Question 586: Current Electricity

Question

A moving coil galvanometer is converted into an ammeter, reading upto $$0.04 \mathrm{~A}$$ by connecting a shunt of resistance '$$3 \mathrm{r}$$' across it and then into an ammeter reading upto $$0.8 \mathrm{~A}$$, when a shunt of resistance '$$r$$' is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used?

Accepted Answer

Shunt resistance is given by
$$S=\frac{I_g G}{I-I_g}$$
In the first case $$: 3 r=\frac{I_g G}{0.04-I_g}$$ ..... (1)
In the second case : $$\quad r=\frac{I_g G}{0.08-I_g}$$ ..... (2)
Dividing $$\mathrm{Eq.}(1)$$ by Eq.(2) we get
$$3=\frac{0.08-\mathrm{I}_{\mathrm{g}}}{0.04-\mathrm{I}_{\mathrm{g}}}$$
Solving, we get $$\mathrm{I}_g=0.02 \mathrm{~A}$$

Answer

$$0.02 \mathrm{~A}$$

Answer

$$0.04 \mathrm{~A}$$

Answer

$$0.08 \mathrm{~A}$$

Answer

$$0.01 \mathrm{~A}$$

Current Electricity

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