Question

A galvanometer of resistance 120 ohms a full scale deflection with a current of 0.005A. How would you convert it an ammeter that reads a maximum current of 5A?
a) connect 2000 ohm in parallel to it
b) connect 200.12 ohms in series to it
c) connect 20.10 ohms in series to it
d) connect 0.012 ohms in parallel to it

Answer 1

Answer on Question #50960, Physics, Electromagnetism

A galvanometer of resistance 120 ohms a full scale deflection with a current of 0.005A. How would you convert it an ammeter that reads a maximum current of 5A?

a) connect 2000 ohm in parallel to it

b) connect 200.12 ohms in series to it

c) connect 20.10 ohms in series to it

d) connect 0.012 ohms in parallel to it

Solution

Shunt resistance $(R_{sh})$ is connected in parallel to an ammeter (see Fig.1)

Fig.1

According to Kirchhoff's current law

$$I = I_{A} + I_{sh}$$

where $I = 5A$, $I_{A} = 0.005A$

According to Kirchhoff's voltage law

$$I_{A} R_{A} = I_{sh} R_{sh}$$

From Eq.(1) and Eq.(2)

$$R_{sh} = \frac{I_{A}}{I_{sh}} R_{A} = \frac{I_{A}}{I - I_{A}} R_{A} = \frac{0.005}{5 - 0.005} \cdot 120 \approx 0.12 \text{ ohms}$$

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