The resistances of the four arms P, Q, R and | vedclass

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(A) For a balanced Wheatstone's bridge, the condition is $\frac{P}{Q} = \frac{R}{S}$.Substituting the given values: $\frac{10 \, \Omega}{30 \, \Omega}

Vedclass · Accepted Answer

$0.2$

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(A) For a balanced Wheatstone's bridge, the condition is $\frac{P}{Q} = \frac{R}{S}$.Substituting the given values: $\frac{10 \, \Omega}{30 \, \Omega} = \frac{30 \, \Omega}{90 \, \Omega}$, which simplifies to $\frac{1}{3} = \frac{1}{3}$.Since the bridge is balanced, no current flows through the galvanometer arm. Therefore, the $50 \, \Omega$ galvanometer resistance is ineffective.The circuit simplifies to two parallel branches: $(P+Q)$ and $(R+S)$ in parallel, connected in series with the internal resistance $r = 5 \, \Omega$.The equivalent resistance of the parallel part is $R_p = \frac{(P+Q)(R+S)}{(P+Q)+(R+S)} = \frac{(10+30)(30+90)}{(10+30)+(30+90)} = \frac{40 	imes 120}{40+120} = \frac{4800}{160} = 30 \, \Omega$.The total equivalent resistance of the circuit is $R_{eq} = r + R_p = 5 \, \Omega + 30 \, \Omega = 35 \, \Omega$.The current drawn from the cell is $I = \frac{E}{R_{eq}} = \frac{7 \, V}{35 \, \Omega} = 0.2 \, A$.

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