In the figure shown below, a resistance of 150.4 is connected in series to an ammeter A of resistance 240 . A shunt resistance of 10 is connected in parallel with the ammeter. The reading of the ammeter is mA. [figure]
Answer
5
Solution
Ammeter (240 Ω) and shunt (10 Ω) in parallel:
Rp = 240 10240 + 10 = 9.6
Total resistance: R = 150.4 + 9.6 = 160
Main current: I = 20160 = 0.125 A
Current through the ammeter (current divides inversely as resistance):
IA = I 10240 + 10 = 0.125 0.04
= 5 10-3 A = 5 mA
Answer: 5