Question 527: Current Electricity

Question

A potentiometer wire of length $4 \mathrm{~m}$ and resistance $5 ~\Omega$ is connected in series with a resistance of $992 ~\Omega$ and a cell of e.m.f. $4 \mathrm{~V}$ with internal resistance $3 ~\Omega$. The length of $0.75 \mathrm{~m}$ on potentiometer wire balances the e.m.f. of

Accepted Answer

$	herefore \quad$ Total Resistance:
$\mathrm{R}=992+5+3=1000 \Omega$
Voltage across $4 \mathrm{~m}$ wire:
$\frac{5}{995+5} 	imes 4=0.02 \mathrm{~V}$
$	herefore \quad$ For one metre wire:
$\frac{0.02}{4}=0.005 \mathrm{~V}$
$	herefore \quad$ For $0.75 \mathrm{~m}$ wire:
$\begin{aligned}
0.004 	imes 0.75 & =0.00375 \
& =3.75 \mathrm{~mV}
\end{aligned}$

Answer

4.00 mV

Answer

3.75 mV

Answer

3.00 mV

Answer

2.50 mV

Current Electricity

A potentiometer wire of length 4 \mathrm{~m} and resistance 5 ~\Omega is connected in series with a resistance of 992 ~\Omega and a cell of e.m.f. 4 \mathrm{~V} with internal resistance 3 ~\Omega. The length of 0.75 \mathrm{~m} on potentiometer wire balances the e.m.f. of

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