Question

How a galvanometer is converted into an ammeter? The resistance of the coil of a galvanometer is 15 Ω and it gives full scale deflection by a current of 4 mA. How it can be converted into an ammeter in order to measure a current upto 6 A?

Hint:

Connect a small shunt in parallel so the coil voltage equals the shunt voltage: \(I_g G = (I-I_g)S\). Solve for \(S\).

Answer 1

Concept: A galvanometer detects only very small currents. To turn it into an ammeter (which reads large currents), a small resistance called a shunt \(S\) is connected in parallel with the galvanometer coil. Most of the line current then passes through the shunt, and only the small full-scale current \(I_g\) passes through the coil.

Step 1 (Given): Galvanometer resistance \(G = 15\ \Omega\); full-scale (maximum) coil current \(I_g = 4\ \text{mA} = 4\times10^{-3}\ \text{A}\); total current to be measured \(I = 6\ \text{A}\).

Step 2 (Set up the parallel condition): Since \(S\) is in parallel with the coil, the potential difference across both is equal. Current through shunt \(= I - I_g\). So:
\[ I_g\,G = (I - I_g)\,S \]

Step 3 (Formula for shunt):
\[ S = \frac{I_g\,G}{I - I_g} \]

Step 4 (Substitute values):
\[ S = \frac{(4\times10^{-3})(15)}{6 - 4\times10^{-3}} = \frac{0.06}{5.996} \]

Step 5 (Arithmetic):
\[ S = 0.01001\ \Omega \approx 0.01\ \Omega \]

Conclusion: Connecting a small shunt resistance of about \(0.01\ \Omega\) in parallel with the galvanometer coil converts it into an ammeter capable of reading up to 6 A.
\[\boxed{S \approx 0.01\ \Omega\ \text{(shunt in parallel)}}\]