Figure shows a simple potentiometer circuit for measuring a small $e.m.f$. produced by a thermocouple. The meter wire $PQ$ has a resistance $5 \,\Omega$ and the driver cell has an e.m.f. of $2\, V$. If a balance point is obtained $0.600\, m$ along $PQ$ when measuring an e.m.f. of $6.00\, mV$, what is the value of resistance $R$ ............... $\Omega$
Medium
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(a) The voltage per unit light of the metre wire $PQ$ is $\left( {\frac{{6.00\,mV}}{{0.600\,m}}} \right)$ i.e. $10\,mV/m$.
Hence potential difference across the metre wire is $10\,mV{\rm{/}}m \times 1\,m = 10\,mV$.
The current drawn from the driver cell is $i = \frac{{10\,mV}}{{5\,\Omega }} = 2\,mA$.
The resistance $R = \frac{{(2\,V - 10\,mV)}}{{2\,mA}} = \frac{{1990\,mV}}{{2\,mA}} = 995\;\Omega $.
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