Question

For a given L - C - R circuit, list 1 - gives voltage across source inductor, capacitor & resistance, and list -2 gives their magnitudes.
$\begin{array}{cccc}& \text{List -1}& & \text{List -II}\\ \left(I\right)& \in & \left(p\right)& 480\\ \left(II\right)& V_L& \left(Q\right)& 400\\ \left(III\right)& V_C& \left(R\right)& 240\\ \left(IV\right)& V_R& \left(S\right)& 200\\ & & \left(T\right)& 160\\ & & \left(U\right)& 0\end{array}$
If current in the circuit at an instant is 6A, then correct match for potential difference at that instant. [Neglect negative sign if any]

• A. $I\to PII\to SIII\to TIV\to U$
• B. $I\to PII\to QIII\to TIV\to R$
• C. $I\to QII\to SIII\to TIV\to R$
• D. $I\to SII\to TIII\to RIV\to U$

Answer 1

The correct option is B $I\to PII\to QIII\to TIV\to R$

$I\to P,II\to Q,III\to T,IV\to R$
For given circuit = $R=40\Omega ,X_C=20\Omega \&X_L=50\Omega$
& $Z=\sqrt{R^2+(X_L-X_C{)}^2}=50\Omega$
$i_0=\frac{{\in }_0}{z}=\frac{500}{50}=10A$

...
$\tan \varphi =\frac{30}{40}\Rightarrow \varphi ={37}^{\circ }$
...
$V_{oL}=i_0\times X_L=10\times 50=500$
$V_{oC}=i_0\times X_C=10\times 20=200$

Say $i=i_0\sin \theta ,$ then $V_L=V_{oL}cos\theta$ and $V_R=V_{oR}sin\theta$
Given,
$i=6=10\sin \theta \Rightarrow \theta ={37}^{\circ }$
$V_L=V_{oL}cos\theta =500\times \frac{4}{5}=400V$
$V_C=-V_{oC}cos\theta =-200\times \frac{4}{5}=-160V$
$V_R=V_{oR}sin\theta =240V$

Since, current lags voltage by $\varphi$, voltage is given by,
$\in ={\in }_0\sin (\theta +\varphi )=480$ V
where $V_{oL}=i_0{X}_L=500V$