Test: The RLC Circuits
20 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Test: The RLC Circuits
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The natural response of an RLC circuit is described by the differential equation
The v(t) is
S2 + 2s + 1 = 0 ⇒ s = -1, -1,
v(t) = (A1 + A2t)e-t
v(0) = 10V, 
A1 = A2 = 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 i1(0) = i2(0) =11A
i1 (1s) = ?
In differential equation putting t = 0 and sovling
Circuit is shown in fig. P.1.6. Initial conditions are (0)i1=i2(0)=11A
i2 (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 vt (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
A2 = -4
In the circuit of fig.the switch is opened at t = 0 after long time. The current iL(t) for t > 0 is
In the circuit shown in fig.all initial condition are zero.
If is (t) = 1 A, then the inductor current iL(t) is
In the circuit shown in fig. all initialcondition are zero
If is(t) = 0.5t A, then iL(t) is
Trying iL (t)= At+ B,
In the circuit of fig. switch is moved from position a to b at t = 0. The iL(t) for t > 0 is
α = ωo 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
iL(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 vo (t) for t >0 is
α = Wo, 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:
- An RLC circuit is an electrical circuit consisting of an inductor (L), Capacitor (C), Resistor (R) it can be connected either parallel or series.
- When the LCR circuit is set to resonate (XL = XC), the resonant frequency is expressed as
⇒f=12π1LC
- Quality factor is
⇒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 × 10-3 H
From the above discussion,
⇒Q=ω0LR=1RLC
⇒Q=11027×10−330××10−6
⇒Q=3×1010=3
- Hence option (1) is correct.
In the circuit of fig. a steady state has been established before switch closed. The i(t) for t > 0 is
α = Wo, critically damped response
s = -2, -2
i(t) = (A + Bt)e-2t, A = -2
At t = 0. ⇒ B = -2
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