Question

You are given many resistors, capacitors and inductors. These are connected to a variable DC voltage source (the first two circuits) or an AC voltage source of $50Hz$ frequency (the next three circuits) in different ways as shown in Column II. When a current $I$ (steady state for DC or rms for AC) flows through the circuit, the corresponding voltages $V_1$ and $V_2$ (indicated in circuits) are related as shown in Column I. Match the two columns.
Column I Column II

i.	$I\ne 0,V_1$ is proportional to $I$	a.
ii.	$I\ne 0,V_2>V_1$	b.
iii.	$V_1=0,V_2=V$	c.
iv.	$I\ne 0,V_2$ is proportional to $I$	d.
		e.

• A. i - c,d,e ; ii - b,c,d,e ; iii - a,b ; iv - b,c,d,e
• B. i - c,d,e ; ii - b,c,e ; iii - a,b ; iv - b,c,d,e
• C. i - c,d ; ii - b,d,e ; iii - a,b ; iv - b,d,e
• D. i - c,e ; ii - b,c,d,e ; iii - a,b ; iv - b,c,e

Answer 1

The correct option is A i - c,d,e ; ii - b,c,d,e ; iii - a,b ; iv - b,c,d,e

(a) In this case, steady state current is zero $(I=0)$ because of the presence of capacitor. Entire potential will be across capacitor in steady state. Hence $V_1=0$ and $V_2=V$ .

(b) In steady state, $V_1=0,$ so $V_2=V.$ Also, $V_2$ is proportional to $I.$

(c) $X_L=\omega L=2\pi fL=2\pi \times 50\times 6\times {10}^{-3}$
$=1.885\Omega$
$R=2\Omega$
Since $R>X_L,$ so $V_2>V_1$
Here $I\ne 0$
$V_1$ and $V_2$ both are proportional to $I$

(d) $X_L=1.885\Omega$
$X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$
$=\frac{1}{2\pi \times 50\times 3\times {10}^{-6}}=1061\Omega$
Hence $X_C>X_L,$ so $V_2>V_1$

(e) $X_C=1.061k\Omega ,R=1k\Omega$
$X_C>R$, so $V_2>V_1$