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 50 $Hz$ 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	$$I\ne 0,V_1\text{is proportional to}I$$	a
ii	$$I\ne 0,V_2>V_1$$	b
iii	$$V_1=0,V_2=V$$	c
iv	$$I\ne 0,V_2\text{is proportional to}I$$	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 = 2Ω
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 $V_2>V_1$