Physics - Current Electricity Question with Solution | TestHub

Current in the primary circuit is given by:

$I=\frac{V}{R+R_{AB}}$

$I=\frac{5}{45+5}=\frac{5}{50}=0.1\text{ A}$

Potential difference across the potentiometer wire AB is:

$V_{AB}=I\times R_{AB}=0.1\times 5=0.5\text{ V}$

Potential gradient (potential drop per unit length) of the wire AB is:

$k=\frac{V_{AB}}{L}=\frac{0.5}{10}=0.05\text{ V/m}$

For balancing an e.m.f. $E=0.4\text{ V}$, the balancing length $l$ is given by:

$E=k\times l$

$0.4=0.05\times l$

$l=\frac{0.4}{0.05}=8\text{ m}$

Answer: $8$ m