JEE Physics Alternating Current MCQs Set 12

Answer: (a) 1, 2 and 3 only

Explanation:
1. Zero current at maximum voltage means a phase difference of 90 degrees.
2. This holds for a pure L, a pure C, or an L-C combination (net reactance only).
3. A resistor keeps current in phase with voltage.

Teacher's Note:
1. A 90 degree phase difference needs purely reactive behaviour.
2. Resistance makes the phase difference less than 90 degrees.

Answer: (a) 1, 3 and 4 only

Explanation:
1. \( \omega=1000 \): \( X_L=100 \), \( X_C=200 \), \( Z=\sqrt{100^2+100^2}=141.4\ \Omega \).
2. With peak emf \( 150\sqrt2 \) the rms emf is \( \dfrac{150\sqrt2}{\sqrt2}=150 \) V, so \( I_{rms}=1.06 \) A and \( P=I^2R=112.5\ \text{W} \), not 225 W.
3. Ideal L and C dissipate no power.

Teacher's Note:
1. Pure reactances never dissipate average power.
2. Power in R is Irms squared times R.

Answer: (b) 1 and 3 only

Explanation:
1. \( \omega=100 \), \( X_C=\dfrac1{100\times25\times10^{-6}}=400\ \Omega \), \( Z=500\ \Omega \).
2. \( I_0=50/500=0.1\ \text{A} \); \( P=\frac12I_0^2R=1.5\ \text{W} \).

Teacher's Note:
1. Average power uses the rms current, i.e. half of the peak squared.
2. Impedance of RC is the root of R squared plus XC squared.

Answer: (d) 1, 2 and 3 only

Explanation:
1. \( f_0=\dfrac1{2\pi\sqrt{LC}}=\dfrac1{2\pi\sqrt{10^{-6}/\pi^2}}=500\ \text{Hz} \).
2. At resonance \( Z=R \), so I = V/R, and VL and VC are opposite in phase.
3. Statement 4 would hold only if the reactances added, which they do not.

Teacher's Note:
1. At resonance XL = XC and they cancel.
2. The capacitor and inductor voltages are always in opposition.

Answer: (c) 1 and 2 only

Explanation:
1. Emf and current are sinusoidal, so their cycle averages are zero.
2. Joule heat \( i^2R \) and stored energy \( \frac12Li^2 \) are never negative, so their averages are positive.

Teacher's Note:
1. Only quantities that change sign can average to zero.
2. Squared quantities cannot.

Answer: (d) 1.2 mJ

Explanation:
1. \( I_{rms}=100/100=1\ \text{A} \).
2. Average energy \( =\dfrac12L_2I_{rms}^2=\dfrac12(2.4\times10^{-3})(1)=1.2\ \text{mJ} \) (since the mean of \( \sin^2 \) is 1/2, equal to \( \tfrac14L_2I_0^2 \)).

Teacher's Note:
1. Average of i squared over a cycle is Irms squared.
2. Energy in an inductor is one half L i squared.

Answer: (b) 5000 rad/s

Explanation:
1. Total inductance \( L=1.6+2.4=4\ \text{mH} \), total capacitance \( C=3+4.5+2.5=10\ \mu\text{F} \).
2. \( \omega_0=\dfrac1{\sqrt{LC}}=\dfrac1{\sqrt{4\times10^{-3}\times10^{-5}}}=5000\ \text{rad/s} \).

Teacher's Note:
1. Parallel capacitances add; series inductances add.
2. Maximum current occurs at resonance.

Answer: (d) 4.8 V

Explanation:
1. \( Z=\sqrt{8^2+6^2}=10\ \Omega \), \( I_0=1\ \text{A} \), current leads by \( 37^\circ \). Voltage across L and C: \( 6\cos(\theta-53.13^\circ) \) with \( \theta=100\pi t \).
2. Condition \( 6\cos(\theta-53.13^\circ)=5\cos\theta \) gives \( \tan\theta=7/24 \), \( \cos\theta=0.9604 \).
3. Required voltage \( 5\cos\theta=4.80\ \text{V} \)

Teacher's Note:
1. Write both voltages as cosines with phases and equate them.
2. The LC pair voltage lags the current by 90 degrees.

