When a resistance of 5 Ω is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of 250 mA , however when 1050 Ω resistance is connected with it in series, it gives

Question

When a resistance of 5 Ω is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of 250 mA , however when 1050 Ω resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is _________Ω .

TestHub · Accepted Answer

The formula to calculate the maximum current through the galvanometer, when a shunt resistance $r_{1}$ is connected is given by ${(i_{G})}_{\max} = \frac{r_{1}}{r_{1} + R_{G}} i_{\max} . . . (1)$
When a series resistance $r_{2}$ is connected, the potential difference across the galvanometer is given by
$V = {(i_{G})}_{\max} [R_{G} + r_{2}] . . . (2)$
From equations (1) and (2), it can be written that
$\begin{array}{rcl} V & = & \frac{r_{1}}{r_{1} + R_{G}} i_{\max} (R_{G} + r_{2}) \\ \Rightarrow & V r_{1} + V R_{G} = i_{\max} r_{1} R_{G} + i_{\max} r_{1} r_{2} \\ \Rightarrow & (V - i_{\max} r_{1}) R_{G} = i_{\max} r_{1} r_{2} - V r_{1} \\ \Rightarrow & R_{G} = \frac{i_{\max} r_{1} r_{2} - V r_{1}}{V - i_{\max} r_{1}} . . . (3) \end{array}$
Substitute the values of the known parameters into equation (3) to calculate the required galvanometer resistance.
$\begin{array}{rcl} R_{G} & = & \frac{0.250 A \times 5 Ω \times 1050 Ω - 25 V \times 5 Ω}{25 V - 0.250 A \times 5 Ω} \\ = & 50 Ω \end{array}$

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