A circuit is constructed with a resistor, two inductors, one capacitor, one batt
ID: 2298176 • Letter: A
Question
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 = 198 mH. The capacitance is C = 179 ?F and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.
1.)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 ?o, the resonant frequency of the circuit just after the switch is opened? (in rad/s)
3.) 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 = 4.18 ms. Q(t1) is defined to be positive if V(a)
Explanation / Answer
just before opening the switch indctor will act as a simple wire , so potentail drop across inductor is zero.
io = current in the circuit just before opening the switch = V/R1 = 12/416 = 0.0288 A
1) UL1(0) = energy stored in L1 = L1*io^2/2 = 0.384 *(0.0288)^2/2 =1.59*10^-4 J
2) W = resonance frequency = sqrt ( 1/[ (L1+L2)*c] ) = sqrt ( 1/[0.582*179*10^-6) =97.97 rad/s
F = frequency = W/(2*pi) = 15.60 Hz
3) Qo = charge on capacitor just before opning the switch = C*V =2.14*10^-3 C
Eo = Energy stored in capacitor just before opning the switch
= Qo^2/(2C)
= CV^2/2
=
Fromonservation of energy ::
(Qmax )^2/(2C) = (L1+L2)*io^2/2 + Eo
( Qmax)^2 = C*0.582 *(0.0288)^2 + C^2V^2
Qmax =2.16 *10^-3 C
4) Q(t) = charge on capacitor
= Qo cos ( wt)
=2.14 *10^-3 *cos (97.97 *t)
at t = 4.18 ms , Q(t) =2.14 *10^-3 *cos (97.97 *4.18) =1.96*10^-3 C
4)
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