Question

The potential difference between points A and B of the following figure is :

• A. $\frac{2}{3}V$
• B. $\frac{8}{9}V$
• C. $\frac{4}{3}V$
• D. $2V$

Answer 1

Answer: (C)

$\frac{4}{3}V$

$\frac{1}{R_{net}}=\frac{1}{15}+\frac{1}{15}$

$R_{net}=\frac{15}{2}=7.5$

$\therefore I=\frac{2}{7.5}=\frac{4}{15}$

Now the current divides equally in both the branches

So current in $ABC\left(branch1\right)=\frac{2}{15}Amp$.

Now $V_A-V_B=\frac{I}{2}\times 10$

$=\frac{2}{15}\times 10$

$=\frac{4}{3}V$.