GATE 2014 EE (EE2) – Question 40
Capacitor voltage in the shown circuit is $v_c(t)=A_1\sin(\omega_1t-\theta_1)+A_2\sin(\omega_2t-\theta_2)$. Find A₁,A₂ respectively.

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Correct answer: (A) 2.0 and 1.98
Explanation
At10rad/s,open the current source;its series inductor branch then carries no current. The1Ω–1Fdivider gives A1=20/√(1+10²)=1.990≈2.0. At5rad/s,short the voltage source. The ideal current source injects10Ainto the parallel1Ωand1Fcombination,regardless of the series inductor voltage. Thus A2=10/|1+j5|=1.961≈1.98. The nearest listed pair is **A**.