Question

In the given circuit, the potential difference $V_A–V_{B}is(V_{A}and{V}_B$ are the potentials at points A and B respectively)

• A. - 2 V
• B. + 3 V
• C. - 3 V
• D. + 2 V

Answer 1

The correct option is A - 2 V


The equivalent resistance can be calculated as:
$R_{eq}=\frac{(4+2)\times (2+2)}{(4+2)(2+2)}=\frac{12}{5}\Omega$
$Current,I=\frac{V}{R_{eq}}=\frac{12}{\left(\frac{12}{5}\right)}=5A$
$I_A=I\times \frac{2+2}{(4+2)+(2+2)}$
$=5\times \frac{4}{10}=2A$
$V_A=12-I_A\times \left(4\right)$
$=12-2\times 4=4V...\left(i\right)$
$I_B=I\times \frac{4+2}{(4+2)+(2+2)}=5\times \frac{6}{10}=3A$
$V_B=12-I_B\times 2=12-3\times 2=6V...\left(ii\right)$
Using (i) and (ii):
$V_A-V_B=(4-6)V=-2V$
Hence, the correct answer is option (d).