Question

A moving coil galvanometer is converted into an ammeter reading upto 0.03 A by connecting a shunt of resistance 4r across it and into an ammeter reading upto 0.06 A when a shunt of resistance r connected across it. What is the maximum current which can be through this galvanometer if no shunt is used

• A. 0.01 A
• B. 0.02 A
• C. 0.03 A
• D. 0.04 A

Answer 1

Answer: (B)

0.02 A

Answer 2

For ammeter, $S = \frac { i _ { g } } { \left( i - i _ { g } \right) } G$

⇒ $i _ { g } G = \left( i - i _ { g } \right) S$ So $i _ { g } G = \left( 0.03 - i _ { g } \right) 4 r$ …… (i)

and $i _ { g } G = \left( 0.06 - i _ { g } \right) r$ …… (ii)

Dividing equation (i) by (ii)

$1 = \frac { \left( 0.03 - i _ { g } \right) 4 } { 0.06 - i _ { g } } \Rightarrow 0.06 - i _ { g } = 0.12 - 4 i _ { g }$⇒ 3i_g = 0.06

⇒ i_g­ = 0.02 A