Physics, asked by arr7157, 1 month ago

A power source of 112 V is in series with two lamps that provide resistance of 12 and 16, respectively. What is the current through the circuit?
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Answered by alisaqulain10

1

Answer:

let R1 = 12 and R2 = 16 , Given battery =112V

Rs = R1 + R2 = 12 +16 = 28 ohm

now according to ohm law

V=IR --- I = V/R = 112V / 28 = 4 ampere

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