# NEET Questions Solved

PMT - 1994

A battery of e.m.f. 10 V is connected to resistance as shown in figure. The potential difference ${V}_{A}-{V}_{B}$  between the points A and B is

(1) –2 V

(2) 2 V

(3) 5 V

(4) $\frac{20}{11}V$

(2) ${R}_{eq}=5\Omega$, Current $i=\frac{10}{5}=2A$ and current in each branch = 1A

Potential difference between C and A,

${V}_{C}-{V}_{A}=1×1=1V$ .......(i)

Potential difference between C and B,

${V}_{C}-{V}_{B}=1×3=3V$ ......(ii)

On solving (i) and (ii) ${V}_{A}-{V}_{B}=2\text{\hspace{0.17em}}volt$

Short Trick : $\left({V}_{A}-{V}_{B}\right)=\frac{i}{2}\left({R}_{2}-{R}_{1}\right)=\frac{2}{2}\left(3-1\right)=2V$

Difficulty Level:

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