Practice energy stored in the inductor and magnetic energy density equations with examples. Find the energy stored in the inductor as a function of time and the maximum power delivered to the inductor.

- Consider a RL circuit, as shown in figure.

- A part of energy supplied by the battery appears as internal energy in the resistance circuit and the remaining is stored in the form of magnetic field in the inductor.
- Let switch S
_{1}is closed and switch S_{2}is at position "a". - Then by applying Kirchhoff's law in the circuit

\(\mathcal{E} - IR - L\dfrac{dI}{dt}\; = 0\)

\(\mathcal{E} =IR +L \dfrac{dI}{dt}\)

- Multiplying both sides by current \(I\)

\(I\mathcal{E}= I^2R + IL \dfrac{dI}{dt}\)

- Here \(I\mathcal{E}\) = Rate at which energy is supplied by the battery

\(I^2R\) = Rate at which energy is delivered

\(LI\dfrac{dI}{dt}\) = Rate at which energy is being stored in the inductor

- Let \(U\) be the energy stored in the inductor and is given by

\(\dfrac{dU}{dt} = LI\;\dfrac{dI}{dt}\)

- To find total energy stored in the inductor

\(U = \int\;dU\)

\(U = \int \limits^I_0\;LI dI\)

\(U = \dfrac{1}{2}\;LI^2\)

A Halved

B Doubled

C Tripled

D Remains the same

\(L = \dfrac{\mu_0\;N^2\;A}{\ell}\)

\(L = \dfrac{\mu_0\;N^2\;}{\ell}\;\dfrac{A\ell}{\ell}\)

\(L = \dfrac{\mu_0\;N^2\;}{\ell^2}\;(A\ell)\)

\(L = \mu _0\;n^2 (V)\)

where,

V = Volume = \(A\ell\)

n = Number of turns per unit length = \(\dfrac{N}{\ell}\)

- The magnetic field of the solenoid is given by

\(B = \mu_0nI\)

\(\Rightarrow\;\;\;I= \dfrac{B}{\mu_0n}\)

- Energy stored in an inductor is given by

\(U = \dfrac{1}{2}\;LI^2\)

\(U = \dfrac{1}{2}(\mu_0n^2V) × \dfrac{B^2}{\mu^2_0 \;n_2}\)

\(U = \dfrac{B^2}{2 \mu _0}V\)

- The magnetic energy density or energy stored per unit volume

\(U_B = \dfrac{U}{V} = \dfrac{B^2}{2 \mu_0}\)

This expression is valid for any region of space.

A 4 × 108 J

B 3 × 108 J

C 5 × 106 J

D 6 × 106 J

- The energy stored in inductor while charging of RL circuit is given by

**\(U_L = \dfrac{1}{2}\;L\;i^2_0\;\left(1-e^{-t/\tau}\right)^2\)**

- The power delivered to inductor is defined as the rate at which energy is stored in inductor.

** **\(P = \dfrac{dU_L}{dt} \)

\(P = \dfrac{d}{dt}\left[\dfrac{L}{2}\;i^2_0\left(1-e^{-t/\tau}\right)^2\right]\)

\(P = \dfrac{L\;i^2_0}{2}\left(e^{-t/\tau }-e ^{-2t/\tau}\right)\)

- For maximum power

\(\dfrac{dP}{dt} = 0\)

\(\dfrac{L\;i_0^2}{\tau}\;\left(\dfrac{-1e\;^{-t/\tau}\;}{\tau}+\dfrac{2\;e^{-2t/\tau}}{\tau}\right) = 0\)

\(e^{-t/\tau} = \dfrac{1}{2}\)

So,

\(P_{max} = \dfrac{L\;i_0^2}{\tau} \;\left[\dfrac{1}{2} - \left(\dfrac{1}{2}\right)^{2}\right]\)

\(P_{max} =\dfrac{L\; i^2_0}{\tau}\; × \dfrac{1}{4}\)

\(P_{max} =\dfrac{RL\;\mathcal{E}^2}{L\;R^2} × \dfrac{1}{4} \;\;\;\left[ \tau = \dfrac {L}{R}\right]\)

\(P_{max} = \dfrac{\mathcal{E}^2}{4 R}\)

A \(\mathcal{E}^2\;/\;2R\)

B \(\dfrac{\mathcal{E}^2}{4 R}\)

C \(\mathcal{E} ^2 / 8 R\)

D \(\dfrac{\mathcal{E}^2}{R}\)

- Consider a circuit in which an inductor of inductance L and a resistor of resistance R, are connected in series with a battery of emf \(\mathcal{E}\).

