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

Step 1 (Idea of range multiplication): An ammeter is made by wiring a low resistance shunt across a galvanometer. Its range is expressed through the multiplying factor \(n = I/I_g\), which tells how many times larger the measurable current is compared with the coil's full-scale current.

Step 2 (Find the factor): With \(I = 6\ \text{A}\) and \(I_g = 4\times10^{-3}\ \text{A}\),
\[ n = \frac{I}{I_g} = \frac{6}{4\times10^{-3}} = 1500 \]

Step 3 (Shunt in terms of the factor): Starting from \(I_g G = (I - I_g)S\) and dividing throughout by \(I_g\), the shunt becomes
\[ S = \frac{G}{\,n - 1\,} \]

Step 4 (Substitute):
\[ S = \frac{15}{1500 - 1} = \frac{15}{1499} \]

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

Result: A parallel shunt of roughly \(0.01\ \Omega\) diverts 1499 parts of the current while 1 part flows through the coil, so the meter now reads up to 6 A.
\[\boxed{S = \frac{G}{n-1} \approx 0.01\ \Omega}\]