An LR circuit having L = 4H,  and R=1 Ω E = 6 V is switched on at t = 0. The power dissipated in joule heating at t = 4 s is

The time constant of the circuit isλ=LR=41=4 sThe current at t = 4 s is thereforei=ER1-e-t/λ=61-1e=6 A×0.63 A=3.8 AThe power dissipated in Joule heating = i2 R=(3.8 A)2×10Ω=140 W