Question

An ammeter is obtained by shunting a 30ω galvanometer with 30ω resistance. What should additional shunt be connected across it to double the range?

Answer 1

Answer:

The additional shunt is 10 ohm.

Explanation:

Given that,

Shunt $S= 30\Omega$

Galvanometer $G= 30\Omega$

For ammeter

$\dfrac{S}{G}=\dfrac{I_{g}}{I-I_{g}}$

$\dfrac{G}{S}=\dfrac{I-I_{g}}{I_{g}}$

$\dfrac{30}{30}=\dfrac{I}{I_{g}}-1$

$1+1=\dfrac{I}{I_{g}}$

$I=2I_{g}$

Now, range is doubled

$I=4I_{g}$

The shunt required

$S=\dfrac{I_{g}}{4I_{g}-I_{g}}G$

$S=\dfrac{I_{g}}{3I_{g}}\times 30$

$S=\dfrac{1}{3}\times30$

$S= 10\Omega$

Hence, The additional shunt is 10 ohm.

Answer 2

Answer:

15Ω

Explanation:

Initially (I-Ig)S= IgG => I-Ig/Ig = 30/30 = 1

Therefore, I - Ig = Ig

Ig = I/2      ----> (1)

To double the range of the ammeter, it means I'= 2I

(I-Ig)S= IgG

(2I-I/2) S = I/2 (30)

(3I/2) S = (I/2) 30

S= 30/3 = 10Ω

Now, the shunt is 10Ω

So, the additional shunt required is,

1/x + 1/30 = 1/10

30 + x = 3x

2x = 30

x = 15Ω

So, the additional shunt required is 15Ω