The galvanometer deflection,when key K_1 is c | vedclass

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(B) Let $R$ be the resistance of the galvanometer. When $K_1$ is closed and $K_2$ is open,the total resistance of the circuit is $220 + R$. The curren

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$22$

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(B) Let $R$ be the resistance of the galvanometer. When $K_1$ is closed and $K_2$ is open,the total resistance of the circuit is $220 + R$. The current through the galvanometer is $i = \frac{V}{220 + R} = k	heta_0$,where $k$ is the galvanometer constant.When $K_2$ is closed,the galvanometer (resistance $R$) is in parallel with the resistor $R_2 = 5\,\Omega$. The new deflection is $\frac{	heta_0}{5}$,so the new current through the galvanometer is $i' = \frac{i}{5} = \frac{V}{5(220 + R)}$.The voltage across the parallel combination is $V_{parallel} = i' R = \frac{iR}{5}$.The current through $R_2$ is $i_2 = \frac{V_{parallel}}{R_2} = \frac{iR/5}{5} = \frac{iR}{25}$.The total current from the battery is $I = i' + i_2 = \frac{i}{5} + \frac{iR}{25} = i \left( \frac{5 + R}{25} ight)$.Applying Kirchhoff's voltage law to the main loop:$V = I(220) + V_{parallel} = i \left( \frac{5 + R}{25} ight) 220 + \frac{iR}{5}$.Since $V = i(220 + R)$,we have:$i(220 + R) = i \left[ \frac{220(5 + R)}{25} + \frac{R}{5} ight]$.$220 + R = \frac{44(5 + R)}{5} + \frac{R}{5} = \frac{220 + 44R + R}{5} = \frac{220 + 45R}{5} = 44 + 9R$.$220 - 44 = 9R - R \Rightarrow 176 = 8R \Rightarrow R = 22\,\Omega$.

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