You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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, $\frac{V}{E} = \frac{n}{n + 1}$

Suggest Corrections

Similar questions

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:

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

^{'} 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?

Join BYJU'S Learning Program

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, $\frac{V}{E} = \frac{n}{n + 1}$