Question

The current sensitivity of a moving coil galvanometer increases by 20% when its resistance increases by a factor 2 while keeping the number of turns constant. The voltage sensitivity changes by _________.
1. 20%
2. 40%
3. 30%
4. 10%

Answer 1

Correct Answer - Option 2 : 40%

CONCEPT:

Galvanometer:

• A galvanometer is a device that is used to detect the current, to find the direction of the current, and also to compare the current.

Current sensitivity of a galvanometer:

• The deflection θ per unit current is known as the current sensitivity of the galvanometer.

Voltage sensitivity of a galvanometer:

• The deflection θ per unit voltage is known as the voltage sensitivity of the galvanometer.

​CALCULATION:

• We know that the voltage sensitivity is given as,

\(⇒ V_{s}=\left ( \frac{I_{s}}{R} \right )\)     -----(1)

Where V_s = voltage sensitivity, I_s = current sensitivity, and R = resistance

• When the resistance is increased by factor 2,

\(⇒ R^{'}=2R\)     -----(2)

\(⇒ I_{s}^{'}=\) I_s + 20% of I_s

\(⇒ I_{s}^{'}=I_{s}+\frac{20}{100}I_{s}\)

\(⇒ I_{s}^{'}=1.2I_{s}\)     -----(3)

\(⇒ V_{s}^{'}=\frac{I_{s}^{'}}{R^{'}}\)

\(⇒ V_{s}^{'}=\frac{1.2I_{s}}{2R}\)     -----(4)

By equation 1 and equation 4,

\(⇒ V_{s}^{'}=0.6V_{s}\)     -----(5)

• The change in voltage sensitivity is given as,

⇒ %\( \Delta V=\frac{V_{s}^{'}-V_{s}}{V_{s}}\times100\)

⇒ %\( \Delta V=\frac{0.6V_{s}-V_{s}}{V_{s}}\times100\)

⇒ %\( \Delta V=-40\)%

Hence, option 2 is correct.