A galvanometer has a resistance of 100 Ω. A potential difference of 100 mV between its terminals gives a full scale deflection. The shunt resistance required to convert it into an ammeter reading upto 5 A is

Question

A galvanometer has a resistance of 100 Ω. A potential difference of 100 mV between its terminals gives a full scale deflection. The shunt resistance required to convert it into an ammeter reading upto 5 A is (A) 0.01 Ω  (B) 0.02 Ω  (C) 0.03 Ω  (D) 0.04 Ω

Tardigrade Admin · Accepted Answer

The current required for full-scale deflection is

i_{g} = \frac{PD across galvanometer}{resistance of galvanometer}

i_{g} = \frac{100 \times 1 0 ^{- 3}}{100Ω} = 1 0^{- 3} A

The shunt resistance required is

S = (\frac{i _{g}}{i - i _{g}}) G = (\frac{1 0 ^{- 3}}{5 - 1 0 ^{- 3}}) \times 100 = 0.02 Ω

Q. A galvanometer has a resistance of $100 Ω$ . A potential difference of $100 mV$ between its terminals gives a full scale deflection. The shunt resistance required to convert it into an ammeter reading upto $5 A$ is

$0.01 Ω$

$0.02 Ω$

$0.03 Ω$

$0.04 Ω$

Q. A galvanometer has a resistance of 100Ω. A potential difference of 100mV between its terminals gives a full scale deflection. The shunt resistance required to convert it into an ammeter reading upto 5A is

0.01Ω

0.02Ω

0.03Ω

0.04Ω

The current required for full-scale deflection is ig​= resistance of galvanometer PD across galvanometer ​ ig​=100Ω100×10−3​=10−3A The shunt resistance required is S=(i−ig​ig​​)G=(5−10−310−3​)×100=0.02Ω

Q. A galvanometer has a resistance of $100 Ω$ . A potential difference of $100 mV$ between its terminals gives a full scale deflection. The shunt resistance required to convert it into an ammeter reading upto $5 A$ is

$0.01 Ω$

$0.02 Ω$

$0.03 Ω$

$0.04 Ω$