Answer: (d) \( 100\sqrt2 \) V

Explanation:
1. With S open the circuit is at resonance: \( \omega^2=\dfrac1{LC}=2500\pi^2 \), \( \omega=50\pi \) rad/s.
2. Then \( X_L=50\pi\times\dfrac1{25\pi}=2\ \Omega \).
3. With S closed the phase is 45 degrees, so \( R=X_L=2\ \Omega \), \( I_{rms}=\dfrac{100/\sqrt2}{2}=\dfrac{50}{\sqrt2} \) A and \( Z=2\sqrt2 \), giving \( V_{rms}=100 \) V.
4. Peak voltage \( =100\sqrt2\ \text{V} \).

Teacher's Note:
1. With the capacitor shorted only R and L remain.
2. Peak = sqrt2 times rms.

Answer: (c) 2 and 4 only

Explanation:
1. Average power \( =V_{rms}I_{rms}\cos\phi=1000\cos\phi \) W.
2. Since \( \cos\phi\le1 \), it can equal 1000 W (resistive circuit) or be smaller, never larger.

Teacher's Note:
1. The power factor lies between 0 and 1.
2. 1000 W is the maximum possible value.

Answer: (a) \( \dfrac{1000}{\sqrt2} \) W and 10 A

Explanation:
1. \( P=\dfrac{100\times20}{2}\cos45^\circ=\dfrac{1000}{\sqrt2}\ \text{W} \).
2. Wattless (rms) current \( \dfrac{20}{\sqrt2}\sin45^\circ=10\ \text{A} \).

Teacher's Note:
1. Use rms values of both emf and current.
2. The wattless part carries sin(phi).

Answer: (d) 242 W

Explanation:
1. Equal angles give \( X_L=X_C \): resonance, \( Z=R \).
2. \( P=\dfrac{V^2}{R}=\dfrac{220^2}{200}=242\ \text{W} \).

Teacher's Note:
1. Equal and opposite phase angles mean XL = XC.
2. At resonance the power is V squared over R.

Answer: (a) 3 and 4 only

Explanation:
1. A pure resistor has phi = 0.
2. An LCR circuit at resonance has XL = XC, so phi = 0.
3. Pure L or pure C have phi = 90 degrees, power factor 0.

Teacher's Note:
1. Power factor is cos(phi) = R/Z.
2. It equals 1 only when net reactance is zero.

Answer: (d) R and C

Explanation:
1. Phase lead in time \( \Delta t=\pi/400 \) s at \( \omega=100 \) gives \( \phi=\omega\,\Delta t=\pi/4 \) (45 degrees).
2. A current that leads by 45 degrees (less than 90) needs a resistor and a capacitor.

Teacher's Note:
1. Current leading means a capacitor is present.
2. A phase angle below 90 degrees means resistance is also present.

Answer: (c) 24 W

Explanation:
1. Capacitive reactance \( X_C=1/(\omega C)=1/(50\times\frac1{50})=1\ \Omega \); coil reactance \( X_L=0.08\times50=4\ \Omega \).
2. Net reactance 3 ohm, \( Z=\sqrt{3^2+3^2}=3\sqrt2\ \Omega \).
3. \( P=\dfrac{V^2R}{Z^2}=\dfrac{144\times3}{18}=24\ \text{W} \) (the 12 V is taken as the rms value, as in the source data).

Teacher's Note:
1. Average power is Vrms squared times R over Z squared.
2. The capacitor reduces the net reactance and so raises the power.

Answer: (b) 1, 2, 3 and 4 only

Explanation:
1. 1 is the basic definition.
2. Since \( R=|z|\cos\phi \), 2 and 4 are the same quantity written differently.
3. For 3: \( I_{rms}^2R=\dfrac{E_{max}^2}{2|z|^2}R \).

Teacher's Note:
1. Power factor cos(phi) equals R over |z|.
2. All four forms are equivalent.

Answer: (b) 1 and 3 only

Explanation:
1. Line current \( I=\dfrac{600\times10^3}{3000}=200\ \text{A} \).
2. Line resistance \( 0.4\times20\times2=16\ \Omega \), loss \( 200^2\times16=640\ \text{kW} \).
3. Plant output \( 600+640=1240\ \text{kW} \).

