Question

In Section \(20.9,\) in order to find the energy stored in an inductor, we assumed that the current was increased from zero at a constant rate. In this problem, you will prove that the energy stored in an inductor is \(U_{\mathrm{L}}=\frac{1}{2} L I^{2}-\) that is, it only depends on the current \(I\) and not on the previous time dependence of the current. (a) If the current in the inductor increases from \(i\) to \(i+\Delta i\) in a very short time \(\Delta t,\) show that the energy added to the inductor is $$\Delta U=L i \Delta i$$ [Hint: Start with \(\Delta U=P \Delta t .]\) (b) Show that, on a graph of \(L i\) versus \(i,\) for any small current interval \(\Delta i,\) the energy added to the inductor can be interpreted as the area under the graph for that interval.(c) Now show that the energy stored in the inductor when a current \(I\) flows is \(U=\frac{1}{2} L I^{2}\)

Short answer

Based on the step by step solution, answer the following: Question: Prove that the energy stored in an inductor when a current I flows through it is U = 0.5 * L * I^2. Answer: In order to prove that the energy stored in an inductor is U = 0.5 * L * I^2, we followed three steps: 1. We calculated the energy added to the inductor when the current increased from i to i + Δi and showed that ΔU = L*i*Δi. 2. We then interpreted the energy added to the inductor on a Li vs. i graph as the area under the graph for any small current interval Δi. 3. Finally, we integrated the expression for ΔU from 0 to I to find the total energy added as the current increases from 0 to I. The result was U = 0.5 * L * I^2, proving that the energy stored in an inductor only depends on the current I and not on the previous time dependence of the current.

Step-by-step solution

Calculate the energy added to the inductor when the current increases

According to the hint, we know that ΔU = P * Δt, where P is the power. We also know that the power through an inductor can be calculated as P = V * i, where V is the voltage. The voltage across an inductor can be expressed as V = L * (di/dt), where di is the change in current and dt is the change in time. Substituting this expression for voltage into the power equation, we get P = L * (di/dt) * i. Now, we multiply this expression by Δt to find the change in energy, ΔU: $$ \Delta U = P \Delta t = L \left( \frac{\Delta i}{\Delta t} \right) i \Delta t $$ Since (Δi/Δt) * Δt = Δi, we get: $$ \Delta U = L i \Delta i $$

Interpret the energy added to the inductor on a Li vs. i graph

On a graph of Li vs. i, ΔU represents the area under the graph for a small current interval Δi. Since the graph is linear, the area of this small interval can be described as the area of a rectangle with sides L * i and Δi. Therefore, the energy added to the inductor can be interpreted as: $$ \Delta U = L i \Delta i $$ This expression shows that the energy added to the inductor can indeed be represented as the area under the graph of Li vs. i for any small current interval Δi.

Calculate the energy stored in the inductor when a current I flows

To find the energy stored in the inductor when a current I flows, we need to find the total energy added as the current increases from 0 to I. We can do this by integrating the expression for ΔU from 0 to I: $$ U = \int_0^I L i \, di $$ Since L is a constant, we can take it out of the integral: $$ U = L \int_0^I i \, di $$ Now we can evaluate the integral: $$ U = L \left[\frac{1}{2} i^2 \right]_0^I $$ So, the energy stored in the inductor when a current I flows is: $$ U = \frac{1}{2} L I^2 $$ This proves that the energy stored in an inductor only depends on the current I and not on the previous time dependence of the current.