A 24.0-V battery is connected in series with a resistor and an inductor, with R = 8.20 Ω and L = 2.00 H, respectively.
(b) Find the energy stored in the inductor at an instant that is a time interval of one time constant after the switch is closed.
To calculate the energy stored in the inductor at one time constant after the switch is closed, we'll need to determine the time constant first. The time constant (τ) for an RL circuit is given by the formula:
τ = L/R
In this case, the inductance (L) is 2.00 H and resistance (R) is 8.20 Ω. Plugging in these values, we can calculate the time constant:
τ = 2.00 H / 8.20 Ω
τ ≈ 0.2439 s
Now, to find the energy stored in the inductor after one time constant, we can use the formula:
Energy stored = (1/2) * L * I^2
where I is the current flowing through the inductor.
At one time constant, the current flowing through the inductor has reached approximately 63.2% of its maximum value. This is given by:
I = Imax * (1 - e^(-t/τ))
where Imax is the maximum current and t is the time.
In this case, since the battery is connected directly to the inductor and resistor in series, the maximum current (Imax) can be calculated using Ohm's law:
Imax = V / R
where V is the battery voltage.
Plugging in the given values, we get:
Imax = 24.0 V / 8.20 Ω
Imax ≈ 2.93 A
Now we can calculate the current at one time constant:
I = 2.93 A * (1 - e^(-(1 * τ)/τ))
I ≈ 2.93 A * (1 - e^(-1))
I ≈ 2.93 A * (1 - 0.368)
I ≈ 1.85 A
Finally, we can substitute the values of Inductance (L) and current (I) into the energy formula:
Energy stored = (1/2) * L * I^2
Energy stored = (1/2) * 2.00 H * (1.85 A)^2
Energy stored ≈ 3.42 J
Therefore, the energy stored in the inductor at one time constant after the switch is closed is approximately 3.42 Joules.