Question

A battery of e.m.f. $10\text{V}$ is connected to resistances as shown in figure. The potential difference $V_A–V_B$ between the points $A$ and $B$ is

• A. $-2\text{V}$
• B. $2\text{V}$
• C. $5\text{V}$
• D. $\frac{20}{11}\text{V}$

Answer 1

The correct option is B $2\text{V}$

Current is equally divided into both the branches because resistance is same i.e. $4\Omega$
$\therefore$ Current is $\left(1\text{A}\right)$
Now, P.D. across $1\Omega$ is $\to 1\times 1=1\text{volt}$
and P.D. across $3\Omega$ is $\to 1\times 3=3\text{volt}$
Now,
P.D. between $A\&B=V_A-V_B=2\text{volt}$