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JEE Advanced 2019 Paper 2 · Physics · Electromagnetic Induction and Alternating Current · medium

JEE Advanced 2019 Paper 2 Physics: Paragraph 1

A series circuit contains a resistor R=2 ΩR=2\ \Omega, an inductor L=10 mHL=10\ \mathrm{mH} and a capacitor CC driven by a source of variable angular frequency ω\omega and constant rms voltage VV. In each question the quantities in List-I are to be evaluated at resonance and matched with the numerical values in List-II (all values are in the SI unit appropriate to the List-I quantity).

List-II: (1) 2.52.5 (2) 55 (3) 1010 (4) 2525 (5) 100100 (6) 200200

Question 1

Single correct option

Take C=100 μFC=100\ \mu\mathrm{F} and V=20 VV=20\ \mathrm{V}. List-I: (P) rms current at resonance (A); (Q) quality factor of the circuit; (R) rms voltage across the capacitor at resonance (V); (S) bandwidth Δω\Delta\omega between half-power points (rad s−1\mathrm{rad\,s^{-1}}). Which option gives the correct match?
  1. A. P→2, Q→3, R→5, S→6
  2. B. P→3, Q→2, R→5, S→6
  3. C. P→3, Q→2, R→6, S→5
  4. D. P→3, Q→5, R→2, S→6
Show answer and solution

Answer: (B) P→3, Q→2, R→5, S→6

Resonance occurs at ω0=1/LC=1/10−2×10−4=103 rad s−1\omega_0=1/\sqrt{LC}=1/\sqrt{10^{-2}\times10^{-4}}=10^3\ \mathrm{rad\,s^{-1}}. At resonance the impedance is just RR, so I=V/R=20/2=10I=V/R=20/2=10 A, giving P to (3). The quality factor is Q=ω0L/R=103×10−2/2=5Q=\omega_0L/R=10^3\times10^{-2}/2=5, so Q to (2). The capacitor voltage is VC=I/(ω0C)=10/(103×10−4)=100V_C=I/(\omega_0C)=10/(10^3\times10^{-4})=100 V (five times the supply voltage), so R to (5). The half-power bandwidth of a series resonant circuit is Δω=R/L=2/10−2=200 rad s−1\Delta\omega=R/L=2/10^{-2}=200\ \mathrm{rad\,s^{-1}}, so S to (6). The correct code is P to 3, Q to 2, R to 5, S to 6. The four values 10, 5, 100 and 200 are all distinct entries of List-II, which also rules out options that repeat or swap entries.

Question 2

Single correct option

Now CC is changed to 400 μF400\ \mu\mathrm{F} and the rms source voltage to V=10 VV=10\ \mathrm{V}, with RR and LL unchanged. List-I: (P) rms current at resonance (A); (Q) quality factor of the circuit; (R) rms voltage across the capacitor at resonance (V); (S) bandwidth Δω\Delta\omega (rad s−1\mathrm{rad\,s^{-1}}). Which option gives the correct match?
  1. A. P→3, Q→2, R→5, S→6
  2. B. P→2, Q→1, R→4, S→6
  3. C. P→2, Q→1, R→5, S→4
  4. D. P→3, Q→1, R→4, S→6
Show answer and solution

Answer: (B) P→2, Q→1, R→4, S→6

With the new capacitor, ω0=1/10−2×4×10−4=500 rad s−1\omega_0=1/\sqrt{10^{-2}\times4\times10^{-4}}=500\ \mathrm{rad\,s^{-1}}. The resonant current is I=V/R=10/2=5I=V/R=10/2=5 A, so P to (2). The quality factor is Q=ω0L/R=500×10−2/2=2.5Q=\omega_0L/R=500\times10^{-2}/2=2.5, so Q to (1). The capacitor voltage is VC=I/(ω0C)=5/(500×4×10−4)=25V_C=I/(\omega_0C)=5/(500\times4\times10^{-4})=25 V, so R to (4). The bandwidth Δω=R/L=200 rad s−1\Delta\omega=R/L=200\ \mathrm{rad\,s^{-1}} depends only on RR and LL, which have not changed, so S still maps to (6). Check: VC=QV=2.5×10=25V_C=QV=2.5\times10=25 V, consistent. The correct code is P to 2, Q to 1, R to 4, S to 6. Option swaps such as P to 3 come from forgetting that the source voltage was halved, while R to 5 repeats the earlier capacitor voltage.
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