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  • In the circuit diagrams (A, B, C and D) shown below, R is high resistance and S is a resistance of the order of galvanometer resistance G. The correct circuit, corresponding to the half deflection

In the circuit diagrams (A, B, C and D) shown below, R is high resistance and S is a resistance of the order of galvanometer resistance G. The correct circuit, corresponding to the half deflection method for finding the resistance and figure of merit of the galvanometer, is the circuit labelled as :

             

                

Option: 1

Circuit A with   G=\frac{RS}{\left ( R-S \right )}


Option: 2

Circuit B with G = S


Option: 3

Circuit C with G = S


Option: 4

Circuit D with G=\frac{RS}{R-S }


Answers (1)

best_answer

In figure D, the current will flow through the circuit when key \mathrm{K}_{1} is closed and  \mathrm{K}_{2} is open. The current flowing through the galvanometer is proportional to the deflection in it.
\mathrm{I}_{1}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{G}}=\mathrm{k} \theta

where, E - emf of the cell

R- resistance from the resistance box

G- galvanometer resistance for current I

\theta - galvanometer deflection for current I

k - proportionality constant.

When \mathrm{K}_{2} is closed and by adjusting the shunt resistance S, we can make galvanometer deflection as \theta / 2.

Then the current in the circuit is :I_{2}=\frac{E}{R+\frac{G S}{G+S}}

Now, a fraction of the current I2 in the circuit flows through the galvanometer, which is given by:

\mathrm{I}^{\prime}=\frac{\mathrm{I}_{2} \mathrm{~S}}{\mathrm{G}+\mathrm{S}}=\mathrm{k} \theta / 2

Now, from the above relations, we can get the resistance of the given galvanometer as:

G=\frac{R S}{R-S}

 

Posted by

Riya

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