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.
\(\mathcal{E} - IR - L\dfrac{dI}{dt}\; = 0\)
\(\mathcal{E} =IR +L \dfrac{dI}{dt}\)
\(I\mathcal{E}= I^2R + IL \dfrac{dI}{dt}\)
\(I^2R\) = Rate at which energy is delivered
\(LI\dfrac{dI}{dt}\) = Rate at which energy is being stored in the inductor
\(\dfrac{dU}{dt} = LI\;\dfrac{dI}{dt}\)
\(U = \int\;dU\)
\(U = \int \limits^I_0\;LI dI\)
\(U = \dfrac{1}{2}\;LI^2\)
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 = \(\dfrac{N}{\ell}\)
\(B = \mu_0nI\)
\(\Rightarrow\;\;\;I= \dfrac{B}{\mu_0n}\)
\(U = \dfrac{1}{2}\;LI^2\)
\(U = \dfrac{1}{2}(\mu_0n^2V) × \dfrac{B^2}{\mu^2_0 \;n_2}\)
\(U = \dfrac{B^2}{2 \mu _0}V\)
\(U_B = \dfrac{U}{V} = \dfrac{B^2}{2 \mu_0}\)
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
\(U_L = \dfrac{1}{2}\;L\;i^2_0\;\left(1-e^{-t/\tau}\right)^2\)
\(P = \dfrac{dU_L}{dt} \)
\(P = \dfrac{d}{dt}\left[\dfrac{L}{2}\;i^2_0\left(1-e^{-t/\tau}\right)^2\right]\)
\(P = \dfrac{L\;i^2_0}{2}\left(e^{-t/\tau }-e ^{-2t/\tau}\right)\)
\(\dfrac{dP}{dt} = 0\)
\(\dfrac{L\;i_0^2}{\tau}\;\left(\dfrac{-1e\;^{-t/\tau}\;}{\tau}+\dfrac{2\;e^{-2t/\tau}}{\tau}\right) = 0\)
\(e^{-t/\tau} = \dfrac{1}{2}\)
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 \(\dfrac{\mathcal{E}^2}{R}\)
\(\mathcal{E}- IR - L\dfrac {dI}{dt}= 0\)
Rearranging the equation
\(\mathcal{E}- IR = L\dfrac {dI}{dt}\)
\(\dfrac {dI}{\mathcal{E}- IR}\;\;=\;\;\dfrac {dt}{L}\)
\(\int\limits^{I}_0\;\dfrac {dI}{\mathcal{E}-IR}\;=\; \int\limits^{t}_0 \dfrac {dt}{L}\)
Let \(x = \mathcal{E}- IR\), then \(dx=-RdI\)
\(\dfrac {-1}{R}\;\;\int\limits^{\mathcal{E}-IR}_\mathcal{E}\dfrac {dx}{x}\;\;=\;\;\dfrac {1}{L}\int\limits^{t}_0dt\)
\(\Rightarrow \;\;\dfrac {-1}{R}[lnx]^{\mathcal{E}-IR}_\mathcal{E} = \dfrac {t}{L}\)
\(\Rightarrow \;\;ln \dfrac {\mathcal{E} -IR}{\mathcal{E}}= \dfrac {-Rt}{L}\)
\(\Rightarrow \;\; \dfrac {\mathcal{E} -IR}{\mathcal{E}} = e^{-Rt/L}\)
\(\Rightarrow \mathcal{E} - IR = \mathcal{E} \;e^{-t(L/R)}\)
\(\Rightarrow\;\;I = \dfrac {\mathcal{E} }{R}[1-e^{-t/(L/R)}]\)
\(I = \dfrac {\mathcal{E}}{R}[1-e^{-t/\tau}]\)
\(I = \dfrac {\mathcal{E}}{R}[1-e^{-t/t}]\)
\(I = \dfrac {\mathcal{E}}{R}[1-e^{-1}]\)
\(I = \dfrac {\mathcal{E}}{R}[1-.37]\)
\(I = \dfrac {\mathcal{E}}{R} × 0.632\)
\( I = 63.2 \) % of its final value
\(I= \dfrac{\mathcal{E}}{R} (1 - e ^{-tR/L})\)
