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Standard XII

Physics

Potentiometer

A cell of emf...

Question

A cell of emf $E$ and internal resistance $r$ is connected in series with an external resistance $n r,$ then the ratio of terminal potential difference to emf is

\frac{1}{n}

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\frac{1}{(1 + n)}

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\frac{n}{(n + 1)}

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\frac{(n + 1)}{n}

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Solution

The correct option is B $\frac{n}{(n + 1)}$
Current drawn from the cell is $I = \frac{E}{R + r} = \frac{E}{n r + r}$
Terminal voltage , $V = I R = \frac{E}{n r + r} . n r = \frac{E n}{(n + 1)}$
Thus, $\frac{V}{E} = \frac{n}{n + 1}$

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Q. Infinite number of cells having emfs and internal resistances

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ara connected in series in same manner across an external resistance of

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Q. A cell of e.m.f. E and internal resistance r is connected in series with an external resistance nr then the ratio of the terminal potential difference to E.M.F. is

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^{'} I^{'}

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I = \frac{N e}{m R + n r}

n identical cells are joined in series with its two cells A and B in the loop with reversed polarities. Emf of each cell is E and internal resistance r. The potential difference across cell A or B is (Here n > 4)

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^{'} e^{'}

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^{'} m^{'}

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A cell of emf E and internal resistance r is connected in series with an external resistance nr, then the ratio of terminal potential difference to emf is

The correct option is B n(n+1)Current drawn from the cell is I=ER+r=Enr+rTerminal voltage , V=IR=Enr+r.nr=En(n+1)Thus, VE=nn+1

A cell of emf $E$ and internal resistance $r$ is connected in series with an external resistance $n r,$ then the ratio of terminal potential difference to emf is

The correct option is B $\frac{n}{(n + 1)}$
Current drawn from the cell is $I = \frac{E}{R + r} = \frac{E}{n r + r}$
Terminal voltage , $V = I R = \frac{E}{n r + r} . n r = \frac{E n}{(n + 1)}$
Thus, $\frac{V}{E} = \frac{n}{n + 1}$