A galvanometer of resistance 100 when connected in series with 400 measures a voltage of upto 10 V. The value of resistance required to convert the galvanometer into ammeter to read upto 10 A is x 10-2 . The value of x is :
- 2
- 20
- 800
- 200
Answer
(2) 20
Solution
Full-scale current: ig = VG + R = 10100 + 400 = 0.02 A
Shunt S in parallel: igG = (I - ig)S
S = 0.02 10010 - 0.02 = 29.98 0.2
= 20 10-2
the galvanometer still carries only ig; the rest (I - ig) passes through S
x = 20
Answer: (2)