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The cell has an emf of `2 V` and the internal resistance of this cell is `0.1 Omega`, it is connected to resistance of `3.9 Omega`, the voltage across the cell will be |
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Answer» Here, `epsilon = 2 V, r = 0.1 Omega, R = 3.9 Omega` Current `I = epsilon/R +r = 2/3.9 + 0.1=0.5A` pot. Diff. across the terminals of the cell, i.e., terminal pot. Diff. `V = epsilon - Ir =2 - 0.5 xx 0.1 = 1.95 V` |
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