Question

2.A shunt resistance of 5 Ω is connected to a galvanometer of resistance of 100 Ω.
The current in the galvanometer is 0.42 A. What is the current in the circuit?

Answer 1

Answer:

8.82A

Explanation:

let Rg,Ig and Vg represent the resistance ,current through galvanometer andterminal voltage of galvanometer and

Rs for shunt resistance, V= circuit. voltage.

total resistance of the circuit

R =( Rg×Rs) / (Rg+Rs)

= (100×5) / (100+5)

= 100/21 ohm

voltage of the circuit

= voltage between terminals of galvanometer

Vg = Ig×Rg

= 0.42×100

= 42 volt

=>current in the circuit = 42/(100/21)

= 882 / 100

= 8.82 A

hope this helps you