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

Question

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

Accepted Answer

Correct answer: 50. (50) The formula to calculate the maximum current through the galvanometer, when a shunt resistance r_1 is connected is given by (i_G )_ max= r_1r_1+R_Gi_ max … (a ) 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)…..(b) From equations (a) and (b), it can be written that V= r_1r_1+R_Gi_ max(R_G+r_2) ⇒Vr_1+VR_G=i_ maxr_1R_G+i_ maxr_1r_2 ⇒(V-i_ maxr_1)R_G=i_ maxr_1r_2-Vr_1 ⇒R_G= i_ maxr_1r_2-Vr_1V-i_ maxr_1 … (c ) Substitute the values of the known parameters into equation (c) to calculate the required galvanometer resistance. R_G= 0.250 A 5 1050 -25 V 5 25 V-0.250 A 5 =50

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

Solution

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