Question

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

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

Answer 1

Answer: (B)

From circuit the current,
$I=\frac{E}{R}$
The equivalent resistance of above circuit
$=3+2+1$
$=6k\Omega$
$=6\times 10^3\Omega$
$E=3$ volt
$I=\frac{3}{6\times 10^3}$
$=0.5\times 10^{-3}A$
Potential ,
$V_{AD}=iR=0.5\times 10^{-3}\times 3\times 10^3$
$=1.5V$
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 $C$
$V_B-0\cdot 5\times 10^{-3}\times 2\times 10^3+\frac{1}{2}=V_C$
$V_B-V_C=1-\frac{1}{2}=0\cdot 5V$