The resistances of the four arms $P, Q, R$ and $S$ in a Wheatstone's bridge are $10 \mathrm{ohm}, 30

Question

The resistances of the four arms $P, Q, R$ and $S$ in a Wheatstone's bridge are $10 \mathrm{ohm}, 30 \mathrm{ohm}, 30 \mathrm{ohm}$\ and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 50 ohm respectively. If the galvanometer resistance is 50 ohm , the current drawn from the cell will be\

Professor PC Thomas Classes · Answer

(d) : The situation is as shown in the figure. For a balanced Wheatstone's bridge $$ \begin{aligned} & \frac{P}{Q}=\frac{R}{S} \ 	herefore \quad & \frac{10 \Omega}{30 \Omega}=\frac{30 \Omega}{90 \Omega} 	ext { or } \frac{1}{3}=\frac{1}{3} \end{aligned} $$ It is a balanced Wheatstone's bridge. Hence no current flows in the galvanometer arm. Hence, resistance $50 \Omega$ becomes ineffective.\ $	herefore \quad$ The equivalent resistance of the circuit is $$ \begin{aligned} R_{\mathrm{eq}} & =5 \Omega+\frac{(10 \Omega+30 \Omega)(30 \Omega+90 \Omega)}{(10 \Omega+30 \Omega)+(30 \Omega+90 \Omega)} \ & =5 \Omega+\frac{(40 \Omega)(120 \Omega)}{40 \Omega+120 \Omega}=5 \Omega+30 \Omega=35 \Omega \end{aligned} $$ Current drawn from the cell is $$ I=\frac{7 \mathrm{~V}}{35 \Omega}=\frac{1}{5} \mathrm{~A}=0.2 \mathrm{~A} $$