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Standard XII

Physics

EMF and EMF Devices

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 fo the terminal potential difference to E.M.F is

1 / n

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\frac{1}{n + 1}

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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 A $\frac{n}{n + 1}$
Applying voltage division rule, to find terminal voltage(v);
Voltage drop $n R = E \cdot \frac{n R}{n R + R} = \frac{E \cdot n R}{R [n + 1]}$
$\Rightarrow \frac{V}{E} = \frac{n}{[n + 1]}$

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Similar questions

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

Q. Infinite number of cells having emfs and internal resistances

(E, r), (\frac{E}{n}, \frac{r}{n}), (\frac{E}{n^{2}}, \frac{r}{n^{2}}), (\frac{E}{n^{3}}, \frac{r}{n^{3}}), . . . .

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

\frac{n r}{n - 1}

. Current flowing through the external resistor is:

^{'} n^{'}

cells, each of emf

^{'} e^{'}

and internal resistance

^{'} r^{'}

are joined in series to form a row.

^{'} m^{'}

such rows are connected in parallel to form a battery of

N = m n

cells. This battery is connected to an external resistance

^{'} R^{'}

Show that current

^{'} I^{'}

flowing through the external resistance

^{'} R^{'}

is given by:

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)

^{'} n^{'}

cells, each of emf

^{'} e^{'}

and internal resistance

^{'} r^{'}

are joined in series to form a row.

^{'} m^{'}

such rows are connected in parallel to form a battery of

N = m n

cells. This battery is connected to an external resistance

^{'} R^{'}

What is the emf of this battery and how much is its internal resistance?

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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 fo the terminal potential difference to E.M.F is

The correct option is A nn+1Applying voltage division rule, to find terminal voltage(v);Voltage drop nR=E⋅nRnR+R=E⋅nRR[n+1]⇒VE=n[n+1]

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

The correct option is A $\frac{n}{n + 1}$
Applying voltage division rule, to find terminal voltage(v);
Voltage drop $n R = E \cdot \frac{n R}{n R + R} = \frac{E \cdot n R}{R [n + 1]}$
$\Rightarrow \frac{V}{E} = \frac{n}{[n + 1]}$