Question

The ratios of the voltage sensitivities, resistances and areas of the coils of two moving coil galvanometers A and B are 4:3, 3:4 and 1:2 respectively. If the number of turns of the coil of galvanometer A is 200, then the number of turns of the coil of galvanometer B is (All other quantities remain same in both the cases)

• A. 100
• B. 150
• C. 200
• D. 400

Hint:

For comparison problems, it's best to start with the main formula ($S_V = NAB/kR$) and derive the proportionality for the quantities that are changing ($S_V \propto NA/R$). Then, set up a ratio equation to solve for the unknown.

Answer 1

The Correct Option is A

The voltage sensitivity ($S_V$) of a moving coil galvanometer is defined as the deflection per unit voltage.
$S_V = \frac{\theta}{V}$, where $\theta$ is the deflection and $V$ is the voltage.
The deflection is given by $\theta = \frac{NAB}{k}I$, where N is the number of turns, A is the area, B is the magnetic field, k is the torsional constant, and I is the current.
Voltage $V = IR$, where R is the resistance of the coil.
Substituting these into the sensitivity formula:
$S_V = \frac{(NAB/k)I}{IR} = \frac{NAB}{kR}$.
We are given that all other quantities (B and k) remain the same. So, we can write the proportionality:
$S_V \propto \frac{NA}{R}$.
We are given the ratios for galvanometer A and B:
$\frac{S_{VA}}{S_{VB}} = \frac{4}{3}$.
$\frac{R_A}{R_B} = \frac{3}{4}$.
$\frac{A_A}{A_B} = \frac{1}{2}$.
We can set up a ratio of the sensitivities using the proportionality:
$\frac{S_{VA}}{S_{VB}} = \frac{N_A A_A / R_A}{N_B A_B / R_B} = \frac{N_A}{N_B} \cdot \frac{A_A}{A_B} \cdot \frac{R_B}{R_A}$.
Now, substitute the given ratios:
$\frac{4}{3} = \frac{N_A}{N_B} \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{4}{3}\right)$.
The term $4/3$ cancels from both sides:
$1 = \frac{N_A}{N_B} \cdot \frac{1}{2}$.
This gives the relationship between the number of turns: $N_B = \frac{N_A}{2}$.
We are given that the number of turns for galvanometer A is $N_A = 200$.
$N_B = \frac{200}{2} = 100$.