Question

A moving coil galvanometer of resistance $20 \Omega$ produces full scale deflection for a current of $50$ mA. How will you convert the galvanometer into-
(i) An ammeter of range $20$ A and
(ii) A voltmeter of range $120$ V.

Answer 1

Hint: An ammeter is a device utilized to measure the current passing within a circuit. The galvanometer can turn into an ammeter by joining a shunt (low resistance) parallel to the galvanometer. A voltmeter is a machine utilized to estimate the potential difference between two locations in a circuit. A galvanometer can turn into a voltmeter by combining a multiplier (high resistance) series with the galvanometer.

Complete step-by-step answer:
Conversion of galvanometer into ammeter-
Let G be the galvanometer's resistance; I be the current measured by ammeter, $I_{g}$ is the current passed through the galvanometer.
G =$20 \Omega$, I =$20$ A, $I_{g}$ =$50$ mA
(I -$I_{g}$ ) is the current passing through the shunt S.
Value of $S = G \dfrac{I_{g}}{I - I_{g}}$
$S = 20 \dfrac{0.05}{20 – 0.05} = 0.05 \Omega$
A $0.05 \Omega$ shunt should be connected in parallel with the galvanometer.
Conversion of galvanometer into voltmeter-
Let G be the galvanometer's resistance; V be the voltage measured by voltmeter, is the current passed through the galvanometer.
G =$20 \Omega$, V=$120$ V, $I_{g}$ =$50$ mA
$R = \dfrac{V}{ I_{g} }- G$
$R = \dfrac{120}{ 0.05 }- 20 = 2380 \Omega$
A $2380 \Omega$ resistance should be connected in series with a galvanometer.

Additional Information: For ammeters, when the shunt is connected in parallel with the galvanometer, net resistance is less than the shunt, and when it is connected in series in the circuit, it will not influence the current passing by the circuit.

Note: R should be high for getting the voltmeter from the galvanometer. R is the high resistance connected in series with the galvanometer. So, the voltmeter's net resistance is high and will not take current from the circuit. Ideal resistance of voltmeter is infinity.