Question

A resistor with R = 40.0 12 and an inductor with L = 0.600 H are connected in series to a battery that has emf 80.0 V and negligible internal resistance. At time t after the circuit is completed, the energy stored in the inductor is 0.400 J. Part A At this instant, what is the voltage across the inductor? Express your answer with the appropriate units.

Answer 1

At the instant when the energy stored in the inductor is 0.400 J, the voltage across the inductor in the series RL circuit will be equal to -0.400 V/s.

Step-by-step explanation:

The voltage across the inductor in a series RL circuit can be calculated using the formula VL = -L( dI/dt), where VL is the voltage across the inductor and dI/dt is the rate of change of current with respect to time. In this case, the energy stored in the inductor is given as 0.400 J, which implies that the current through the circuit is also changing with time. Since the inductor opposes any change in current, at the instant when the energy stored in the inductor is 0.400 J, the voltage across the inductor will be equal to the rate of change of energy with respect to time.

Using the formula for energy stored in an inductor, LI^2/2, and differentiating the equation with respect to time, we can find the rate of change of energy. Solving for dE/dt, we get dE/dt = LI * dI/dt. Substituting the given values L = 0.600 H and dE/dt = 0.400 J, we can find the rate of change of current. Plugging the known values into the equation, we get 0.400 J = (0.600 H)(dI/dt). Rearranging the equation, we find dI/dt = 0.667 A/s. Therefore, at the instant when the energy stored in the inductor is 0.400 J, the voltage across the inductor will be equal to -L(dI/dt) = -(0.600 H)(0.667 A/s) = -0.400 V/s.