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JEE Advanced 2016 Paper 1 · Physics · Electromagnetic Induction and Alternating Current · easy

A series circuit consists of a resistor of 30 Ω, an ideal inductor of 15 mH and an ideal capacitor of 200/3 muF. It is connected to a source of rms…

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A series circuit consists of a resistor of 30 Ω30\ \Omega, an ideal inductor of 15 mH15\ \mathrm{mH} and an ideal capacitor of 2003 μF\tfrac{200}{3}\ \mu\mathrm{F}. It is connected to a source of rms voltage 100 V100\ \mathrm{V} whose angular frequency is set to exactly three times the resonant angular frequency of the circuit. The reading, in amperes, of an ideal rms ammeter in the circuit is
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Answer: 2

The resonant angular frequency is ω0=1LC=115×10−3×2003×10−6=110−6=1000 rad s−1\omega_0=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{15\times10^{-3}\times\frac{200}{3}\times10^{-6}}}=\dfrac{1}{\sqrt{10^{-6}}}=1000\ \mathrm{rad\,s^{-1}}, so the source runs at ω=3000 rad s−1\omega=3000\ \mathrm{rad\,s^{-1}}. The reactances are XL=ωL=3000×15×10−3=45 ΩX_L=\omega L=3000\times15\times10^{-3}=45\ \Omega and XC=1ωC=13000×2003×10−6=5 ΩX_C=\dfrac{1}{\omega C}=\dfrac{1}{3000\times\frac{200}{3}\times10^{-6}}=5\ \Omega. The net reactance is X=XL−XC=40 ΩX=X_L-X_C=40\ \Omega, so the impedance is Z=R2+X2=302+402=50 ΩZ=\sqrt{R^2+X^2}=\sqrt{30^2+40^2}=50\ \Omega. The rms current is I=V/Z=100/50=2 AI=V/Z=100/50=2\ \mathrm{A}, and since the ammeter reads rms values the reading is 22. The circuit is inductive with cos⁡ϕ=0.6\cos\phi=0.6, a useful check. As a further check, the resonant current would have been 100/30=3.33 A100/30=3.33\ \mathrm{A}, larger than the 2 A2\ \mathrm{A} obtained away from resonance, as expected.
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