Transients And Steady State Response Practice Questions
Generate GATE-level questions on Transient and Steady-state response. Focus on:
1. First-order RL and RC circuits.
2. Second-order RLC circuits (overdamped, underdamped, critically damped).
3. Step, impulse, and sinusoidal response.
30 questions · 10 PYQs · 0 AI practice · GATE EE 2027
Q21GATE 2006MCQ
In the circuit shown in the figure, the current source I=1A, the voltage source V=5 V, R1=R2=R3=1Ω,L1=L2=L3=1H,C1=C2=1F The currents (in A) through R3 and through the voltage source V respectively will be
In steady state, inductor behaves as short-circuit and capacitor behaves as open-circuit Voltage across, R3=V=5V Current through, R3=I1=R3V=15=5A Apply KCL, −I+i1−IV=0 Current through voltage source =IV=I1−1=5−1=4A
Q22GATE 2006
MCQ
An ideal capacitor is charged to a voltage V0 and connected at t=0 across an ideal inductor L. (The circuit now consists of a capacitor and inductor alone). If we let ω0=LC1 , the voltage across the capacitor at time t>0 is given by
Voltage across capacitor will discharge through inductor upto voltage across the capcitor becomes zero. During thos period , electrostatic energy stored in capacitor is transferred into magnetic energy which is stored in inductor. Now, inductor will start charging capacitor, magnetic energy in inductor is converted into electostatic energy in capacitor. Expression for Vc(t) can be obtained in s-domain. As capacitor is charged initially to voltage V0 , then representation of capacitor in s-domain As current through the inductor is zero at t=0, then The circuit at t>0 [latex]I(s)=\frac{V_0/s}{\frac{1}{sC}+sL}=\frac{V_0}{s^2L+\frac{1}{C}} I(s)=\frac{V_0}{L}\left ( \frac{1}{s^{2+}\frac{1}{LC}} \right ) Voltage across capacitor =voltage across inductor =V(s),V(s)=I(s)×(sL)=LV0(s2+LC11)×(sL)=V0(s2+LC1s) As, ω0=LC1V(s)=V0(s2+ω02s) Voltage across the capacitor =V(t)=L−1[V(s)]=L−1(s2+ω02V0s)V(t)=V0cosω0t where, ω0=LC1
Q23GATE 2005MCQ
A coil of inductance 10 H and resistance 40 Ω is connected as shown in the figure. After the switch S has been in contact ith point 1 for a very long time, it is moved to point 2 at, t=0. For the value of R=40 Ω , the time taken for 95% of the stored energy to be dissipated is close to
The circuit (in s-domain) I(s)=(20+40+40)+10s20=10s+10020=s+102i(t)=L−1[I(s)]=l−1[s+102]=2e−10ti(t)=iL(0+)e−LRefft=2e−10(20+40+40)t=2e−10t Initial stored energy in inductor W0=21LiL2(0+)=21×10×22=20Joules Remaining energy in inductor W1=0.05W0=0.05×20=1Joules21Li12=121×10×i12=1i1=51=0.4472Ai=2e−10T Let at t=T , current decrease to i10.4472=2e−10TT≈0.15sec
Q24GATE 2005MCQ
The circuit shown in the figure is in steady state, when the switch is closed at t = 0. Assuming that the inductance is ideal, the current through the inductor at t = 0+ equals
Before closing the switch, at t=0− , the circuit is in steady state. So, inductor behaves as short-circuit iL(0−)=1010=1A After closing the switch, at t=0+ Current through inductor can not change abruptly. ∴iL(0+)=iL(0−)=1A
Q25GATE 2005MCQ
In the figure given, the initial capacitor voltage is zero. The switch is closed at t = 0. The final steady-state voltage across the capacitor is
At (t→0+) . The capacitor act as short-circuit. At (t→∞) , the capacitor will become open circuit. Therefore, voltage across capacitor =10+1020×10=10V
Q26GATE 2004MCQ
In figure, the capacitor initially has a charge of 10 Coulomb. The current in the circuit one second after the switch S is closed will be
Before closing the switch, the circuit was not energized, therefore, current through inductor and voltage across capacitor are zero. After closing the switch, at t=0+ inductor acts as open-circuit and capacitor acts as shortc ircuit. Equivalent circuit at t=0+I=3+4∣∣410=2AVL(0+)I×(4∣∣4)=2×2=4V
Q28GATE 2002MCQ
Consider the circuit shown in figure. If the frequency of the source is 50 Hz, then a value of t0 which results in a transient free response is
For transient free response, tan(ωt0)=RωLtan(2π×50×t0)=52π×50×0.012π×50×t0=tan−1(5π)=32.14∘=0.561radt0=100π0.561=1.786ms
Q29GATE 2002MCQ
An 11 V pulse of 10 μs duration is applied to the circuit shown in figure. Assuming that the capacitor is completely discharged prior to applying the pulse, the peak value of the capacitor voltage is
Vc(t)=Vc(∞)−[Vc(∞)−Vc(0)]e−t/RCVc(peak)=10−(10−0)e1110×103×11×10−910×10−6=10(1−e−1)=6.32V(WhereRnet=10∣∣1=1110kΩ)Vc(∞)=11×10+110=10V and ∵ pulse of duration 10μs is applies. Hence, capacitor charges till 10μs and then starts discharging, so Vc will be maximum at t=10μs .
Q30GATE 2001MCQ
A unit step voltage is applied at t = 0 to a series RL circuit with zero initial conditions.
At t=0+ inductor works as open circuit. Hence, complete source voltage drops across it and consequently, current through the resistor R is zero. Hence, voltage across the resistor at t=0+ is zero. And further with time it rises accroding to VR(t)=(1−e−Rt/L)u(t) .
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