First slide

Moving coil Galvanometer

Question

Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is

Moderate

A

40 $Ω$
B

25 $Ω$
C

250 $Ω$
D

500 $Ω$

Solution

Let N=number of turns in galvanometer
A=Area, B= magnetic field
k=the restoring torque per unit twist.
$Current sensitivity, I_{S} = \frac{N B A}{k}$
$Voltage sensitivity, V_{S} = \frac{N B A}{k R_{G}}$
So, resistance of galvanometer
$R_{G} = \frac{I_{S}}{V_{S}} = \frac{5 \times 1}{20 \times 10^{- 3}} = \frac{5000}{20} = 250 Ω$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App