Calculate the total energy of the system at that instant, after the switch is closed. Practice derivative of total energy in an LC circuit and find the maximum value of charge on the capacitor.
\(U_C = \dfrac {Q^{2}}{2C}\)
Q is an instantaneous charge.
\(U_L = \dfrac {1}{2} LI^{2}\)
\(U_C=\dfrac {Q^{2}}{2C}+\dfrac {1}{2} LI^{2}\)
\(U_C = \dfrac {Q^{2}}{2C}\)
Q is an instantaneous charge.
\(U_L = \dfrac {1}{2} LI^{2}\)
\(U=\dfrac {Q^{2}}{2C}+\dfrac {1}{2} LI^{2}\)
\(\dfrac{dU}{dt} = 0\)
\(\dfrac {d}{dt} \left(\dfrac {Q^{2}}{2C}\;+\;\dfrac{1}{2} LI^{2}\right ) = 0\)
\(\dfrac {Q}{C}\;\dfrac {dQ}{dt} + LI \dfrac {dI}{dt} = 0\;\;\;\;\;\;\;—(1)\)
We know that, \(I = \dfrac {dQ}{dt}\;\;\;\;\;\;—(2)\)
\(\dfrac{dI}{dt}\;=\;\dfrac{d^{2}Q}{dt^{2}}\;\;\;\;\;\;\;\;—(3)\)
Put (2) and (3) in (1),
\(\dfrac {Q}{C} + \dfrac {Ld^{2}Q}{dt^{2}}=0\)
This is a differential equation of second order whose degree is 1.
\(\dfrac {d^{2}Q}{dt^{2}}\;=\;\dfrac{-1}{LC}Q\)
Order → The highest derivative in the differential equation is known as order.
Degree → The power of the highest derivative is known as degree.
A \(\dfrac {d^{2}Q}{dt^{2}} +Q = 0\)
B \(\dfrac {d^{2}Q}{dt^{2}} +\dfrac {1}{2}\;Q = 0\)
C \(\dfrac {d^{2}Q}{dt^{2}} +2Q = 0\)
D \(\dfrac {d^{2}Q}{dt^{2}} -Q = 0\)
A \(\dfrac{100}{\pi} Hz\)
B \(\dfrac {60}{\pi} Hz\)
C \(\dfrac {50}{\pi} Hz\)
D \(\dfrac {250}{\pi}Hz\)
Maximum charge across capacitor in an LC circuit is given as,
\(Q_{max}=C\Delta V\)
where,
C = Capacitance
\(\Delta V\) = Voltage
\(U=K.E.+U_{PE}\)
For spring mass system \(U=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\)
For an LC circuit \(U=\dfrac{Q^2}{2C}+\dfrac{1}{2}LI^2\)
Case 1
Also, at t = 0, the current in an LC circuit is zero as capacitor is fully charged. So, all the energy is stored as electric potential energy in the capacitor.
Case 2
Case 3
Case 4
For LC circuit \(I=I_{max} \,\,,\,\,Q=0\)
For spring mass system \(x=0\,\,,\,\,v=v_{max}\)
Case 5
For LC circuit : \(I=0\,,Q=Q_{max}\)
For spring mass system : \(x=x_0\,,\,v=0\)
A Potential energy in capacitor
B Kinetic energy of spring
C Magnetic energy stored in inductor
D Half of potential energy stored in capacitor
\(U_C = \dfrac {Q^{2}}{2C}\)
Q is an instantaneous charge .
\(U_L = \dfrac {1}{2} LI^{2}\)
\(U_C=\dfrac {Q^{2}}{2C}+\dfrac {1}{2} LI^{2}\)
Since there is no resistance in the circuit, so no energy is radiated away from circuit and also by ignoring the electromagnetic radiation, the total energy of the system remains constant. (U = Constant)
\(\dfrac{dU}{dt} = 0\)
\(\dfrac {d}{dt} \left(\dfrac {Q^{2}}{2C}\;+\;\dfrac{1}{2} LI^{2}\right ) = 0\)
\(\dfrac {Q}{C}\;\dfrac {dQ}{dt} + LI \dfrac {dI}{dt} = 0\;\;\;\;\;\;\;—(1)\)
We know that, \(I = \dfrac {dQ}{dt}\;\;\;\;\;\;—(2)\)
\(\dfrac{dI}{dt}=\dfrac{d^{2}Q}{dt^{2}}\;\;\;\;\;\;\;\;—(3)\)
Put (2) and (3) in (1),
\(\dfrac {Q}{C} + \dfrac {Ld^{2}Q}{dt^{2}}=0\)
This is the differential equation of second order whose degree is 1.
\(\dfrac {d^{2}Q}{dt^{2}}=\dfrac {-1\,Q} {LC}\)
Order → The highest derivative in the differential equation is known as order.
Degree → The power of the highest derivative is known as degree.
\(\dfrac {d^{2}Q}{dt^{2}} = \dfrac {-1}{L} \dfrac {Q}{C}\)
where \(Q_0\) is the amplitude of charge and \(\phi\) is the phase.
The corresponding current,
\(I = \dfrac {dQ}{dt} = \dfrac{d}{dt}(Q_0\;cos\;(\omega_0t\;+\phi ))\)
\(I(t) = - \omega_0 Q_0 \;sin(\omega_0t + \phi)\)
\(I(t) = -I_0\; sin (\omega_0t+\phi)\)
Here \(\phi = 0\),
\(Q(t) = Q_0\; cos \;\omega_0t\)
Conclusion:- The charge lags the current by \(\dfrac{\pi}{2}\) phase.
A \(Q= \dfrac {I_0}{\omega }cos \left(\omega t -\dfrac {\pi}{2}\right)\)
B \(Q= \dfrac {I_0}{\omega }cos \left(\omega t -\dfrac {\pi}{4}\right)\)
C \(Q= \dfrac {I_0}{\omega }cos \left(\omega t -\dfrac {\pi}{3}\right)\)
D \(Q= \dfrac {I_0}{\omega }sin \left(\omega t -\dfrac {\pi}{2}\right)\)