The correct option (B) 2 calorie per sec
Explanation:
H = I2 Rt
(H/t) = I2R
for 5Ω, (H/t) = 10 cal
∴ 10 = I12 × 5
I1 = √2A
Now V = I1(5) = I2(4 + 6)
∴ 5I1 = 10I2
∴ I2 = (I1/2) = (√2/2) = (1/√2)A
Hence heat in 4Ω resistor per second = (H/t) = I22 R
= (1/√2)2 × 4 = 2 cal