The resistances in the two arms of the meter bridge are $5 \Omega$ and $R \Omega$ respectively. When

Question

The resistances in the two arms of the meter bridge are $5 \Omega$ and $R \Omega$ respectively. When the resistance $R$ is shunted with an equal resistance, the new balance point is at $1.6 l_{1}$. The resistance $R$ is\

Professor PC Thomas Classes · Answer

(b) : In the first case,\ At balance point\ $\frac{5}{R}=\frac{l_{1}}{100-l_{1}}$\ In the second case,\ At balance point\ $\frac{5}{(R / 2)}=\frac{1.6 l_{1}}{100-1.6 l_{1}}$\ Divide eqn. (i) by eqn. (ii), we get\ $$ \frac{1}{2}=\frac{100-1.6 l_{1}}{1.6\left(100-l_{1}ight)} $$ or $160-1.6 l_{1}=200-3.2 l_{1}$ $$ 6 l_{1}=40 	ext { or } l_{1}=\frac{40}{1.6}=25 \mathrm{~cm} $$ Substituting this value in eqn. (i), we get $$ \frac{5}{R}=\frac{25}{75} 	ext { or } \quad R=\frac{375}{25} \Omega=15 \Omega $$