Nine cells, each of emf ε and internal resistance r, are connected in series to form a loop. If one cell is connected wrongly, then potential difference across this cell will be

Question

Nine cells, each of emf ε and internal resistance r, are connected in series to form a loop. If one cell is connected wrongly, then potential difference across this cell will be (A)  (16/9) ϵ (B) Zero (C)  (17/9) ϵ (D)  (8/9) ϵ

Tardigrade Admin · Accepted Answer

i = (8 ϵ -ϵ/9r)= (7ϵ/9r) V = ϵ +ir = ϵ + (7ϵ/9) = (16ϵ /9)

Q. Nine cells, each of emf ε and internal resistance r, are connected in series to form a loop. If one cell is connected wrongly, then potential difference across this cell will be

$\frac{16}{9} ϵ$

Zero

$\frac{17}{9} ϵ$

$\frac{8}{9} ϵ$

$i = \frac{8 ϵ - ϵ}{9 r} = \frac{7 ϵ}{9 r}$
$V = ϵ + i r$
= $ϵ + \frac{7 ϵ}{9} = \frac{16 ϵ}{9}$

Q. Nine cells, each of emf ε and internal resistance r, are connected in series to form a loop. If one cell is connected wrongly, then potential difference across this cell will be

916​ϵ

Zero

917​ϵ

98​ϵ

i=9r8ϵ−ϵ​=9r7ϵ​ V=ϵ+ir = ϵ+97ϵ​=916ϵ​

$\frac{16}{9} ϵ$

$\frac{17}{9} ϵ$

$\frac{8}{9} ϵ$

$i = \frac{8 ϵ - ϵ}{9 r} = \frac{7 ϵ}{9 r}$
$V = ϵ + i r$
= $ϵ + \frac{7 ϵ}{9} = \frac{16 ϵ}{9}$