# Moving Coil Galvanometer, Voltmeter, Ammeter, and their Conversions

## Problems from IIT JEE

**Problem (IIT JEE 2008): **
*Statement 1:* The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as core inside the coil.

*Statement:* Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.

- Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
- Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
- Statement 1 is true, statement 2 is false.
- Statement 1 is false, statement 2 is true.

**Solution: **
The sensitivity of a moving-coil galvanometer is defined as angular deflection $\theta$ per unit current $i$. It is given as $\theta/i=nAB/k$, where $n$ is number of turns, $A$ is coil area, $B$ is magnetic field, and $k$ is torsional constant of suspension wire. To increase sensitivity, $B$ is increased by placing a soft iron core inside the coil. The soft iron has a high magnetic permeability and can be easily magnetized or demagnetized.

**Problem (IIT JEE 2005):**
A moving coil galvanometer of resistance ${100}\;{\Omega}$ is used as an ammeter using a resistance ${0.1}\;{\Omega}$. The maximum deflection current in the galvanometer is ${100}\;{\mu A}$. Find the current in the circuit, so that the ammeter shows maximum deflection,

- 100.1 mA
- 1000.1 mA
- 10.01 mA
- 1.01 mA

**Solution: **
A galvanometer of resistance $G$ is converted to an ammeter by connecting a small shunt resistance $S$ in parallel (see figure). Kirchhoff's loop law gives,
\begin{align}
&i_gG-(i-i_g)S=0,\quad \implies \quad i=i_g(G+S)/S. \nonumber
\end{align}
The maximum deflection current of galvanometer sets upper limit on the current measured by this ammeter. Substitute the values to get,
\begin{alignat}{2}
&i=i_g({G+S})/{S}=(100\times{10}^{-6})\left(({100+0.1})/{0.1})\right)={100.1}\;\mathrm{mA}. \nonumber
\end{alignat}