Question

A galvanometer of resistance
120Ω
a full scale deflection with a current of 0.0005A. How would you convert it to an ammeter that reads a maximum current of 5A?
connect
2000Ω
in parallel to it
connect
200.12Ω
in series to it
connect
20.10Ω
in series to it
connect
0.012Ω
in parallel to it

Answer 1

Answer on Question 55888, Physics, Electromagnetism

Question:

A galvanometer of resistance $120\Omega$ shows a full scale deflection with a current of 0.0005A. How would you convert it to an ammeter that reads a maximum current of 5A?

a) connect $2000\Omega$ in parallel to it

b) connect $200.12\Omega$ in series to it

c) connect $20.10\Omega$ in series to it

d) connect $0.012\Omega$ in parallel to it

Solution:

Galvanometer can be converted into an ammeter by shunting it with a very small resistance as we can see in the scheme below:

Let the resistance of galvanometer be $R_{g}$ and it gives full scale deflection when current $I_{g}$ is passed through it. Then, from the Ohm's law we can write the potential difference across the galvanometer:

$$V _ {g} = I _ {g} R _ {g}.$$

Let a shunt of resistance $R_{s}$ is connected in parallel to galvanometer and total current through the circuit is $I$ . Then, from the Kirchoff's first law we can write the current through the shunt:

$$I _ {s} = I - I _ {g}.$$

Then, the potential difference across the shunt will be:

$$V _ {s} = I _ {s} R _ {s} = \left(I - I _ {g}\right) R _ {s}.$$

But, potential difference across the galvanometer and shunt resistance are equal, so we can write:

$$\left(I - I _ {g}\right) R _ {s} = I _ {g} R _ {g}.$$

From this formula, we can find the shunt resistance $R_{s}$ :

$$R _ {s} = \frac {I _ {g}}{I - I _ {g}} R _ {g} = \frac {0 . 0 0 0 5 A}{5 A - 0 . 0 0 0 5 A} \cdot 1 2 0 \Omega = 0. 0 1 2 \Omega .$$

**Answer:**

d) connect $0.012\Omega$ in parallel to it

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