• 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 S₁ is closed and switch S₂ 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 

              \(U = \int\;dU\)

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

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

•  Consider a solenoid whose inductance is given as 

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

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

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

             \(L = \mu _0\;n^2 (V)\)

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.

• 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,

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

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

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

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

• 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 S₂ switch is thrown to point "a" and S₁ 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 1. When battery is connected 

• 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 S₂ switch is thrown to point "a" and S₁ 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 S₂ is at position "a" for a long time due to allow the current to reach its equilibrium value \(\dfrac {\mathcal{E}}{R}\).
• Now, switch S₂ 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 

         \(U = \dfrac{1}{2}\; LI^2\)

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