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Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are r1 and r2 (r1 > r2). If the potential difference across the source of internal resistance r1 is zero, then the value of R will be:

Option 1 -

r1 – r2

Option 2 -

r 1 r 2 r 1 + r 2

Option 3 -

r 1 + r 2 2

Option 4 -

r 2 − r 1

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 1


    Detailed Solution:

    VAB = 0

    o r ,         ε − i r 1 = 0

    ⇒ ε − 2 ε r 1 + r 2 + R r 1 = 0 ⇒ 1 = 2 r 1 r 1 + r 2 + R

    ⇒ R + r 1 + r 2 = 2 r 1

    R = r 1 − r 2

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