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

$$U = \int\;dU$$

$$U = \int \limits^I_0\;LI dI$$

$$U = \dfrac{1}{2}\;LI^2$$

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.

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

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,

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

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

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

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

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

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

$$U = \dfrac{1}{2} LI^2$$

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

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

$$U = \dfrac{1}{2}\; LI^2$$

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

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