- Suppose S
_{2}switch is thrown to point "a" and S_{1}is closed at time t = 0, the current starts increasing in the circuit.

- By applying Kirchhoff's law

\(\mathcal{E}- IR - L\dfrac {dI}{dt}= 0\)

Rearranging the equation

\(\mathcal{E}- IR = L\dfrac {dI}{dt}\)

\(\dfrac {dI}{\mathcal{E}- IR}\;\;=\;\;\dfrac {dt}{L}\)

- Integrating both sides and applying limits

\(\int\limits^{I}_0\;\dfrac {dI}{\mathcal{E}-IR}\;=\; \int\limits^{t}_0 \dfrac {dt}{L}\)

Let \(x = \mathcal{E}- IR\), then \(dx=-RdI\)

\(\dfrac {-1}{R}\;\;\int\limits^{\mathcal{E}-IR}_\mathcal{E}\dfrac {dx}{x}\;\;=\;\;\dfrac {1}{L}\int\limits^{t}_0dt\)

\(\Rightarrow \;\;\dfrac {-1}{R}[lnx]^{\mathcal{E}-IR}_\mathcal{E} = \dfrac {t}{L}\)

\(\Rightarrow \;\;ln \dfrac {\mathcal{E} -IR}{\mathcal{E}}= \dfrac {-Rt}{L}\)

\(\Rightarrow \;\; \dfrac {\mathcal{E} -IR}{\mathcal{E}} = e^{-Rt/L}\)

\(\Rightarrow \mathcal{E} - IR = \mathcal{E} \;e^{-t(L/R)}\)

\(\Rightarrow\;\;I = \dfrac {\mathcal{E} }{R}[1-e^{-t/(L/R)}]\)

- This expression shows that current does not increase instantly to its final equilibrium value as switch is closed. It takes infinite time to reach steady state.
- By time constant \(\tau = \dfrac {L}{R}\)

\(I = \dfrac {\mathcal{E}}{R}[1-e^{-t/\tau}]\)

- When time constant \(\tau = t\)

\(I = \dfrac {\mathcal{E}}{R}[1-e^{-t/t}]\)

\(I = \dfrac {\mathcal{E}}{R}[1-e^{-1}]\)

\(I = \dfrac {\mathcal{E}}{R}[1-.37]\)

\(I = \dfrac {\mathcal{E}}{R} × 0.632\)

\( I = 63.2 \) % of its final value

\(I= \dfrac{\mathcal{E}}{R} (1 - e ^{-tR/L})\)

\( I = \dfrac{\mathcal{E}}{R} (1 - e ^{-t/\tau})\)

where \(\tau = \dfrac{L}{R}\)

- So, energy stored in the inductor after time "t" when battery is connected is given as

\(U = \dfrac{1}{2} LI^2\)

\(U_t = \dfrac{1}{2}L \left(\dfrac{\mathcal{E}}{R}\right)^2\;\;\left(1-e^{t/\tau}\right)^2\)

A 23.62 mJ

B 28.57 mJ

C 50.72 mJ

D 20.46 mJ

**Case 1. **When battery is connected

- Suppose S
_{2}switch is thrown to point "a" and S_{1}is closed at time t = 0, the current starts increasing in the circuit.

- By applying Kirchhoff's law

\(\mathcal{E} - IR -L \dfrac {dI}{dt}= 0\)

Rearranging the equation

\(\mathcal{E} - IR = L\dfrac {dI}{dt}\)

\(\dfrac {dI}{\mathcal{E}- IR}\;\;=\;\;\dfrac {dt}{L}\)

- Integrating both sides and applying limits

\(\int\limits^{I}_0\;\dfrac {dI}{\mathcal{E}-IR}\;=\; \int\limits^{t}_0 \dfrac {dt}{L}\)

Let \(x = \mathcal{E}- IR\), then \(dx=-RdI\)