Teacher's Note:
1. A two-wire line has twice the length of resistance.
2. Plant power is load plus line loss.

Answer: (a) \( I_0/\sqrt2 \)

Explanation:
1. Originally \( X_C=R \), so \( Z=\sqrt2R \) and \( I_0=V_0/(\sqrt2R) \).
2. At \( \omega/\sqrt3 \): \( X_C'=\sqrt3R \), \( Z'=\sqrt{R^2+3R^2}=2R \).
3. \( I'=\dfrac{V_0}{2R}=\dfrac{I_0\sqrt2}{2}=\dfrac{I_0}{\sqrt2} \).

Teacher's Note:
1. Lower frequency raises capacitive reactance.
2. Compare impedances before and after to get the ratio of currents.

Answer: (c) 0.5 A

Explanation:
1. \( X_L=2\pi\times50\times\frac{2.2}{\pi}=220\ \Omega \), so \( Z=220\sqrt2\ \Omega \) and \( I_{rms}=1/\sqrt2=0.707\ \text{A} \).
2. Phase angle is 45 degrees, so wattless component \( I\sin\phi=0.707\times0.707=0.5\ \text{A} \).

Teacher's Note:
1. The wattless component is the part of the current 90 degrees out of phase with the voltage.
2. It equals I sin(phi), while the power component is I cos(phi).

Answer: (a) 2, 3 and 4 only

Explanation:
1. \( V_R=\dfrac{V_sR}{\sqrt{R^2+X_C^2}} \) gives \( \dfrac{V_s^2}{V_R^2}=1+\dfrac{1}{4\pi^2f^2C^2R^2} \).
2. So \( 1/f^2 \) is linear in \( V_s^2/V_R^2 \) and in \( 1/V_R^2 \) (Vs fixed).
3. Also \( \dfrac{V_R}{\sqrt{V_s^2-V_R^2}}=\dfrac{R}{X_C}=2\pi fCR \), linear in f.
4. f^2 against VR^2 is not linear.

Teacher's Note:
1. Rearrange the voltage divider relation until one side is the inverse square of the frequency.
2. The ratio VR over VC equals 2 pi f C R.

Answer: (d) 1 and 2 only

Explanation:
1. \( \omega=100 \) rad/s, \( X_L=100\ \Omega \), \( X_C=\dfrac1{100\times2\times10^{-5}}=500\ \Omega \).
2. \( Z=\sqrt{300^2+400^2}=500\ \Omega \), \( I=50/500=0.1\ \text{A} \), \( V_C=0.1\times500=50\ \text{V} \).

Teacher's Note:
1. Find omega from the frequency, then both reactances.
2. Voltage across a part is current times its reactance.

Answer: (d) 4

Explanation:
1. \( R^2+X_C^2=1.25R^2 \) gives \( X_C=0.5R \).
2. \( \dfrac{1}{\omega C}=0.5R \) gives \( RC=\dfrac2\omega=\dfrac{2}{500}=4\ \text{ms} \).

Teacher's Note:
1. Time constant of an RC circuit is RC.
2. Use XC in terms of R to eliminate both.

Answer: (b) 1 and 2 only

Explanation:
1. \( X_L=2\pi fL=2\times\frac{22}{7}\times70=440\ \Omega \).
2. \( I=110/440=0.25\ \text{A} \).

Teacher's Note:
1. Reactance is 2 pi f L.
2. Rms current is rms voltage over reactance.

Answer: (d) 1 and 3 only

Explanation:
1. Branch 1: \( Z_1=100-100j \), \( I_1=20/141.4=0.141\ \text{A} \) at +45 degrees; branch 2: \( Z_2=50+50j \), \( I_2=0.283\ \text{A} \) at -45 degrees.
2. Sum \( =(0.1+0.2)+(0.1-0.2)j \), magnitude 0.316 A, about 0.3 A.
3. Voltage across 100 ohm \( =0.141\times100=10\sqrt2 \) V; across 50 ohm \( 0.283\times50=10\sqrt2 \) V, not 10 V.

Teacher's Note:
1. Treat each branch as a complex impedance and add the branch currents as phasors.
2. Both branches see the full source voltage.