NEET Class neet Answered
solve the problem
Asked by pankajpankaj55224 | 16 Jun, 2022, 04:37: PM
Expert Answer
Power dissipated in resistor = i2 R
where i is the current passing through resistor of resistance R
if 36 W is dissipated in 9 Ω resistor , then current i through resistor is calculated as follows
i2 × 9 = 36
hence we get current i = 
A = 2 A
A = 2 AVoltage across 9 Ω resistor = i × R = 2 × 9 = 18 V
Since 6 Ω resistor is in parallel to 9 Ω resistor , voltage across 6 Ω resistor also 18 V
Hence current through 6 Ω resistor = voltage / resistance = (18 / 6 ) A = 3 A
Hence current drawn from battery = ( 3 + 2 ) A = 5 A
Voltage across 2 Ω resistor = i × R = 5 × 2 = 10 V
Answered by Thiyagarajan K | 16 Jun, 2022, 08:36: PM
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