In this circuit, $\epsilon$₀ = 12 V, R = 6.0 $\Omega$, and L = 0.48 H. The switch is closed at time t = 0. At time t = 0.25 s, the rate of Joule heating is approximately

Question

Quizplus · Accepted Answer

After the switch is closed, the current in the circuit will begin to increase from 0 A. The rate of change of current is given by 
$$\frac{dI}{dt}=\frac{\varepsilon}{L/R}= \frac{12\ V}{0.48\ H/6.0\ \Omega}=150\ A/s$$
The power dissipated by Joule heating is given by
$$P=I^2R$$
At time t = 0.25 s, the current in the circuit is
$$I=\frac{\varepsilon}{R}(1-e^{-Rt/L})=\frac{12\ V}{6.0\ \Omega}(1-e^{-6.0\ \Omega \cdot 0.25\ s/0.48\ H})\approx 2.82\ A$$
Therefore, the rate of Joule heating at time t = 0.25 s is
$$P=I^2R=(2.82\ A)^2\cdot6.0\ \Omega\approx 45.4\ W$$
We can also calculate the rate of change of power as
$$\frac{dP}{dt}=2IR\frac{dI}{dt}=2(2.82\ A)(6.0\ \Omega)\frac{150\ A}{s}\approx 50.7\ W/s$$
Since the rate of change of power is positive, the rate of Joule heating is increasing at this time. Therefore, the closest answer is B) 22 W.

In This Circuit ϵ\epsilonϵ 0 = 12 V, R = 6 Ω\OmegaΩ

In This Circuit $\epsilon$ ₀ = 12 V, R = 6 $\Omega$