JEE Physics Alternating Current MCQs Set 11

Answer: (d) 70.7 ohm; 0.707

Explanation:
1. \( X_L=100\pi\cdot\frac3\pi=300\ \Omega,\ X_C=\frac{1}{100\pi\cdot40\times10^{-6}/\pi}=250\ \Omega \).
2. \( Z=\sqrt{50^2+50^2}=70.7\ \Omega,\ \cos\phi=R/Z=0.707 \).

Teacher's Note:
1. The pi in L and C cancels with the pi in omega.
2. Power factor = R/Z.

Answer: (b) 1

Explanation:
1. At resonance \( X_L=X_C \), so \( Z=R \) and \( \cos\phi=R/Z=1 \).

Teacher's Note:
1. At resonance the circuit is purely resistive.
2. Current and voltage are in phase.

Answer: (a) 311 V and 2.5 ms

Explanation:
1. Peak voltage \( V_0=\sqrt2\,V_{rms}=1.414\times220\approx311\ \text{V} \).
2. The rms value occurs at phase 45 degrees, so going to zero takes one-eighth of a period: \( t=\frac T8=\frac{20\ \text{ms}}{8}=2.5\ \text{ms} \).

Teacher's Note:
1. V = V0 sin(wt) equals V0/sqrt2 when wt = 45 degrees, which is T/8 after zero.
2. Time from rms value to zero is therefore T/8, not T/4.

Answer: (b) 100 V

Explanation:
1. \( V=\sqrt{60^2+80^2}=100\ \text{V} \).

Teacher's Note:
1. R and L voltages are 90 degrees apart.
2. 3-4-5 again.

Answer: (a) 100%

Explanation:
1. \( \cos\phi=R/Z\Rightarrow Z_1=2R,\ Z_2=4R \).
2. Impedance goes from 2R to 4R, an increase of 100%.

Teacher's Note:
1. With R fixed, Z = R/cos phi.
2. Percentage increase = (4R - 2R)/2R.

Answer: (d) 23 A and zero

Explanation:
1. \( \omega=200,\ X_L=20\ \Omega,\ X_C=\frac1{200\times500\times10^{-6}}=10\ \Omega \).
2. \( I_{rms}=230/10=23\ \text{A} \); with no resistance the average power is zero.

Teacher's Note:
1. The current is wattless when R = 0.
2. Net reactance is 20 - 10 = 10 ohm.

Answer: (d) 20 A

Explanation:
1. \( X_L=\omega L=100\times5\times10^{-3}=0.5\ \Omega \).
2. \( I_0=10/0.5=20\ \text{A} \).

Teacher's Note:
1. Convert mH to H.
2. Here omega = 100 rad/s.

Answer: (c) may be zero

Explanation:
1. If the interval starts at t = 0 the mean is \( 2E_0/\pi \), but if it starts at a quarter period the positive and negative parts cancel.
2. So it depends on the starting instant, and may be zero.

Teacher's Note:
1. Half a cycle averages to zero only when it is centred on a zero crossing.
2. Always check the start of the averaging interval.

Answer: (b) 125 ohm and 288 J

Explanation:
1. \( X_L=200\ \Omega,\ X_C=100\ \Omega,\ Z=\sqrt{75^2+100^2}=125\ \Omega \).
2. \( I_{rms}=\frac{10/\sqrt2}{125}=0.0566\ \text{A},\ P=I^2R=0.24\ \text{W} \).
3. \( E=0.24\times1200=288\ \text{J} \).

Teacher's Note:
1. Dissipation takes place only in the resistor.
2. Remember to convert 20 minutes to 1200 s.

Answer: (d) high inductance and low resistance

Explanation:
1. A choke reduces ac current by its reactance \( \omega L \) (high L) without wasting power, so its resistance should be small.

Teacher's Note:
1. Its reactance does the limiting.
2. Low R means little heat.

Answer: (d) \( \tfrac12I_p^2R \)

Explanation:
1. \( I_{rms}=I_p/\sqrt2\Rightarrow P=I_{rms}^2R=\tfrac12I_p^2R \).

