Question 630: Current Electricity

Question

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E
0
and a resistance r
1
. An unknown e.m.f. E is balanced at a length $$l$$ of the potentiometer wire. The e.m.f. E will be  given by

Accepted Answer

The current through the potentiometer wire is
$$I = {{{E_0}} \over {\left( {r + {r_1}} ight)}}$$
and the potential difference across the wire is
$$V = Ir = {{{E_0}r} \over {\left( {r + {r_1}} ight)}}$$
The potential gradient along the potentiometer wire is
$$k = {V \over L} = {{{E_0}r} \over {\left( {r + {r_1}} ight)L}}$$
As the unknown e.m.f. E is balanced against length l of the potentiometer wire,
$$ 	herefore $$ $$E = kl = {{{E_0}r} \over {\left( {r + {r_1}} ight)}}{l \over L}$$

Answer

$${{{E_0}l} \over L}$$

Answer

$${{L{E_0}r} \over {\left( {r + {r_1}} \right)l}}$$

Answer

$${{L{E_0}r} \over {l{r_1}}}$$

Answer

$${{{E_0}r} \over {\left( {r + {r_1}} \right)}}.{l \over L}$$

Current Electricity

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E 0 and a resistance r 1 . An unknown e.m.f. E is balanced at a length $l$ of the potentiometer wire. The e.m.f. E will be given by

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