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\mathrm{N} identical cells, each of emf \varepsilon and internal resistance r, are joined in series out of these,\mathrm{n} cells are wrongly connected, i.e. their terminals are connected in reverse of that required for series connection,n<N / 2. Let \varepsilon_0 be the emf of the resulting battery and r_0 be its internal resistance, then

Option: 1

\varepsilon_0=(N-n) \varepsilon, r_0=(N-n) r


Option: 2

\varepsilon_0=(N-2 n) \varepsilon \quad r_0=(N-2 r) r


Option: 3

\varepsilon_0=(N-2 n) \varepsilon, \quad r_0=N r


Option: 4

\varepsilon_0=(\mathrm{N}-\mathrm{n}) \varepsilon, \quad \mathrm{r}_0=\mathrm{Nr}


Answers (1)

best_answer

For every cell that is wrongly connected the emf decreases by 2 \varepsilon. However, internal resistance does not depend on direction and therefore remains the same for all cells.

\varepsilon_0=(N-2 n) \varepsilon

\mathrm{r}_0=\mathrm{Nr}

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Riya

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