You visited us 8 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Parallel Combination of Cells

n identical c...

Question

$n$ identical cells each of electromotive force $e$ and internal resistance $r$ are connected in series of this combination. The current through $V$ is:

\frac{2 e}{n}

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

\frac{n e}{n R + r}

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

\frac{e}{R + n r}

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

\frac{n e}{R + r}

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

Open in App

Solution

The correct option is A $\frac{2 e}{n}$
$I = (\frac{n e - 2 e}{n r})$
$I = \frac{(n - 2) e}{n r}$
$V = e - I r$
$V = e - \frac{(n - 2) e}{n}$
$V = \frac{(n e - n e + 2 e)}{n}$
$V = \frac{2 e}{n}$
Hence,
option $(A)$ is correct answer.

Suggest Corrections

Similar questions

^{'} 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}

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:

When $n$ resistors, each of resistance $R$ , are connected in series, the equivalent resistance is ___. When $n$ resistors, each of resistance $R$ , are connected in parallel, their equivalent resistance is ___.

N

cells each of e.m.f.

E

& internal resistance

r

are grouped into sets of

K

cells connected in series. The

(\frac{N}{K})

sets are connected in parallel to a load of resistance

R

, then:

Q. Two identical cells, each of emf

E

and internal resistance

r

, when connected in series or parallel across a resistance

R

are found to give the same current through

R

. In this case

Join BYJU'S Learning Program

n identical cells each of electromotive force e and internal resistance r are connected in series of this combination. The current through V is:

The correct option is A 2enI=(ne−2enr)I=(n−2)enrV=e−IrV=e−(n−2)enV=(ne−ne+2e)nV=2enHence,option (A)is correct answer.

$n$ identical cells each of electromotive force $e$ and internal resistance $r$ are connected in series of this combination. The current through $V$ is:

The correct option is A $\frac{2 e}{n}$
$I = (\frac{n e - 2 e}{n r})$
$I = \frac{(n - 2) e}{n r}$
$V = e - I r$
$V = e - \frac{(n - 2) e}{n}$
$V = \frac{(n e - n e + 2 e)}{n}$
$V = \frac{2 e}{n}$
Hence,
option $(A)$ is correct answer.