A 24-V battery is connected in series with a resistor and an inductor, with R = 5.6 Ω and L = 2.0 H, respectively.

(a) Find the energy stored in the inductor when the current reaches its maximum value.


(b) Find the energy stored in the inductor one time constant after the switch is closed.

V = iR + L di/dt

i = a(1-e^-kt)
for large t
i =24/5.6 = a
so
a = 4.29
i = 4.29(1-e^-kt)
di/dt = 4.29 k e^-kt
24 = 24-24e^-kt + 2(4.29)k e^-kt
24 = 2(4.29) k
k = 24/(2*4.29) = R/L
so
i = 4.29(1-e^-(Rt/L))
current is max at great t
i max = 4.29 - 0
energy = (1/2) L i^2 =(1/2)(2)4.29^2
= 18.4 Joules

one time constant T =L/R and e^-(Rt/L) = 1/e = .368
i = 4.29 (1-.368) = 4.29*.632 = 2.71 amps
energy = (1/2)(2)2.71^2 = 7.35 Joules

(a) Well, you know what they say, maximum current, maximum fun! To find the energy stored in the inductor when the current reaches its maximum value, you can use the formula:

Energy = 0.5 * L * I^2

Where L is the inductance and I is the maximum current. Plugging in the values, you get:

Energy = 0.5 * 2.0 H * (I_max)^2

Now, I don't have the exact value of I_max, but I can tell you that the energy stored in the inductor is going to be electrifying!

(b) Now, one time constant after the switch is closed, things might have settled down a bit. To calculate the energy stored in the inductor, you can use the formula:

Energy = 0.5 * L * I^2

But this time, the current I will have changed after one time constant. To find I, you can use the formula:

I = I_max * e^(-t/τ)

Where I_max is the maximum current, e is the mathematical constant, t is the time, and τ is the time constant (which is equal to L/R in this case).

Once you have the value of I after one time constant, you can plug it back into the energy formula:

Energy = 0.5 * L * (I_after_one_time_constant)^2

And voila! You'll have the energy stored in the inductor, ready to enlighten your circuits.

To find the energy stored in the inductor when the current reaches its maximum value, we need to use the formula for the energy stored in an inductor:

E = (1/2) * L * I^2

Where:
E = energy stored in the inductor in joules
L = inductance in henries
I = maximum current in amperes

(a) Maximum current is the current that flows through the circuit when the inductor is fully charged. In this case, the battery voltage is 24 V and the resistance is 5.6 Ω. Using Ohm's Law, we can calculate the current:

I = V / R
I = 24 V / 5.6 Ω
I ≈ 4.29 A

Now we can calculate the energy stored in the inductor when the current reaches its maximum value:

E = (1/2) * L * I^2
E = (1/2) * 2.0 H * (4.29 A)^2
E ≈ 38.61 J

Therefore, the energy stored in the inductor when the current reaches its maximum value is approximately 38.61 joules.

To find the energy stored in the inductor one time constant after the switch is closed, we need to determine the time constant of the circuit. The time constant (τ) of an RL circuit is given by:

τ = L / R

(b) Plugging in the values for the resistor and inductor, we get:

τ = 2.0 H / 5.6 Ω
τ ≈ 0.36 s

One time constant (1τ) is the amount of time it takes for the current in the circuit to reach approximately 63.2% of its maximum value. So, one time constant after the switch is closed, the current will have reached 63.2% of its maximum value. Let's calculate the current:

I_1τ = 0.632 * I_max
I_1τ = 0.632 * 4.29 A
I_1τ ≈ 2.71 A

Now we can calculate the energy stored in the inductor one time constant after the switch is closed:

E = (1/2) * L * I_1τ^2
E = (1/2) * 2.0 H * (2.71 A)^2
E ≈ 7.68 J

Therefore, the energy stored in the inductor one time constant after the switch is closed is approximately 7.68 joules.

To find the energy stored in the inductor at different instances, we need to use the formulas related to energy in an inductor.

(a) To find the energy stored in the inductor when the current reaches its maximum value, we first need to find the maximum current. In an RL circuit, the maximum current is given by:

Imax = V / sqrt(R^2 + (ωL)^2)

Where V is the voltage of the battery (24 V), R is the resistance (5.6 Ω), ω is the angular frequency (ω = 2πf, where f is the frequency), and L is the inductance (2.0 H).

Let's assume the frequency is not mentioned and use a commonly used value of 50 Hz. Then, ω = 2πf = 2π(50) = 100π rad/s.

Plugging these values into the formula, we have:

Imax = 24 / sqrt((5.6^2) + (100π * 2)^2)
Imax ≈ 1.196 A (rounded to three decimal places)

The energy stored in an inductor is given by:

E = (1/2) * L * (Imax^2)

Plugging in the calculated values, we have:

E = (1/2) * 2.0 * (1.196^2)
E ≈ 1.429 Joules (rounded to three decimal places)

Therefore, the energy stored in the inductor when the current reaches its maximum value is approximately 1.429 Joules.

(b) To find the energy stored in the inductor one time constant after the switch is closed, we need to find the time constant first. The time constant (τ) of an RL circuit is given by:

τ = L / R

Plugging in the given values, we have:

τ = 2.0 / 5.6
τ ≈ 0.357 seconds (rounded to three decimal places)

One time constant after the switch is closed, the current will decay to approximately 1/e (about 36.8%) of its initial value. Let's calculate the current at that time using the formula:

I = Imax * e^(-t/τ)

Where t is the time in seconds.

Plugging in the values, we have:

I = 1.196 * e^(-1/0.357)
I ≈ 0.437 A (rounded to three decimal places)

The energy stored in the inductor at this time is given by:

E = (1/2) * L * (I^2)

Plugging in the calculated values, we have:

E = (1/2) * 2.0 * (0.437^2)
E ≈ 0.190 Joules (rounded to three decimal places)

Therefore, the energy stored in the inductor one time constant after the switch is closed is approximately 0.190 Joules.