Question

A moving coil galvanometer is converted into an ammeter reads up to $0.03\text{A}$ by connecting a shunt of resistance $4r$ across it and ammeter reads up to $0.06\text{A}$, 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. $0.04\text{A}$
• B. $0.03\text{A}$
• C. $0.02\text{A}$
• D. $0.01\text{A}$

Answer 1

The correct option is C $0.02\text{A}$

Let R be the resistance of the ammeter & I be the maximum current passing through the ammeter.
In the first case $0.03A$ is main current and in the second case $0.06A$ is main current.
$I=\frac{0.03\times 4r}{(R+4r)}$ for Ist case
$I=\frac{0.06\times r}{(R+r)}$ for second case
$\frac{0.03\times 4r}{R+4r}=\frac{0.06\times r}{(R+r)}$
$R+4r=2R+2r\Rightarrow R=2r$
The value of
$I=\frac{0.03\times 4r}{2r+4r}=\frac{0.03\times 4r}{6r}$
$=0.02\text{A}$
This is the maximum value of current in ammeter.