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 at $t=\frac{397\pi }{180}\times {10}^{-3}sec$


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

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

Answer 1

The correct option is B I $\to$ R II $\to$ T III $\to$ Q IV$\to$ Q

$X_L=\omega L={10}^3\times 20\times {10}^{-3}=20\Omega$
$X_C=\frac{1}{wc}=\frac{{10}^6}{{10}^3\times 50}=20\Omega$
R = 20 $\Omega$
$Z=\sqrt{R^2+(X_L-X_c{)}^2}=20\Omega$
$i_0=\frac{{ϵ}_0}{z}=\frac{20}{20}=1A$
So, $i=i_0\sin \left(\omega t\right)=1\sin \left(\omega t\right)$

At $t=\frac{397\pi \times {10}^{-3}}{180}s,i=20\sin (\text{/}{10}^3\times \frac{397}{180}\times \pi \times \text{/}{10}^{-3})$
$\frac{397\pi }{180}={397}^{\circ }=(360+37{)}^{\circ }$
So $i=1\sin {37}^{\circ }=1\times \frac{3}{5}=0.6A$

$V_R=iR=0.6\times 20=12V$
$V_L=L\frac{di}{dt}=L{i}_0\omega \cos \omega t=20\times {10}^{-3}\times 1\times {10}^3\times \frac{4}{5}=16V$
Since $X_C=X_L$, the circuit is purely resistive. So,
$V_C=-V_L=-16V$
${ϵ}_0=20\sin {37}^{\circ }=12V$