Question 178: Current Electricity

Question

The current sensitivity of moving coil galvanometer is increased by $$25 \%$$. This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:

Accepted Answer

The current sensitivity ($I_s$) of a moving coil galvanometer is given by:
$$I_s = \frac{nBA}{C}$$
where $$n$$ is the number of turns in the coil, $$B$$ is the magnetic field, $$A$$ is the area of the coil, and $$C$$ is the torsional constant of the suspension wire.
We are told that the current sensitivity is increased by 25% by changing only the number of turns and the area of the cross section while keeping the resistance of the galvanometer coil constant. Let the new number of turns be $$n'$$, and the new area be $$A'$$.
Since the current sensitivity is increased by 25%, we have:
$$I_s' = 1.25I_s = \frac{n' B A'}{C}$$
The resistance ($R$) of the galvanometer coil is given by:
$$R = ho \frac{l}{A}$$
where $$ho$$ is the resistivity of the wire and $$l$$ is the length of the wire. Since the resistance is kept constant, we can write:
$$R' = ho \frac{l'}{A'} = R$$
Now, the voltage sensitivity ($V_s$) of the galvanometer is given by:
$$V_s = I_s 	imes R$$
We want to find the percentage change in the voltage sensitivity. Let the new voltage sensitivity be $$V_s'$$:
$$V_s' = I_s' 	imes R'$$
Since $$R = R'$$, we can write:
$$V_s' = 1.25I_s 	imes R = 1.25V_s$$
The percentage change in the voltage sensitivity is:
$$\frac{V_s' - V_s}{V_s} 	imes 100\% = \frac{1.25V_s - V_s}{V_s} 	imes 100\% = 25\%$$

Answer

+25%

Answer

$$-$$50%

Answer

$$-$$25%

Answer

Zero

Current Electricity

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