Two coils of resistance R1 = 3Ω and R2 = 9Ω are connected in series across a battery of potential difference 14 V. Draw the circuit diagram. Find the electrical energy consumed in 1 min in each resistance.
R1 = 3Ω, R2 = 9Ω Rs = R1 + R2 = 9 + 3 = 12 Ω Now, I = VR=1412 = 1.167 Amp. [I in series remains constant.] Electric energy consumed in R1 H1 = I2R1t = (1.167)2 × 3 × 60 = 245.14 J Electric energy consumed in R2 H2 = I2R2t = 735.42J
