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Two cells of emfs \[\varepsilon_1\] and \[\varepsilon_2\] (with \[\varepsilon_1 > \varepsilon_2\]) are connected in series, but one of the two cells is connected in the opposite direction. The equivalent emf of the combination is:

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Question

Two cells of emfs \[\varepsilon_1\] and \[\varepsilon_2\] (with \[\varepsilon_1 > \varepsilon_2\]) are connected in series, but one of the two cells is connected in the opposite direction. The equivalent emf of the combination is:

Options

  • \[\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2\]

  • \[\varepsilon_{eq} = \varepsilon_1 - \varepsilon_2\]

  • \[\varepsilon_{eq} = 0\]

  • \[\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}\]

MCQ
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Solution

If one of the two cells is connected in the opposite direction, the equivalent emf decreases and equals \[\varepsilon_1 - \varepsilon_2\] (given \[\varepsilon_1 > \varepsilon_2\]). The sum \[\varepsilon_1 + \varepsilon_2\] applies only when both cells are in the same sense.

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