A silver voltameter of resistance 2, Omega an | vedclass

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(D) Initially, the total resistance of the circuit is $R_1 = 2\, \Omega + 3\, \Omega = 5\, \Omega$. The current through the voltameter is $i_1 = \frac

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Decreases by $37.5\, \%$

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(D) Initially, the total resistance of the circuit is $R_1 = 2\, \Omega + 3\, \Omega = 5\, \Omega$. The current through the voltameter is $i_1 = \frac{V}{5}$.When a $2\, \Omega$ resistor is connected in parallel with the $2\, \Omega$ voltameter, the equivalent resistance of this parallel combination is $R_p = \frac{2 	imes 2}{2 + 2} = 1\, \Omega$.The new total resistance of the circuit is $R_2 = 1\, \Omega + 3\, \Omega = 4\, \Omega$. The main current in the circuit is $i = \frac{V}{4}$.Since the parallel resistors are equal ($2\, \Omega$ each), the current $i$ splits equally. Thus, the current through the voltameter is $i_2 = \frac{i}{2} = \frac{V}{8}$.The rate of deposition of silver is proportional to the current, $R \propto i$. The percentage change in the rate is $\frac{i_2 - i_1}{i_1} 	imes 100$.Percentage change $= \frac{\frac{V}{8} - \frac{V}{5}}{\frac{V}{5}} 	imes 100 = \frac{\frac{5V - 8V}{40}}{\frac{V}{5}} 	imes 100 = \frac{-3V}{40} 	imes \frac{5}{V} 	imes 100 = -37.5\, \%$.Therefore, the rate of deposition decreases by $37.5\, \%$.

$A$ silver voltameter of resistance $2\, \Omega$ and a $3\, \Omega$ resistor are connected in series across a cell. If a resistance of $2\, \Omega$ is connected in parallel with the voltameter, then the rate of deposition of silver

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