# NEET and AIPMT NEET Physics Alternating Current MCQ Questions Solved

NEET - 2016

A filament bulb (500 W,100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is

(a) 230$\Omega$          (b) 46$\Omega$

(c) 26$\Omega$            (d) 13$\Omega$

(c) If a rated voltage and power are given,

then   ${P}_{rated}=\frac{{V}_{rated}^{2}}{R}$

$\therefore$Current in the bulb, $i=\frac{P}{V}$

$i=\frac{500}{100}=5A$

$\therefore$Resistance of bulb, ${R}_{b}=\frac{100×100}{500}=20\Omega$

$\because$Resistance R is connected in series.

$\therefore$ Current$i=\frac{E}{{R}_{net}}=\frac{230}{R+{R}_{0}}$

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NEET - 2016

Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication ?

(a)

(b)

(c)

(d)

(c) For better tuning, peak of current growth must be sharp. This is ensured by a high value of quality factor Q.

Now, quality factor is given by Q=$\frac{1}{R}\sqrt{\frac{L}{C}}$

From the given options highest value of Q is associated with $R=15\Omega$, L=3.5H and C=30 $\mu F$

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NEET - 2016

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

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

$\frac{80}{\sqrt{{\left(l{X}_{L}-l{X}_{C}\right)}^{2}+{\left(lR\right)}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{80}{\sqrt{{\left(100-40\right)}^{2}+{\left(80\right)}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{80}{\sqrt{{\left(60\right)}^{2}+{\left(80\right)}^{2}}}=\frac{80}{100}=0.8$

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NEET - 2016

A 100 $\Omega$ resistanceand a capacitor of 100 $\Omega$ 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

(a) Impedence of the R-C circuit, Z= $\sqrt{{R}^{2}+{X}_{c}^{2}}$

where, R= 100$\Omega$ and ${X}_{C}$=100$\Omega$

Peak value of the current,

${l}_{max}=\frac{{V}_{max}}{Z}=\frac{220\sqrt{2}}{100\sqrt{2}}=2.2A$

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NEET - 2016

A small signal voltage V(t)=Vo sinωt is applied across an ideal capacitor C
(a) over a full cycle the capacitor C does not consume any energy from the voltage source
(b) current I(t) is in phase with voltage V(t)
(c) current I(t) leads voltage V(t) by 180°
(d) current I(t) lags voltage V(t) by 90°

Concept Videos :-

#1 | Some Important Integrals
#3 | Av value of Alternating Current & Voltage
#4 | RMS Value of Alternating Current & Voltage
#5 | Solved Problems on RMS & Average Value
#23 | Solved Problems: Set 3

Concept Questions :-

Rms and average values

(a) For an AC circuit containing capacitor only, the phase difference between current and voltage will be π /2 (i.e.  90°).

In this case current is ahead of voltage by π /2.Hence,power in this case is given by
P=VI cosφ                             (φ= phase difference between voltage and current)
P=VI cos 90°=0

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NEET - 2016

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) 0.51 W

(d) Given
L=20mH C=50μf R=40 Ω   emf, V=10sin340t

∴ Power loss in AC circuit will be given as
Pav=I2VR=[EV/Z]2 R

=(10/√2)2 40 [1/402+(340x20x103-1/340x50x10-6)]2

=100/2x40x1/1600+(68-588)2
=2000/1600+2704=0.46w≈0.51W

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NEET - 2015

A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z' the power drawn will be

(a)P(R/Z)2

(b)P √R/Z

(c)P(R/Z)

(d)P

When a resistor is connected to an AC source. The power drawn will be
P= Vrms x Irms = Vrms x Vrms/R

=> V2rms=PR

When an inductor is connected in series with the resistor, then the power drawn will be
P'=Vrms.Irms cos φ

where,φ=phase difference

∴  P'=V2rms/R x R2/Z2=P x R x R/Z2

=>P'=P x R2/Z2=P(R/Z)2

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NEET - 2015

A series R-C circuit is connected to an alternating voltage source.Consider two situations:

1. When capacitor is air filled.

2. When capacitor is mica filled.

Current through resistor is i and voltage across capacitor is V then

(a)Va<Vb

(b)Va>Vb

(c)ia>ib

(d)Va=Vb

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NEET - 2013

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

(a) frequency of the AC source is decreased

(b) number of turns in the coil is reduced

(c) a capacitance of reactance Xc=XL is included in the same circuit

(d) an iron rod is inserted in the coil

(d) As Z=√R2+X2L=√R2+(2πfvL)2

As I=V/Z, P=I2R

i.e., V↑,L↑=>Z↑,I↓ and P↓

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NEET - 2012

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 $\mathrm{\pi }/3.$ If instead, C is removed from the circuit, the phase difference is again $\mathrm{\pi }/3.$ The power factor of the circuit is

(a)1/2                                            (b)1/$\sqrt{2}$

(c)1                                                (d)$\sqrt{3}/2$

Here, phase difference

$\mathrm{tan}\varphi =\frac{{X}_{L}-{X}_{C}}{R}$

$\mathrm{tan}\frac{\mathrm{\pi }}{3}=\frac{{X}_{L}-{X}_{C}}{R}$

When, L is removed

$\sqrt{3}=\frac{{X}_{C}}{R}$

${X}_{C}=\sqrt{3}R$

When C is removed

$tan\frac{\mathrm{\pi }}{3}=\sqrt{3}=\frac{{X}_{L}}{R}$

${X}_{L}=R\sqrt{3}$

Hence, in resonant circuit

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

$\varphi =0$

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

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