A battery of emf 10 V is connected to resistance as shown in figure. The potential difference $V_A-V_B$ between the points $A$ and $B$ is

• A

  $-2 \mathrm{~V}$
• Correct answer
  B

  2 V
• C

  5 V
• D

  $\frac{20}{11} \mathrm{~V}$

Solution

$R_{\mathrm{eq}}=5 \Omega$, current $i=\frac{10}{5}=2 \mathrm{~A}$ and current in each branch $=1 \mathrm{~A}$. Potential difference between $C$ and $A$,
$V_C-V_A=1 \times 1=1 \mathrm{~V}...(i)$ Potential difference between $C$ and $B$ $V_C-V_B=1 \times 3=3 V...(ii)$ On solving Eqs. (i) and (ii) we get $V_A-V_B=2 \mathrm{~V}$

Hence, the correct answer is option: 'B'.