Question

A 35.0-V battery with negligible internal resistance, a 50.0 ? resistor, and a 1.25-mH inductor with negligible resistance are all connected in series with an open switch. The switch is suddenly closed.

(a) How long after closing the switch will the current through the inductor reach half of its maximum value?
(b) How long after closing the switch will the energy stored in the inductor reach half of its maximum value?

Answer 1

The current through the inductor will reach half of its maximum value after one time constant, which is 0.000025 s. The energy stored in the inductor will reach half of its maximum value after one time constant, which is also 0.000025 s.

Step-by-step explanation:

The time constant for an RL circuit can be calculated using the formula τ = L/R, where τ is the time constant, L is the inductance, and R is the resistance. In this case, the inductance is given as 1.25 mH (or 0.00125 H) and the resistance is 50.0 Ω. Therefore, the time constant is:

τ = 0.00125 H / 50.0 Ω = 0.000025 s

(a) The current through the inductor will reach half of its maximum value after one time constant. So, the time required is:

0.000025 s

(b) The energy stored in an inductor is given by the formula E = (1/2)LI^2, where E is the energy, L is the inductance, and I is the current. The maximum current is determined by the voltage supplied by the battery and the resistance of the circuit. In this case, the voltage is 35.0 V and the resistance is 50.0 Ω. Therefore, the maximum current is:

I = V/R = 35.0 V / 50.0 Ω = 0.7 A

The initial energy stored in the inductor is:

E = (1/2)(0.00125 H)(0.7 A)^2 = 0.00030625 J

Therefore, the energy stored in the inductor will reach half of its maximum value after one time constant, which is:

0.000025 s