Calculate the reading shown by the ammeter [Electricity Class 10]
Attachments:
Answers
Answered by
0
Answer:
emf of each cell is 2 V. Hence potential difference across the battery is 2×2 = 4V. There are two resistors of resistance 4Ω each connected in parallel. Hence equivalenet resistance is 2Ω. Hence current in the circuit as shown by ammeter is 4 / 2 = 2A.
Electrical power across P and Q is i2×R = 2×2×2 = 8 W
Similar questions