Question

Five resistance are connected as shown in figure. If total current flowing is $0.5A$, then the potential difference $V_A-A_B$ is:

• A. $8V$
• B. $6V$
• C. $2V$
• D. $4V$

Answer 1

Answer: (D)

$4V$

In section PQ : Two $6\Omega$ resistances are connected in series

$\therefore$ Equivalent resistance between PQ, $R_{PQ}=6\Omega +6\Omega =12\Omega$

This $12\Omega$ is connected in parallel with another $12\Omega$ resistance.

$\therefore$ Equivalent resistance between $BC$, $\frac{1}{R_{BC}}=\frac{1}{12}+\frac{1}{12}$

$⟹$ $R_{BC}=6\Omega$ ...........(1)

Equivalent resistance between AC $\frac{1}{R_{AC}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$

$⟹$ $R_{AC}=2\Omega$ ...........(2)

Total resistance between AB, $R_{AB}=R_{AC}+R_{BC}=2+6=8\Omega$

$\therefore$ Potential difference between A and B, $V_A-V_B=I{R}_{AB}=0.5\times 8=4V$