_{1}}{A_{1}}) . n^{2} = n^{2} . R$

$[S o, R = (ρ \frac{l_{1}}{A_{1}})]$

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