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Answer 1

Answer:

see the attachment given....

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Answer 2

Answer:

Given that 5 mA flows through a galvanometer of resistance 100 Ω.

Hence potential drop across the galvanometer is 100Ω × 5mA = 0.5 V

According to the circuit, the galvanometer is in parallel connection with the shunt resistance. Hence, the potential drop would remain the same which is 0.5V. Also it is given that, 5 A is the total current flowing in the circuit.

Hence if 5 mA flowed through galvanometer, then the remaining current which is (5A - 5mA) will flow through the shunt.

⇒ V at shunt = Current through shunt × Shunt resistance

⇒ 0.5 V = ( 4.995 A ) × R

⇒ R = 0.5 V / 4.995 A

⇒ R = 0.10 Ω

Hence the value of shunt resistance is 0.10 Ω.

(Refer to the attachment for diagram)