Question

The ratio of heats generated through shunt and galvanometer is 7:5 when they are connected to make an ammeter.If the resistance of the galvanometer is 112Ohm then the resistance of the shunt is

Answer 1

As per the question,

The ratio in the heat generated in the shunt and the galvanometer =$\dfrac{7}{5}$

The resistance of the galvanometer $R_g=112\Omega$

As we know that in the galvanometer, the shunt is connected parallel to the galvanometer.

Hence the potential along the both the shunt and the galvanometer will be same.

Let the resistance of the shunt is $R_s$

Now, $H=\dfrac{v^2}{R}$

So, $\dfrac{H_1}{H_2}=\dfrac{R_g}{R_s}$

$R_s=\dfrac{R_g\times H_2}{H_1}$

$R_s=\dfrac{112\times 5}{7}=80\Omega$

Hence, the resistance of the shunt=80$\Omega$