Teacher's Note:
1. Average of sin^2 over a cycle is 1/2.
2. Half the peak-current power.

Answer: (c) \( 20\ \text{mH},\ \tfrac{25}8\ \mu\text{F} \)

Explanation:
1. \( I=60/120=0.5\ \text{A},\ X_L=40/0.5=80\ \Omega,\ L=80/4000=20\ \text{mH} \).
2. \( C=\frac1{\omega_0^2L}=\frac1{16\times10^6\times0.02}=3.125\ \mu\text{F}=\tfrac{25}8\ \mu\text{F} \).

Teacher's Note:
1. The current at resonance is V_R/R.
2. Use omega0^2 = 1/(LC).

Answer: (b) 849 V and 707 V

Explanation:
1. \( I=200/70.7=2.83\ \text{A} \).
2. \( V_L=2.83\times300=849\ \text{V},\ V_C=2.83\times250=707\ \text{V} \).

Teacher's Note:
1. Individual voltages in a series LCR circuit can exceed the supply voltage.
2. Use I times reactance for each element.

Answer: (a) \( \sqrt{17}\ \text{A} \)

Explanation:
1. \( I_{rms}^2=3^2+\tfrac12(4)^2=17 \).
2. So \( I_{rms}=\sqrt{17}\ \text{A} \).

Teacher's Note:
1. The dc part contributes its square, the ac part half the square of its amplitude.
2. Never add rms values linearly.

Answer: (b) 10 W and 5 W

Explanation:
1. At t = 2/3 ms: \( 250\pi t=\pi/6 \), so \( i=2\sin30^\circ=1\ \text{A} \) and \( \varepsilon=10\sin90^\circ=10\ \text{V} \); instantaneous power = 10 W.
2. Average power \( =\tfrac12\varepsilon_0i_0\cos\phi=\tfrac12(10)(2)\cos60^\circ=5\ \text{W} \).

Teacher's Note:
1. Instantaneous power is simply e times i at that instant.
2. Average power uses the phase difference between e and i.

Answer: (c) \( 2\sqrt3\ \text{A} \)

Explanation:
1. \( I_{rms}^2R=3\,(2^2R)\Rightarrow I_{rms}^2=12 \).
2. \( I_{rms}=2\sqrt3\ \text{A} \).

Teacher's Note:
1. Heat scales with the square of the current.
2. Three times the heat means sqrt3 times the current.

Answer: (d) 5 A

Explanation:
1. \( I_{rms}=\sqrt{I_{dc}^2+I_{ac,rms}^2}=\sqrt{9+16}=5\ \text{A} \).

Teacher's Note:
1. Squares add, because heating effects add.
2. 3-4-5.

Answer: (c) Zero

Explanation:
1. In each period the voltage is linear and symmetric about zero, so the positive area equals the negative area.
2. Hence the average over any whole number of periods is zero.

Teacher's Note:
1. The average of a function over a period is the net signed area divided by the time.
2. Symmetric positive and negative areas cancel.

Answer: (d) 2000 Hz, 2300 W and 500/23 (about 21.7)

Explanation:
1. \( \omega_0=\frac1{\sqrt{LC}}=4000\pi,\ f_0=2000\ \text{Hz} \).
2. \( P_{max}=\frac{V^2}{R}=\frac{230^2}{23}=2300\ \text{W} \).
3. \( Q=\frac{\omega_0L}R=\frac{4000\pi\cdot0.125/\pi}{23}=\frac{500}{23} \).

Teacher's Note:
1. At resonance Z = R.
2. Q = omega0 L / R.

Answer: (b) none of these is true

Explanation:
1. The source voltage is \( \sqrt{V_3^2+(V_1-V_2)^2} \).
2. This equals none of the three expressions listed.

Teacher's Note:
1. The voltages are phasors with different phases.
2. Phasor sum, not arithmetic sum.

Answer: (c) 25 ohm

Explanation:
1. Power is dissipated only in R: \( P=I^2R\Rightarrow R=100/4=25\ \Omega \).

Teacher's Note:
1. Use P = I^2 R, not V^2/R.
2. The voltage across R is not 200 V.