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Physics NEET Questions — Free MCQ Practice

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The potential differences across the resistance, capacitance and inductance are 80V, 40Vand 100V respectively in an L-C-R circuit. The power factor of this circuit is

A.

0.4

B.

0.5

C.

0.8

D.

1.0

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Explanation

(c) Power factor of the L-C-R circuit

$= \cos ϕ = \frac{R}{Z} = \frac{I R}{I Z} = \frac{80}{\sqrt{{(X_{L} - X_{C})}^{2} + R^{2}}}$

= $\frac{80 I}{\sqrt{{(l X_{L} - l X_{C})}^{2} + {(l R)}^{2}}} = \frac{80}{\sqrt{{(100 - 40)}^{2} + {(80)}^{2}}} = \frac{80}{\sqrt{{(60)}^{2} + {(80)}^{2}}} = \frac{80}{100} = 0.8$

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A 100 $Ω$ resistance and a capacitor of 100 $Ω$ reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is

A.

2.2 A

B.

11A

C.

4.4A

D.

11 $\sqrt{2}$ A

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Explanation

(a) The impedance of the R-C circuit, Z= $\sqrt{R^{2} + X_{c}^{2}}$

where, R= 100 $Ω$ and $X_{C}$ =100 $Ω$

$\Rightarrow Z = \sqrt{{(100)}^{2} + {(100)}^{2}} = 100 \sqrt{2} Ω$

The peak value of the current,

$l_{m a x} = \frac{V_{m a x}}{Z} = \frac{220 \sqrt{2}}{100 \sqrt{2}} = 2.2 A$

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An inductor 20 mH, a capacitor 50μF, and a resistor 40Ω are connected in series across a source of emf V=10sin340t. The power loss in the AC circuit is:

A.

0.67 W

B.

0.78W

C.

0.89 W

D.

0.46 W

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Explanation

(d)
$G i v e n : L = 20 m H, C = 50 μ F a n d V = 10 \sin 340 t P o w e r l o s s i n A C c i r c u i t, P_{a v} = {I_{v}}^{2} R = {(\frac{E_{v}}{Z})}^{2} R P_{a v} = {(\frac{\frac{10}{\sqrt{2}}}{\sqrt{{(40)}^{2} + {(340 \times 20 \times 10^{- 3} - \frac{1}{340 \times 50 \times 10^{- 6}})}^{2}}})}^{2} \times 40 P_{a v} = 0.46 W$

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A coil of self-inductance L is connected in series with a bulb B and an AC source. The brightness of the bulb decreases when

A.

frequency of the AC source is decreased

B.

the number of turns in the coil is reduced

C.

a capacitance of reactance X_c=X_L is included in the same circuit

D.

an iron rod is inserted in the coil

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Explanation

(d) As Z=√R²+X²L=√R²+(2πfvL)²

As I=V/Z, P=I²R

i.e., V↑,L↑=>Z↑,I↓ and P↓

$A s w e k n o w; Z = \sqrt{R^{2} + {X_{L}}^{2}} = = \sqrt{R^{2} + {(2 πfVL)}^{2}} A s V a n d L i n c r e a s e s, Z a l s o i n c r e a s e s . \Rightarrow I = \frac{V}{Z} d e c r e a s e s a n d P = I^{2} R a l s o d e c r e a s e s .$

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In an electrical circuit R, L, C, and an AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is $π / 3 .$ If instead, C is removed from the circuit, the phase difference is again $π / 3 .$ The power factor of the circuit is

A.

1/2

B.

1/ $\sqrt{2}$

C.

1

D.

$\sqrt{3} / 2$

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Explanation

Here, phase difference

$\tan ϕ = \frac{X_{L} - X_{C}}{R}$

$\tan \frac{π}{3} = \frac{X_{L} - X_{C}}{R}$

When, L is removed

$\sqrt{3} = \frac{X_{C}}{R}$

$X_{C} = \sqrt{3} R$

When C is removed

$t a n \frac{π}{3} = \sqrt{3} = \frac{X_{L}}{R}$

$X_{L} = R \sqrt{3}$

Hence, in resonant circuit

$\tan ϕ = \frac{\sqrt{3} R - \sqrt{3} R}{R} = 0$

$ϕ = 0$

$∴ P o w e r f a c t o r \cos ϕ = 1$

It is the condition of resonance therefore phase difference between voltage and current is zero and power factor is cos $ϕ$ =1.

