Question

A moving coil galvanometer has resistance $50\mathrm{\Omega }$ and it indicates full deflection at 4mA current. A voltmeter is made using this galvanometer and a $5k\mathrm{\Omega }$ resistance. The maximum voltage that can be measured using this voltmeter, will be close to:
A. 10 V
B. 20 V
C. 40 V
D. 15 V

Answer 1

Hint: This problem can be solved using Ohm’s law. Ohm’s law gives a relationship between current, potential difference and resistance. Add the resistance of the galvanometer and the resistance given. Substitute this value in place of resistance and the current in the formula for Ohm’s law. Solve this expression and thus the voltage is obtained. This obtained voltage is the maximum voltage that can be measured using the voltmeter,
Formula used:
$V=IR$

Complete answer:
Given: I = 4mA
           G= $50\mathrm{\Omega }$
           R= $5k\mathrm{\Omega }$= $5000\mathrm{\Omega }$
According to Ohm’s Law,
$V=IR$
Where, V is the potential difference across the coil
             I is the current flowing through the coil
             R is the resistance
$V=I(G+R)$
Substituting the values in above equation we get,
$V=4\times {10}^{-3}(50+5000)$
$\Rightarrow V=4\times {10}^{-3}\times 5050$
$\Rightarrow V=20200\times {10}^{-3}$
$\Rightarrow V=20.2V$
$V\approx 20V$
Hence, the maximum voltage that can be measured using the voltmeter will be close to 20 V.

So, the correct answer is “Option B”.

Note:
A galvanometer can be converted into an ammeter by connecting a low resistance R in parallel to the galvanometer of resistance G. While a galvanometer can be converted into a voltmeter by connecting a large resistance R in series with the galvanometer of resistance G. Moving coil galvanometer is used to detect alternating current in the circuit because on changing the polarity of the current, the torque also changes its direction.