Consider a circuit with an RL series with L=0.09 H and R=0.05 Ohm. At t=0 the circuit is connected to a battery which provides V0=12 V.

(a)How long does it take to the current to equal a fraction 0.95 of the steady state current?

(b) What is the energy stored (in Joules) in the magnetic field when the current equals a fraction 0.95 of the steady state current?

(c) What is the total energy delivered (in Joules) by the battery up to the time t found in part (a) ?

How much energy (in Joules) has been dissipated in the resistor?

b. I = E/R = 12/0.05 = 240A, =

Please ignore my 1st response. My

computer malfunctioned. I will have to
repeat the entire process.

a. I = E/R = 12/0.05 = 240A.

0.95I = 0.95*240 = 228A.
Vr + Vc = 12
228*0.05 + 12/e^(t/T) = 12
11.4 + 12/e^(t/T) = 12
12/e^(t/T) = 12-11.4 = 0.6
e^(t/T) = 12/0.6 = 20
t/T = 3.
T = L/R = 0.09/0.05 = 1.8=Time constant.
t/1.8 = 3
t = 5.4 s.

b. Energy = 0.5L*I^2=.5*0.09*(228)^2 =
2339.3 J.

thanks, do you have the c?

To answer these questions, we need to analyze the RL circuit's behavior in response to the input voltage. We will use the equations governing the circuit's response to find the required values.

First, let's derive the equations that describe the behavior of an RL circuit. The voltage across an inductor (V_L) is given by:

V_L(t) = L * di(t)/dt

And the voltage across a resistor (V_R) is given by Ohm's law:

V_R(t) = R * i(t)

where i(t) is the current at time t.

(a) How long does it take for the current to reach 0.95 of the steady state current?

To find this time, we need to consider the time constant (τ) of the RL circuit, which is defined as L/R. The time constant represents the time it takes for the current to reach approximately 63.2% of its final (steady state) value.

In this case, τ = L/R = 0.09 H / 0.05 Ω = 1.8 seconds.

To find the time it takes for the current to reach 0.95 of the steady state value, we can use the approximation that after 4 time constants, the current reaches 0.95 of the steady state value.

So, the required time (t) can be calculated as:

t = 4 * τ = 4 * 1.8 seconds = 7.2 seconds

Therefore, it takes approximately 7.2 seconds for the current to reach 0.95 of the steady state current.

(b) What is the energy stored in the magnetic field when the current equals 0.95 of the steady state current?

The energy stored in the magnetic field of an inductor is given by the formula:

E = 1/2 * L * i^2

Substituting the given values, we have:

E = 1/2 * 0.09 H * (0.95 * i_steady)^2

Since we want to find the energy when the current equals 0.95 of the steady state current, we can substitute 0.95 * i_steady for i in the equation.

(c) What is the total energy delivered by the battery up to the time found in part (a)?

The total energy delivered by the battery is equal to the energy stored in the magnetic field of the inductor at time t. We can use the same formula as in part (b) to calculate this energy.

E_total = 1/2 * 0.09 H * (0.95 * i_steady)^2

(d) How much energy has been dissipated in the resistor?

The energy dissipated in the resistor is equal to the work done by the resistor over time t. It can be calculated using the formula:

E_resistor = ∫[0 to t] (V_R * i) dt

Substituting V_R = R * i, we have:

E_resistor = ∫[0 to t] (R * i^2) dt

Since the question doesn't specify the value of time (t) at which we need to calculate the dissipated energy, we cannot provide an exact value without additional information.