100.96 un ntil Sunday, April 2 LC Circuit 2 A circuit is constructed with a resi
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Question
100.96 un ntil Sunday, April 2 LC Circuit 2 A circuit is constructed with a resistor, two inductors, one capacitor, one battery and a switch as shown. The value of the resistance is R1 = 416 The values for the inductances are: L1 = 384 mH and L2 199 mH. The capacitance is C 180 F and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a sign 2 11 The switch has been closed for a long time when at time t = 0, the switch is opened. What is UL1(0), the magnitude of the energy stored in inductor L1 just after the switch is opened? 2) What is 0, the resonant frequency of the circuit just after the switch is opened? radians/s Suemit What is Qmax, the magnitude of the maximum charge on the capacitor after the switch is opened? 4) What is Q(t1), the charge on the capacitor at time t-t1 = 4.2 ms. Q(t1) is defined to be positive if V(a) V(b) is positive 5) what is t2, the first time after the switch is opened that the energy stored in the capacitor is a maximum? 6) what is the total energy stored in the inductors plus the capacitor at time t = t2?Explanation / Answer
a) In steady state condition, capacitor behaves as open circuit and inductor as normal wire.
So, Current, I (t=0) = V/R = 12/416 = 0.028846 A (This will be same through L1 and L2 as capacitor is open)
Energy in L1 = 0.5*L1 * I * I = 0.16*10^-3 J
b) 2 inductor will have total inductance = 384+199 = 583 mH
Capcitance = 180 uF
Resonant frequency = sqrt(1/LC) = 97.61 rad/sec
c) Conserving energy. (When all the energy of inductor gets stored into capacitor, it will have maximum charge)
0.5* L1 * I * I + 0.5* L2 * I * I = Q * Q / 2C ; Q is Q max
Q max = 295.5 uC
d) V across inductors are defined as V(t) = -L* dI/dt
I(t) = Io cos wt
V(t) = L*Io*w* sin wt
Now this is applied across capacitor (L is the equivalent resistance)
Q(t) = C*V(t) = LC*Io*w* sin wt = 583*10^-3*180*10^-6*0.028846 *97.61 sin (97.61*4.2*10^-3) ; {Angle is in radian}
Q (t=4.2 ms) = 117.786 uC
e) Maximum Q will be when sin is maximum
wt = pi/2
t = pi/2w = 16.09 ms
f) Total energy will always remain the same. (Many ways to calculate this, I am using the quickest and simpliest)
E total = Qmax*Qmax / 2*C = 0.2426* 10^-3 J
Other ways are to put t=4.2 in the I(t) equation and and calculate Inductor energy Einductor = 0.5*Leq*I(t)*I(t)
and Ecapacitor = Q(t)*Q(t)/2C and add them. You will get same answer.
comment for doubts.
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