MCQ
For the circuit shown, with ${R_1} = 1.0\,\Omega $, ${R_2} = 2.0\,\Omega $, ${E_1} = 2\,V$ and ${E_2} = {E_3} = 4\,V$, the potential difference between the points $‘a’$ and $‘b’$ is approximately (in $V$)
- ✓$3.3$
- B$2.3$
- C$3.7$
- D$2.7$
✓
Answer
Correct option: A.
$3.3$
a
$ E_{e q}= \frac{\frac{E_{1}}{2 R_{1}}+\frac{E_{2}}{R_{2}}+\frac{E_{3}}{2 R_{1}}}{\frac{1}{2 R_{1}}+\frac{1}{R_{2}}+\frac{1}{2 R_{1}}} $
$ E_{e q}= \frac{\frac{E_{1}}{2 R_{1}}+\frac{E_{2}}{R_{2}}+\frac{E_{3}}{2 R_{1}}}{\frac{1}{2 R_{1}}+\frac{1}{R_{2}}+\frac{1}{2 R_{1}}} $
$ = \frac{{\frac{2}{2} + \frac{4}{2} + \frac{4}{2}}}{{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}}} = \frac{5}{{\frac{3}{2}}} = \frac{{10}}{3} = 3.3$
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