Informative line

Energy Stored

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.

Energy Stored in 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 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 

              \(U = \int\;dU\)

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

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

Illustration Questions

What happens to the energy stored in the inductor, if the inductance of the inductor coil is doubled?

A Halved

B Doubled

C Tripled 

D Remains the same 

×

Since energy is directly proportional to inductance and is given by 

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

where \(L\) = Inductance 

        \(I\) = current 

When the inductance is doubled 

\(L^{'}=\;2L\)

Then energy stored in inductor  \(U^{'}= \dfrac{1}{2} L{'}\;I^2\)

\(U^{'} = \dfrac{1}{2} (2L) I^2\)

\(\Rightarrow \;\;\;U^{'} = LI^2\)

\(\Rightarrow \;\;\;U^{'} = 2 \;U\)

So, energy is doubled.

What happens to the energy stored in the inductor, if the inductance of the inductor coil is doubled?

A

Halved

.

B

Doubled

C

Tripled 

D

Remains the same 

Option B is Correct

Magnetic Energy Density

  •  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.

Illustration Questions

Determine the energy stored in a solenoid, if magnetic field of solenoid \(B = 1\,T \) and volume of region \(V=4\,\pi\,m^3\).

A 4 × 108 J

B 3 × 108 J

C 5 × 106 J

D 6 × 106 J

×

The magnetic energy density or energy stored per unit volume is given as 

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

\(U = \dfrac{B^2}{2\;\mu_0} × V\)

where,

U = Magnetic energy 

B = Magnetic field 

V = Volume 

Given: \(B = 1\, T\),  \(V=4\,\pi\,m^3\)

Energy \(U = \dfrac{(1)^2}{2 × 4 \pi × 10 ^{-7}}\;× \; 4 \pi\)

             \( U = 5 × 10^6 \,J\) 

Determine the energy stored in a solenoid, if magnetic field of solenoid \(B = 1\,T \) and volume of region \(V=4\,\pi\,m^3\).

A

4 × 10J

.

B

3 × 10J

C

5 × 10J

D

6 × 10J

Option C is Correct

Growth of Current in an RL Circuit

  • 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 

       \(U = \dfrac{1}{2} LI^2\)

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

 

 

Illustration Questions

An inductor of inductance \(L = 4\;mH\) and a resistor of resistance \(R = 2\;\Omega\), are connected in series with a battery of emf \(\mathcal{E}= 12\; V\). Find the energy stored in the inductor after time \(t = 2\;msec\)  after connection of battery. \([e^{-1} = 0.37 ]\) 

A 23.62 mJ

B 28.57 mJ

C 50.72 mJ

D 20.46 mJ

×

Growth of current in a RL circuit is given as 

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

where \(\tau = \dfrac{L}{R}\)

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

\(I= \dfrac{12}{2}\;\left(1-e^{-\left (\dfrac{2 × 10^{-3} × 2}{4 × 10 ^{-3}}\right)}\right)\)

\(I = 6 \left(1 - e^{-1}\right)\;=\; 6(1 -0.37)\)

\(I= 6 × 0.63 = 3.78 \;A\)

Energy stored in inductor 

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

\(U = \dfrac{1}{2} × (4 × 10^{-3}) × (3.78)^2\)

\(U = 28.57 \;mJ\)

An inductor of inductance \(L = 4\;mH\) and a resistor of resistance \(R = 2\;\Omega\), are connected in series with a battery of emf \(\mathcal{E}= 12\; V\). Find the energy stored in the inductor after time \(t = 2\;msec\)  after connection of battery. \([e^{-1} = 0.37 ]\) 

A

23.62 mJ

.

B

28.57 mJ

C

50.72 mJ

D

20.46 mJ

Option B is Correct

Decay of Current in RL Circuit 

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 

         \(U = \dfrac{1}{2}\; LI^2\)

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

 

Illustration Questions

An inductor of inductance \(L = 2\;mH\) and a resistor of resistance \(R = 2\;\Omega\), are connected in series with a battery of emf \(\mathcal{E} = 10\; V\). Find the energy stored in the inductor as a function of time.

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

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

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

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

×

Total energy stored in RL circuit given as \(U = \dfrac{1}{2} LI^2\)

where 

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

Here, \(\mathcal{E}\)= E.m.f. of battery 

\(R\) = Resistance 

\(L\) = Inductance  

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

Given: \(L = 2\, mH\)\(R=2\,\Omega\),  \(\mathcal{E}=10\,V\)

 \(U= \dfrac{1}{2} × 2 × 10 ^{-3}\left(\dfrac{10}{2}\right)^2\;\left(1-e^{-t\dfrac{2}{2 × 10 ^{-3}}}\right)^2\)

\(U = 25 \left(1-e^{-10^3t}\right)^2\;mJ\)

An inductor of inductance \(L = 2\;mH\) and a resistor of resistance \(R = 2\;\Omega\), are connected in series with a battery of emf \(\mathcal{E} = 10\; V\). Find the energy stored in the inductor as a function of time.

A

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

.

B

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

C

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

D

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

Option C is Correct

Illustration Questions

Consider a solenoid of radius r =0 .5 m, number of turns N = 500 and length \(\ell\)= 9.8596 m. The current in the solenoid is 2 A. Calculate the energy stored in the solenoid. 

A 50 mJ

B 100 mJ

C 200 mJ

D 150 mJ

×

Inductance of the solenoid is given as 

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

\(L= \dfrac {4 \pi × 10 ^{-7}× (500)^2 × \pi × (.5 )^2}{9.8596}\)

\(L = 250000 × 10 ^{-7}\)

\(L = 25 × 10 ^{-3} H \)

\(L = 25\; mH\)

Energy stored in the inductor is given by 

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

\(U =\dfrac{1}{2}× 25 × 10^{-3} × (2)^2\)

\(U = 50 \;mJ\)

 

 

Consider a solenoid of radius r =0 .5 m, number of turns N = 500 and length \(\ell\)= 9.8596 m. The current in the solenoid is 2 A. Calculate the energy stored in the solenoid. 

A

50 mJ

.

B

100 mJ

C

200 mJ

D

150 mJ

Option A is Correct

Maximum Power Delivered to Inductor in an RL Circuit 

  • 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,

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

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

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

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

 

Illustration Questions

An inductor of inductance L and a resistor of resistance R are  connected in series with a battery of emf \(\mathcal{E}\). Find the maximum power delivered to the inductor.   

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

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

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

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

×

The energy stored in the magnetic field at time 't' is given as

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

\(U = \dfrac{1}{2}\; L\;i^2_0\;\left(1-e^{-t/\tau}\right)^2\;\;\;\;\;\;\;\;\left[i = i_0 \left(1-e^{-t/\tau}\right)\right]\)

Power is the rate of change of energy. 

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

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

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,

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

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

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

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

An inductor of inductance L and a resistor of resistance R are  connected in series with a battery of emf \(\mathcal{E}\). Find the maximum power delivered to the inductor.   

A

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

.

B

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

C

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

D

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

Option B is Correct

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