Two sources of equal emf are connected to an external resistance R . The internal resistance of the two sources are R_1 and R_2 ( R_2 > R_1 ). If the - Physics MCQ AIEEE 2026 | ExamTest.live

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Two sources of equal emf are connected to an external resistance R . The internal resistance of the two sources are R_{1} and R_{2} (R_{2} > R_{1}). If the potential difference across the source having internal resistance R_{2} is zero, then:

Solution & Step-by-step Explanation

Total EMF = E + E = 2 E (assuming series connection to an external R).Total resistance = R + R_{1} + R_{2} .Current I = \frac{2 E}{R + R _{1} + R _{2}} .The potential difference across source 2 : V_{2} = E - I R_{2} = 0 . E = I R_{2} E = (\frac{2 E}{R + R _{1} + R _{2}}) R_{2} 1 = \frac{2 R _{2}}{R + R _{1} + R _{2}} R + R_{1} + R_{2} = 2 R_{2} R = R_{2} - R_{1}

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Two sources of equal emf are connected to an external resistance R . The internal resistance of the two sources are R_{1} and R_{2} (R_{2} > R_{1}). If the potential difference across the source having internal resistance R_{2} is zero, then:

R = R_{2} \times (R_{1} + R_{2}) / (R_{2} - R_{1})

R = R_{2} - R_{1}

R = R_{1} R_{2} / (R_{1} + R_{2})

R = R_{1} R_{2} / (R_{2} - R_{1})

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