Question

A milliammeter of range 10mA has a coil of resistance $1\Omega$. To use it as an ammeter of range 1A. The required shunt must have a resistance of

& \text{A}\text{. }\dfrac{1}{101}\Omega \\\ & \text{B}\text{. }\dfrac{1}{100}\Omega \\\ & \text{C}\text{. }\dfrac{1}{99}\Omega \\\ & \text{D}\text{. }\dfrac{1}{9}\Omega \\\ \end{aligned}$$

Answer 1

We should know the basic concept of conversion of the galvanometer to the ammeter. An ideal ammeter should have zero resistance. We will use the formula of the shunt when the galvanometer is converted into an ammeter to find the required value of resistance that the shunt must-have.

Formula used:

& {{I}_{g}}G=(I-{{I}_{g}})S \\\ & S=\dfrac{{{I}_{g}}}{(I-{{I}_{g}})}\times G \\\ \end{aligned}$$ **Complete answer:** Firstly we will have a glimpse of the ammeter and shunt. Ammeter : We know that to measure the electric current through the circuit element .we connect a device in series with that element known as an ammeter. Shunt : To measure large currents with it, a small resistance is connected in parallel with the galvanometer coil. The resistance connected in the way is called the shunt. As we know galvanometer and shunt are connected in parallel P.D across the galvanometer = P.D across the shunt $$\begin{aligned} & {{I}_{g}}G=(I-{{I}_{g}})S \\\ & S=\dfrac{{{I}_{g}}}{(I-{{I}_{g}})}\times G \\\ \end{aligned}$$ We will put the value and obtain that $$\dfrac{0.01\times 1}{1-0.01}=\dfrac{1}{99}\Omega $$ By connecting a shunt of resistance S across the given galvanometer, we get an ammeter of desired range. The obtained shunt must have a resistance of $\dfrac{1}{99}\Omega $ **So the correct option is (C).** **Additional Information:** To convert a galvanometer into a voltmeter we connect a high resistance called multiplier in series to the galvanometer. A resistance R, which connects in series to the galvanometer, is converted into a voltmeter of range 0 – V volts. The voltmeter which is obtained by converting also has some errors. It is not an ideal voltmeter. **Note:** We know the deflection in the galvanometer is proportional to ${{I}_{g}}$ and hence to I so the scale can be graduated to read the value of current I directly.