\( I = \dfrac{\mathcal{E}}{R} (1 - e ^{-t/\tau})\)
where \(\tau = \dfrac{L}{R}\)
\(U = \dfrac{1}{2} LI^2\)
\(U_t = \dfrac{1}{2}L \left(\dfrac{\mathcal{E}}{R}\right)^2\;\;\left(1-e^{t/\tau}\right)^2\)
A 23.62 mJ
B 28.57 mJ
C 50.72 mJ
D 20.46 mJ
Case 1. When battery is connected
\(\mathcal{E} - IR -L \dfrac {dI}{dt}= 0\)
Rearranging the equation
\(\mathcal{E} - IR = L\dfrac {dI}{dt}\)
\(\dfrac {dI}{\mathcal{E}- IR}\;\;=\;\;\dfrac {dt}{L}\)
\(\int\limits^{I}_0\;\dfrac {dI}{\mathcal{E}-IR}\;=\; \int\limits^{t}_0 \dfrac {dt}{L}\)
Let \(x = \mathcal{E}- IR\), then \(dx=-RdI\)
\(\dfrac {-1}{R}\;\;\int\limits^{\mathcal{E}-IR}_\mathcal{E}\dfrac {dx}{x}\;\;=\;\;\dfrac {1}{L}\int\limits^{t}_0dt\)
\(\Rightarrow \;\;\dfrac {-1}{R}[lnx]^{\mathcal{E}-IR}_\mathcal{E} = \dfrac {t}{L}\)
\(\Rightarrow \;\;ln \dfrac {\mathcal{E} -IR}{\mathcal{E}}= \dfrac {-Rt}{L}\)
\(\Rightarrow \;\; \dfrac {\mathcal{E} -IR}{\mathcal{E}} = e^{-Rt/L}\)
\(\Rightarrow \mathcal{E} - IR = \mathcal{E}\;e^{-t(L/R)}\)
\(\Rightarrow\;\;I = \dfrac {\mathcal{E} }{R}[1-e^{-t/(L/R)}]\)
\(I = \dfrac {\mathcal{E}}{R}[1-e^{-t/\tau}]\)
\(I = \dfrac {\mathcal{E}}{R}[1-e^{-t/t}]\)
\(I = \dfrac{\mathcal{E}}{R}[1-e^{-1}]\)
\(I = \dfrac {\mathcal{E}}{R}[1-.37]\)
\(I = \dfrac {\mathcal{E}}{R} × 0.632\)
\(I\) = 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 \(\tau = \dfrac{L}{R}\)
\(U = \dfrac{1}{2} LI^2\)
\(U_t = \dfrac{1}{2}L \left(\dfrac{\mathcal{E}}{R}\right)^2\;\;\left(1-e^{-t/\tau}\right)^2\)
Case 2. When battery is disconnected
\(IR + L\dfrac {dI}{dt} = 0\)
\(L\dfrac {dI}{dt} = -IR\)
\(\dfrac {dI}{I} = \dfrac {-R}{L}\;\;dt\)
\(\Rightarrow \;\;\int\limits^{I}_{\mathcal{E}/R} \;\dfrac {dI}{I}\;\;=\;\;\dfrac {-R}{L}\;\int \limits^t_0 dt\)
\(\Rightarrow\;\;[lnI]^{I}_{\mathcal{E}/R}\;\;=\;\;\dfrac {-R}{L}t\)
\(\Rightarrow\;\;ln \left(\dfrac {I}{\mathcal{E}/R}\right) = \dfrac {-R}{L}t\)
\(\Rightarrow \dfrac {I}{\mathcal{E}/R}\;\;e^{-Rt/L}\)
\(I = \dfrac {\mathcal{E}}{R}\;\;e^{-t/(L/_R)}\)
\(I = I_0\;\;e^{-t/(L/_R)}\)
Time constant \(\tau = \dfrac {L}{R}\)
So, \(I= I_0 \;\;e^{-t/z}\)
\(I = \dfrac{\mathcal{E}}{R} e^{-t/\tau}\)
\(U = \dfrac{1}{2}\; LI^2\)
\(U_{t_0} = \dfrac{1}{2}\; L \left(\dfrac{\mathcal{E}}{R}e^{-t/\tau}\right)^2\)
A \(2 (1-e^{-3t})^2\;mJ\)
B \(3 (1-e^{-4t})^2\;mJ\)
C \(25 (1-e^{-10^3t})^2\;mJ\)
D \(50 (1-e^{-10^3t})^2\;mJ\)