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

Question

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

TestHub · Accepted Answer

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

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