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) ?
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To solve these problems, we need to consider the behavior of a simple RL circuit. Recall that in an RL circuit, when a battery is connected, the current rises from 0 to its steady state value over time.
(a) We are asked to find the time it takes for the current to equal 0.95 times the steady-state current. Let's start by calculating the time constant, denoted as τ, which is given by the expression τ = L / R. Substituting the given values, we have τ = 0.09 H / 0.05 Ohm = 1.8 seconds.
The time constant represents the time it takes for the current to increase by a factor of e (about 2.718). Therefore, we can say that when t = τ, the current will have reached approximately 63.2% (1 - 1/e) of its steady-state value.
To find the time it takes for the current to equal 0.95 times the steady-state current, we can use the exponential decay formula for an RL circuit:
I(t) = I0 * (1 - e^(-t/τ))
We know that I(t) = 0.95 * I0 (where I0 is the steady-state current), so we can solve for t:
0.95 * I0 = I0 * (1 - e^(-t/τ))
Simplifying the equation:
0.95 = 1 - e^(-t/τ)
Rearranging and solving for t:
e^(-t/τ) = 1 - 0.95
e^(-t/τ) = 0.05
Taking the natural logarithm of both sides:
-t/τ = ln(0.05)
Solving for t:
t = -τ * ln(0.05)
Substituting the value of τ:
t = -1.8 * ln(0.05) ≈ 6.39 seconds
So, it takes approximately 6.39 seconds for the current to equal 0.95 times the steady-state current.
(b) The energy stored in the magnetic field in an inductor is given by the expression:
E = 0.5 * L * I^2
where E is the energy, L is the inductance, and I is the current.
We are asked to find the energy stored when the current equals 0.95 times the steady-state current. To calculate this, we substitute the values:
E = 0.5 * 0.09 H * (0.95 * I0)^2
Simplifying:
E = 0.5 * 0.09 H * 0.9025 * I0^2
E = 0.0406125 * I0^2
Therefore, the energy stored in the magnetic field when the current equals 0.95 times the steady-state current is approximately 0.0406125 times the energy stored at steady state.
(c) The total energy delivered by the battery up to the time found in part (a) can be calculated using the formula:
E_total = 0.5 * L * I^2 + V0 * I * t
where E_total is the total energy delivered, L is the inductance, I is the current, V0 is the battery voltage, and t is the time.
Substituting the given values:
E_total = 0.5 * 0.09 H * (0.95 * I0)^2 + 12 V * 0.95 * I0 * 6.39 s
E_total = 0.0406125 * I0^2 + 73.5864 * I0
The total energy delivered by the battery up to the time found in part (a) is given by the equation 0.0406125 * I0^2 + 73.5864 * I0 Joules.