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The instantaneous values of alternating

current and voltages in a circuit are given

as

$i = \frac{1}{\sqrt{2}} \sin (100 πt) ampere e = \frac{1}{\sqrt{2}} \sin (100 πt + π / 3) volt$

The average power in Watts consumed in the

circuit is

A.

$\frac{1}{4}$

B.

$\frac{\sqrt{3}}{4}$

C.

$\frac{1}{2}$

D.

$\frac{1}{8}$

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Explanation

Given equations

$i = \frac{1}{\sqrt{2}} \sin (100 πt)$

and $e = \frac{1}{\sqrt{2}} \sin (100 πt + π / 3) so, i_{e} = \frac{1}{\sqrt{2}} and V_{e} = \frac{1}{\sqrt{2}}$

We know that average power

$P_{av} = V_{rms} \times i_{rms} cosϕ = \frac{1}{2} \times \frac{1}{2} \times \cos 60 °$

$[since, i_{rms} = \frac{i_{o}}{\sqrt{2}} and V_{rms} = \frac{V_{o}}{\sqrt{2}}]$

$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} W$

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In an AC circuit an alternating voltage $e = 200 \sqrt{2} \sin 100 t$ volt is connected to a capacitor $1 μF .$ The $rms$ value of the current in the circuit is

A.

100 mA

B.

200 mA

C.

20 mA

D.

10 mA

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Explanation

$Given, e = 200 \sqrt{2} \sin 100 t and C = 1 μF$

$E_{rms} = 200 V X_{c} = \frac{1}{ωC} = \frac{1}{1 \times 10^{- 6} \times 100} = 10^{4} Ω i_{rms} = \frac{E_{rms}}{X_{C} \cdot} i_{rms} = \frac{200}{10^{4}} = 2 \times 10^{- 2} A = 20 mA$

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An AC voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to $3 Ω$ , the phase difference between the applied voltage and the current in the circuit is

A.

$π / 4$

B.

$π / 2$

C.

zero

D.

$π / 6$

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Explanation

$\tan ϕ = \frac{X_{L}}{R} = \frac{Lω}{R} \tan ϕ = \frac{3 Ω}{3 Ω} \tan ϕ = 1 ϕ = \tan^{- 1} (1) ϕ = 45 ° ϕ = \frac{π}{4} rad$

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Power dissipated in an L-C-R series circuit connected to an AC source of emf $ε$ is

A.

$\frac{ε^{2} R}{[R^{2} + {(L ω - \frac{1}{C ω})}^{2}]}$

B.

$\frac{ε^{2} {\sqrt{R^{2} + (L ω - \frac{1}{C ω})}}^{2}}{R}$

C.

$\frac{ε^{2} [R^{2} + {(L ω - \frac{1}{C ω})}^{2}]}{R}$

D.

$\frac{ε^{2} R}{\sqrt{R^{2} + {(L ω - \frac{1}{C ω})}^{2}}}$

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Explanation

Power dissipated in series L-C-R.

$P = I_{r m s}^{2} R = \frac{ε_{r m s}^{2} R}{{|Z|}^{2}} = \frac{ε^{2} R}{[R^{2} + {(ω L - \frac{1}{ω C})}^{2}]}$

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In AC circuit the emf (e) and the current (i) at any instant are given respectively by

$e = E_{o} \sin ω t$

$i = I_{o} \sin (ω t - ϕ)$

The average power in the circuit over one cycle of AC is

A.

$\frac{E_{o} I_{o}}{2}$

B.

$\frac{E_{o} I_{o}}{2} \sin ϕ$

C.

$\frac{E_{o} I_{o}}{2} \cos ϕ$

D.

$E_{o} I_{o}$

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Explanation

The power is defined as the rate at which work is being done in the circuit.

Power= rate of work done in one complete cycle.

or $P_{a v} = \frac{W}{T}$

or $P_{a v} = \frac{(E_{o} I_{o} \cos ϕ) T / 2}{T}$

or $P_{a v} = \frac{E_{o} I_{o}}{2} ϕ$

where cos $ϕ$ is called the power factor of an AC circuit.

Physics MCQs for NEET — Practice Questions with Answers

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