Question

A galvanometer has a full scale defection of 1.0A and internal resistance of 0.5ohms, if it is connected to a current of 30 A, calculate the value of resistance of shunt required with the meter without damaging it

Answer 1

The maximum voltage in the galvanometer resistor is given by:

$V_{G}=R_{in}*I_{G}$

Where;

• Electric current $I_{G}=1.0A$
• electric resistance $R_{in}=0.5\Omega$

Assessing numerically.

$V_{G}=1.0A*0.5\Omega=0.5V$

Parallel resistance has the same potential difference.

$V_{G}=V_{shunt}=0.5V$

The current flowing must be equal to:$I_{shunt}=30A-1.0A=29A$

Therefore the value of R is calculated with Ohm's Law

$R_{shunt}=\frac{V_{shunt}}{I_{shunt}}$

Numerically evaluating

$R_{shunt}=\frac{0.5V}{29A}=1.72*10^{-2}\Omega$

the value of resistance of shunt required with the meter without damaging it is

$\boxed{R_{shunt}=1.72*10^{-2}\Omega}$