Question

In the circuit shown below, the cell has an e.m.f. of $10 \,V$ and internal resistance of $1 \,\Omega$. The other resistances are shown in the figure. The potential difference $V_{A}-V_{B}$ is

• A. 6 V
• B. 4 V
• C. 2 V
• D. -2 V

Answer 1

Answer: (D)

Equivalent external resistance of the given circuit, $R_{e q}=4\, \Omega$
Current given by the cell, $i=\frac{E}{R_{e q}+r}=\frac{10}{(4+1)}=2 A$
Hence, $\left(V_{A}-V_{B}\right)=\frac{i}{2} \times\left(R_{2}-R_{1}\right)=\frac{2}{2}(2-4)=-2 \,V$