If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Inductance L=20 mH=20×10−3H
Capacitance of the capacitor
C=50 μF=50×10−6F
Initial charge on the capacitor,
Qi=10 mC=10×10−3C
The total energy stored initially is given by, E=q2C
E=Q2i2C
Substituting the values,
E=12×(10×10−3)250×10−6=1 J
Yes, this energy is conserved during LC oscillations.
Final Answer: 1 J, Yes
(2) Given,
Inductance L=20 mH=20×10−3H
Capacitance of the capacitor
C=50 μF=50×10−6F
Natural frequency of the circuit is given byQi=10 mC=10×10−3C
f=12π√LC
Substituting the values, we get
f=12×3.14√20×10−3×50×10−6
f=159.2 Hz
Natural angular frequency of the circuit is given by
ω=2πf
Substituting the values, we get
ω=2×3.14×159.2=1000 rad/s
Final Answer:
f=159.2 Hz
ω=1000 rad/s
(3)Step 1: Completely electrical (i.e., stored in the capacitor)
Let at any instant the energy stored is completely electrical.
The charge on the capacitor
Q=Q0cos ωt
Q=Q0cos 2πT.t ....(i)
Q is maximum when it is equal to Q0, only if
cos 2πTt=±1=cos nπ
Or
t=nT2, where n=0,1,2,3.......
Then t=0,T2,T,3T2,2T....
Thus, the energy stored is completely electrical at t=0,T2,T,3T2....
Step 2: Completely magnetic (i.e., stored in the inductor)
Let at any instant, the energy stored is completely magnetic as when the electrical energy across the capacitor is zero.
Q=0
Q0cos2TT=0
∴cos2TT.t=0
t=nπ2 or t=nπ4, where n=1,3,5....
It happens if t=T4,3T4,5T4....
Thus, the energy stored is completely magnetic at
t=T4,3T4,5T4....
Final Answer: (i)t=0,T2,T,3T2....
(ii)t=T4,3T4,5T4....
(4) Let Q be the charge on the capacitor when total energy is equally shared between the capacitor and the inductor at time t.
When total energy is equally shared between the inductor and capacitor,
Energy stored in the capacitor =12 (maximum energy )
=12Q2C=12(12Q20C)
⇒Q=Q0√2
Using the equation,
Q=Q0cos2πTt
⇒cos2πTt=1√2
⇒cos2πTt=cos(2n+1)π4
⇒t=(2n+1)T8
⇒t=T8,3T8,5T8.... etc
Final Answer : t=T8,3T8,5T8....
(5) Given,
Inductance L=20 mH=20×10−3H
Capacitance of the capacitor
C=50 μF=50×10−6F
Initial charge on the capacitor,
Qi=10 mC=10×10−3C
The total energy stored initially is given by, E=q22C
E=Q2i2C
Substituting the values,
E=12×(10×10−3)250×10−6=1 J
R damps out the LC oscillations eventually. So if a resistor is inserted in the circuit, then total energy is eventually dissipated as heat.
Thus, the energy dissipated as heat is 1 J
Final Answer: 1 J