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The natural response of an RLC circuit is described by the differential equation
The v(t) is
S^{2} + 2s + 1 = 0 ⇒ s = 1, 1,
v(t) = (A_{1} + A_{2}t)e^{t}
v(0) = 10V,
A_{1} = A_{2} = 10
In the circuit of fig. v_{∞} = 0 for t > 0. The initial condition are v(0) = 6V and dv(0) /dt =3000 V s. The v(t) for t > 0 is
⇒
⇒
The circuit shown in fig. P1.6.5 has been open for a long time before closing at t = 0. The initial condition is v(0) = 2V. The v(t) for t > is
The characteristic equation is
After putting the values,
v(t) = Ae ^{t} + Be^{3t},
Circuit is shown in fig. Initial conditions are i_{1}(0) = i_{2}(0) =11A
i_{1} (1s) = ?
In differential equation putting t = 0 and sovling
Circuit is shown in fig. P.1.6. Initial conditions are (0)i_{1=}i_{2}(0)=11A
i_{2} (1 s)= ?
C = 1 and D = 12
The circuit shown in fig is in steady state with switch open. At t = 0 the switch is closed. Theoutput voltage v_{t} (c) for t > 0 is
The switch of the circuit shown in fig. is opened at t = 0 after long time. The v(t) , for t > 0 is
A_{2} = 4
In the circuit of fig.the switch is opened at t = 0 after long time. The current i_{L}(t) for t > 0 is
In the circuit shown in fig.all initial condition are zero.
If i_{s} (t) = 1 A, then the inductor current i_{L}(t) is
In the circuit shown in fig. all initialcondition are zero
If i_{s}(t) = 0.5t A, then i_{L}(t) is
Trying i_{L} (t)= At+ B,
In the circuit of fig. switch is moved from position a to b at t = 0. The i_{L}(t) for t > 0 is
α = ω_{o} critically damped
v(t) = 12 + (A + Bt)e^{5t}
0 = 12 + A, 150 = 5A + B A = 12, B = 90
v(t) =12 + (90t 12)e^{5t}
i_{L}(t) = 0.02(5) e^{5t}(90t 12) +0.02(90)e^{5t} = (3 9t)e^{5t}
In the circuit shown in fig. a steady state has been established before switch closed. The i(t) for t > 0 is
The switch is closed after long time in the circuit of fig. The v(t) for t > 0 is
In the circuit of fig. i(0) = 1A and v(0) = 0. The current i(t) for t > 0 is
In the circuit of fig. a steady state has been established before switch closed. The v_{o} (t) for t >0 is
α = W_{o}, So critically damped respones
s = 10, 10
Calculate the quality factor Q for an RLC circuit having R = 10 Ω, C = 30μF, and L = 27mH.
CONCEPT:
RLC CIRCUIT:
⇒f=12π1LC
⇒Q=ω0LR=1RLC
Where, XL & XC = Impedance of inductor and capacitor, L, R & C = Inductance, resistance, and capacitance, f = frequency and, ω0 = angular resonance frequency
CALCULATION:
Given that: R = 10 Ω, C = 30 μF = 30 × 10^{6} F, and L = 27mH = 27 × 103 H
From the above discussion,
⇒Q=ω0LR=1RLC
⇒Q=11027×10−330××10−6
⇒Q=3×1010=3
In the circuit of fig. a steady state has been established before switch closed. The i(t) for t > 0 is
α = W_{o}, critically damped response
s = 2, 2
i(t) = (A + Bt)e^{2t}, A = 2
At t = 0. ⇒ B = 2
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