\(\dfrac {-1}{R}\;\;\int\limits^{\mathcal{E}-IR}_\mathcal{E}\dfrac {dx}{x}\;\;=\;\;\dfrac {1}{L}\int\limits^{t}_0dt\)

\(\Rightarrow \;\;\dfrac {-1}{R}[lnx]^{\mathcal{E}-IR}_\mathcal{E} = \dfrac {t}{L}\)

\(\Rightarrow \;\;ln \dfrac {\mathcal{E} -IR}{\mathcal{E}}= \dfrac {-Rt}{L}\)

\(\Rightarrow \;\; \dfrac {\mathcal{E} -IR}{\mathcal{E}} = e^{-Rt/L}\)

\(\Rightarrow \mathcal{E} - IR = \mathcal{E}\;e^{-t(L/R)}\)

\(\Rightarrow\;\;I = \dfrac {\mathcal{E} }{R}[1-e^{-t/(L/R)}]\)

- This expression shows that current does not increase instantly to its final equilibrium value as switch is closed. It takes infinite time to reach steady state.
- By time constant \(\tau = \dfrac {L}{R}\)

\(I = \dfrac {\mathcal{E}}{R}[1-e^{-t/\tau}]\)

- When time constant \(\tau = t\)

\(I = \dfrac {\mathcal{E}}{R}[1-e^{-t/t}]\)

\(I = \dfrac{\mathcal{E}}{R}[1-e^{-1}]\)

\(I = \dfrac {\mathcal{E}}{R}[1-.37]\)

\(I = \dfrac {\mathcal{E}}{R} × 0.632\)

\(I\) = 63.2 % of its final value

\(I = \dfrac{\mathcal{E}}{R} (1 - e ^{-tR/L})\)

\(I = \dfrac{\mathcal{E}}{R} (1 - e ^{-t/\tau})\)

where \(\tau = \dfrac{L}{R}\)

- So, energy stored in the inductor after time t when battery is connected is given as

\(U = \dfrac{1}{2} LI^2\)

\(U_t = \dfrac{1}{2}L \left(\dfrac{\mathcal{E}}{R}\right)^2\;\;\left(1-e^{-t/\tau}\right)^2\)

**Case 2.** When battery is disconnected

- Consider a RL circuit as show in figure.

- A switch S
_{2}is at position "a" for a long time due to allow the current to reach its equilibrium value \(\dfrac {\mathcal{E}}{R}\). - Now, switch S
_{2}is thrown from a to b. - Applying Kirchhoff's law,

\(IR + L\dfrac {dI}{dt} = 0\)

\(L\dfrac {dI}{dt} = -IR\)

\(\dfrac {dI}{I} = \dfrac {-R}{L}\;\;dt\)

- Integrating both sides

\(\Rightarrow \;\;\int\limits^{I}_{\mathcal{E}/R} \;\dfrac {dI}{I}\;\;=\;\;\dfrac {-R}{L}\;\int \limits^t_0 dt\)

\(\Rightarrow\;\;[lnI]^{I}_{\mathcal{E}/R}\;\;=\;\;\dfrac {-R}{L}t\)

\(\Rightarrow\;\;ln \left(\dfrac {I}{\mathcal{E}/R}\right) = \dfrac {-R}{L}t\)

\(\Rightarrow \dfrac {I}{\mathcal{E}/R}\;\;e^{-Rt/L}\)

\(I = \dfrac {\mathcal{E}}{R}\;\;e^{-t/(L/_R)}\)

\(I = I_0\;\;e^{-t/(L/_R)}\)

- We know

Time constant \(\tau = \dfrac {L}{R}\)

So, \(I= I_0 \;\;e^{-t/z}\)

\(I = \dfrac{\mathcal{E}}{R} e^{-t/\tau}\)

- Energy in the inductor

\(U = \dfrac{1}{2}\; LI^2\)

\(U_{t_0} = \dfrac{1}{2}\; L \left(\dfrac{\mathcal{E}}{R}e^{-t/\tau}\right)^2\)

A \(2 (1-e^{-3t})^2\;mJ\)

B \(3 (1-e^{-4t})^2\;mJ\)

C \(25 (1-e^{-10^3t})^2\;mJ\)

D \(50 (1-e^{-10^3t})^2\;mJ\)