Question

Consider the circuit given here. The potential difference $V_{BC}$ between the points $B$ and $C$ is

• A. $1\, V$
• B. $0.5\, V$
• C. $0\,V$
• D. $-1\, V$

Answer 1

Answer: (B)

From circuit the current,
$I=\frac{E}{R}$
The equivalent resistance of above circuit
$=3+2+1 $
$=6 k \Omega$
$=6 \times 10^{3} \Omega$
$E =3 \, $ volt
$ I = \frac{3}{ 6 \times 10^3}$
$ = 0.5 \times 10^{-3} A$
Potential ,
$V_{AD} = iR = 0.5 \times 10^{-3} \times 3\times 10^3$
$ = 1.5 \,V$
Charge,
$ \frac{1}{C} = \frac{1}{1} + \frac{1}{2} = \frac{3}{2}$
$ C = \frac{2}{3}$
$Q = C \cdot V_{AD}$
$ = \frac{2}{3} \times 1.5$
$ = 1\,\mu C$
Applying KVL (Kirchhoff's Voltage Law) from $B$ to $C$
$V_{B}-0 \cdot 5 \times 10^{-3} \times 2 \times 10^{3}+\frac{1}{2}=V_{C}$
$V_{B}-V_{C}=1-\frac{1}{2}=0 \cdot 5\, V$