Question

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

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

Answer 1

Answer: (B)

$0.5V$

$i=\frac{3}{6\times {10}^3}=0.5\times {10}^{-3}$
$V_{AD}=iR=0.5\times {10}^{-3}\times 3\times {10}^3=1.5V$
$Q=\frac{2\times 1.5}{3}=2\times 0.5=1\mu C$
KVL from $B$ to $C$
$V_B-0.5\times {10}^{-3}\times {10}^3+\frac{1}{2}=V_C$
$V_B-V_C=1-\frac{1}{2}=0.5V$