Question 660: Current Electricity

Question

A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10 $\Omega $. Its internal resistance is

Accepted Answer

Here $E > {{ER} \over {R + r}}$, hence the lengths 110 cm and 100 cm are interchanged.
Without being short-circuited through R, only the battery E is balanced.
$E = {V \over L} 	imes {l_1} = {V \over L} 	imes 110$   ...(i)
When R is connected across E,  $Ri = {V \over L} 	imes {l_2}$
$ \Rightarrow R\left( {{E \over {R + r}}} ight) = {V \over L} 	imes 100$   ...(ii)
Dividing (i) by (ii), we get
${{R + r} \over R} = {{110} \over {100}}$
$ \Rightarrow $ 100 R + 100 r = 110 R
$ \Rightarrow $ 10 R = 100 r
$ 	herefore $ $r = {{10R} \over {100}} = {{10 	imes 10} \over {100}}$  ($ \because $ R = 10$\Omega $)
$ \Rightarrow $ r = 1$\Omega $.

Answer

2.0 ohm

Answer

zero

Answer

1.0 ohm

Answer

0.5 ohm

Current Electricity

A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10 \Omega . Its internal resistance is

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