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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2019
  • Medium
  • 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 \(V\))

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Answer & Solution

Correct Answer

(A) \(3.3\)

Step-by-step Solution

Detailed explanation

\( E_{e q}= \frac{\frac{E_{1}}{2 R_{1}}+\frac{E_{2}}{R_{2}}+\frac{E_{3}}{2 R_{1}}}{\frac{1}{2 R_{1}}+\frac{1}{R_{2}}+\frac{1}{2 R_{1}}} \)

\( = \frac{{\frac{2}{2} + \frac{4}{2} + \frac{4}{2}}}{{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}}} = \frac{5}{{\frac{3}{2}}} = \frac{{10}}{3} = 3.3\)

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