Question 371: Current Electricity

Question

The length of a wire of a potentiometer is $$100$$ $$cm$$, and the $$e.$$ $$m.$$ $$f.$$ of its standard cell is $$E$$ volt. It is employed to measure the $$e.m.f.$$ of a battery whose internal resistance in $$0.5\Omega .$$ If the balance point is obtained at $$1=30$$ $$cm$$ from the positive end, the $$e.m.f.$$ of the battery is 
where $$i$$ is the current in the potentiometer wire.

Accepted Answer

Potential gradient along wire, K = $${E \over {100}}$$ volt/cm
For battery V = E' Ã¢ÂÂ ir, where E' is emf of battery.
or K ÃÂ 30 = E' Ã¢ÂÂ ir, where current i is drawn from battery
or $${{E 	imes 30} \over {100}}$$ = E' + 0.5i
or E' = $${{30E} \over {100}} - 0.5i$$

Answer

$${{30E} \over {100.5}}$$

Answer

$${{30E} \over {\left( {100 - 0.5} \right)}}$$

Answer

$${{30\left( {E - 0.5i} \right)} \over {100}}$$

Answer

$${{30E} \over {100}} - 0.5i$$,  where i is the current in the potentiometer
wire

Current Electricity

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