Question

A moving coil galvanometer converted into an ammeter reads up to $0.03 \,A$ by connecting a shunt of resistance $4 \,r$ across it and ammeter reads up to $0.06 \,A$, when a shunt of resistance $r$ is used. What is the maximum current which can be sent through this galvanometer if no shunt is used?

• A. 0.03 A
• B. 0 .04 A
• C. 0.02 A
• D. 0.01 A

Answer 1

Answer: (C)

Let $G$ be the resistance of galvanometer and $i_{g}$ the current which on passing through the galvanometer produces full-scale deflection.

Since $G$ and $S$ are in parallel, the potential difference across them will be
$i_{g} \times G=\left(i-i_{g}\right) \times S$
$\frac{i_{g}}{i}=\frac{S}{S+G}$
Hence, $i_{g}=\frac{4 r}{4 r+G} \times 0.03$ ......(1)
$i_{g}=\frac{r}{r+G} \times 0.06$ ..........(2)
Equating Eqs. (1) and (2), we get
$4 r+G=2(r+G)$
$\Rightarrow G=2 r$
$\therefore i_{g}=\frac{4 r}{4 r+2 r} \times 0.03=0.02 A [$ from Eq. (1)]