Question

For a given series LCR circuit, list-I shows instantaneous voltage across inductor, capacitor, resistance and source and list-II shows values corresponding to quantites of list-I in SI units.

Match List-I with List -2 when current in circuit is half of its maximum value for the first time.

$\begin{array}{cccc}& \text{List- 1}& & \text{List - 2}\\ \text{I)}& V_L& \text{P)}& 10\\ \text{II)}& V_c& \text{Q)}& 12\\ \text{III)}& V_R& \text{R)}& 16\\ \text{IV)}& ϵ& \text{S)}& 17.3\\ & & \text{T)}& -16\\ & & \text{U)}& -17.3\end{array}$

• A. I $\to$ S II $\to$ U III $\to$ P IV$\to$ P
• B. I $\to$ P II $\to$ U III $\to$ S IV$\to$ Q
• C. I $\to$ P II $\to$ P III $\to$ Q IV$\to$ R
• D. I $\to$ S II $\to$ U III $\to$ Q IV$\to$ P

Answer 1

The correct option is A I $\to$ S II $\to$ U III $\to$ P IV$\to$ P

$i=\frac{i_0}{2}\therefore \theta ={30}^{\circ }ϵ={ϵ}_0\sin \theta =20\times \frac{1}{2}=10Vi=i_0\sin \theta =1\times \frac{1}{2}=0.5A{V}_R=iR=0.5\times 20=10A$

$V_L=L\frac{di}{dt}=L{i}_0\omega \cos \omega t=20\times {10}^{-3}\times 1\times {10}^3\times \frac{\sqrt{3}}{2}=17.3V$
Since $X_C=X_L$, the circuit is purely resistive. So,
$V_C=-V_L=-17.3V$
${ϵ}_0=20\sin {37}^{\circ }=10V$