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A cell of emf E is connected across an external resistance R. When current 'I' is drawn from the cell, the potential difference across the electrodes of the cell drops to V.

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Question

A cell of emf E is connected across an external resistance R. When current 'I' is drawn from the cell, the potential difference across the electrodes of the cell drops to V. The internal resistance 'r' of the cell is ______.

Options

  • `((E - V)/E)R`

  • `((E - V)/R)`

  • `((E - V)R)/I`

  • `((E - V)/V)R`

MCQ
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Solution

A cell of emf E is connected across an external resistance R. When current 'I' is drawn from the cell, the potential difference across the electrodes of the cell drops to V. The internal resistance 'r' of the cell is `underlinebb(((E - V)/V)R)`.

Explanation:

The electromotive force (e) or e.m.f. is the energy provided by a cell or battery per coulomb of charge passing through it. It is measured in volts (V). It is equal to the potential difference across the terminals of the cell when no current is flowing through it. E = I(R + r), where E = electromotive force in volts, I = current in amperes, R = resistance of the load in the circuit in ohms, and r = internal resistance of the cell in ohms. That is, E = IR + Ir, E = V + Ir.

Rearranging the equation we get,

r = `((E - V))/I`

Substituting `I = V/R` in the above equation, we get,

r = `((E - V)R)/V`

∴ Internal resistance, r = `((E - V)R)/V`

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