Transient Analysis in SDomain MCQ Quiz  Objective Question with Answer for Transient Analysis in SDomain  Download Free PDF
Last updated on Aug 12, 2024
Latest Transient Analysis in SDomain MCQ Objective Questions
Transient Analysis in SDomain Question 1:
What is the value of v_{0}(t) for the circuit shown in the figure, assuming zero initial conditions?
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 1 Detailed Solution
Given that, R=\(4\Omega\), L = 1H, C = \(\frac{1}{4} F\)
The response for a series RLC circuit with source will be, V(t) = V_{ss} + (A_{1} + A_{2}t)e^{at} for a = W_{0.}
\(\therefore\) a = \(R\over2L \)= \(4\over 2\times1\)= 2
\(\omega\)_{0} = \(1\over\sqrt LC\) = \(1\over \sqrt1\times \frac {1}{4}\)= 2.
a = \(\omega\)_{0 }; system is critically damped.
At t = \(\propto\); inductor is short circuit and capacitor is open circuit.
V(t) = Vss + (A1 + A2t)eat
= 40 + A_{1}e^{2t }+ A_{2}te^{2t}
At t = 0, voltage across capacitor is zero V(t) = 0.
\(\therefore\) 0 = 40 + A_{1}e^{2} + A_{2}te^{2}
^{\(\therefore\)} A_{1} + A_{2} = 40
\(\frac{dv(0)}{dt}\)= 0 + 2 A_{1}e^{2t}  2A_{2}te^{2t }+ A_{2}e^{2t}
0 = 2 \(\times\)(40)  0 + A_{2}
\(\therefore\) A_{2 }= 80
Here, option 2 is correct.
Transient Analysis in SDomain Question 2:
An RLC series circuit has R = 4 Ω, L = 2 H and C = 2 F. What type of transient current response is offered by the circuit for step function voltage input?
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 2 Detailed Solution
Concept:
The characteristic equation defining the behavior of a sourcefree series RLC circuit is given by:
\({s^2} + 2α s + ω _c^2 = 0\)
Where:
α = damping factor given by
\(α = \frac{R}{{2L}}\)
ω0 = Undamped natural frequency or resonant frequency is given by:
\({ω _0} = \frac{1}{{\sqrt {LC} }}\)
Analysis:
The roots of the characteristic equation will be:
\(s = \frac{{  2α \pm \sqrt {{{\left( {2α } \right)}^2}  4\left( 1 \right)\left( {ω _c^2} \right)} }}{2}\)
\(s = \frac{{  2α \pm \sqrt {4{α ^2}  4ω _c^2} }}{2}\)
\(s = \frac{{  2α \pm \sqrt {{α ^2}  ω _c^2} }}{2}\)
\(s =  α \pm \sqrt {{α ^2}  ω _0^2} \)
\({s_1} =  α + \sqrt {{α ^2}  ω _0^2} \)
\({s_2} =  α  \sqrt {{α ^2}  ω _0^2} \)
From the above two equations, we can have three types of solutions:
1) If α > ω0, we have the overdamped case
2) If α = ω0, we have the critically damped case
3) If α < ω0, we have the underdamped case.
Calculation:
With R = 4 Ω, L = 2 H, and C = 2 F, we get:
\(α = \frac{R}{{2L}} = \frac{4}{2 \times 2}\)
α = 1
Also:
\({ω _0} = \frac{1}{{\sqrt {LC} }} = \frac{1}{\sqrt{2\times2}}\)
ω0 = 0.5 rad/sec
Since α > ω0, we conclude that the resulting current response will be overdamped.
Transient Analysis in SDomain Question 3:
Here:
R = 5 Ω
C_{1} = C_{2} = 10 μF
V_{1} = 10 V
V_{2} = 5 V
Which of the following is/are correct?
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 3 Detailed Solution
Concept:
First order RC network:
Voltage or current can be calculated from the formula :
f(t) = f(∞) + [f(0^{+}) f(∞)]e^{t/τ}
Where;
f → It can be a function of voltage or current
f(0+) → Value of the function just after the operation of the switch
f(∞) → Value of the function just after the operation of the switch and at t = ∞
τ → Time constant
For RC network;
τ = RC
R → Equivalent resistance across the capacitor
Note:
→ Capacitor does not allow sudden change of voltage [V_{C}(0^{}) = VC(0^{+})]
→ Capacitor behaves as an open circuit at a steady state.
Calculation:
Before the operation of the switch (t = 0^{} )
VC1(0) = VC1(0+) = 10 V
VC2(0) = VC2(0+) = 5 V
After the operation of the switch (t = 0^{+} )
VC1(0) = VC1(0+) = 10 V
VC2(0) = VC2(0+) = 5 V
\(i(0^{+}) = \frac{V_{C1}(0^{+}) V_{C2}(0^{+})}{R} = \frac{105}{5} =1\,A\)
at t =∞ → capacitors will be open circuit
i(∞) = 0 A
τ = RC
R= 5 Ω
\(C_{eq} = \frac{C_{1} × C_{2}}{C_{1} + C_{2}} = 5 \,\, μ F\)
τ = RC = 5 × 5μF = 25 × 10^{6} sec
\(i(t) = i(\infty) + [i(0^{+}  i(\infty)]e^{t/RC}\)
\(\Rightarrow i(t) = 0 + [1  0]e^{t/RC}\)
\(\therefore i(t) = e^{t/RC} = e^{t\times 10^{6}/25} \)
Voltage across capacitors:
\(V_{C1}(t) = \frac{1}{C_{1}}\int_{0}^{t} I_{C_{1}}dt\)
\(\Rightarrow V_{C1}(t) = \frac{1}{C_{1}}\int_{0}^{t} i(t)dt \,\,[\because I_{C_{1}} = i(t) ] \)
\(\therefore V_{C1}(t) = \frac{1}{C_{1}}\int_{0}^{t} e^{t/\tau} dt = \frac{\tau}{C_{1}}(1e^{t\times 10^{6}/25})= 2.5(1e^{t\times 10^{6}/25}) V\)
\(V_{C_{2}}(t) = \frac{1}{C_{2}}\int_{0}^{t} I_{C_{2}}dt\)
\(\Rightarrow V_{C_{2}}(t) = \frac{1}{C_{2}}\int_{0}^{t} i(t)dt \,\,[\because I_{C_{2}} = i(t) ] \)
\(\therefore V_{C_{2}}(t) = \frac{1}{C_{2}}\int_{0}^{t} e^{t/\tau} dt = \frac{\tau}{C_{2}}(1e^{t\times 10^{6}/25}) \\= 2.5(1e^{t\times 10^{6}/25}) V\)
Transient Analysis in SDomain Question 4:
In the circuit shown below, the switch S is closed at t = 0 and opened again at t = π sec. Prior to closing the switch at t = 0, V_{c1} = 10 V while L and C_{2} do not have any stored energy. Find the voltages V_{c1} and V_{c2} at t = π sec. C_{1} and C_{2} = 1 F, L = 2 H
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 4 Detailed Solution
Concept:
The transient equation is not applicable for the circuit with an order greater than 1. For such circuits, Laplace transform is used.
\(2sI(s) + {I(s) \over s} + {I(s)\over s} = {10\over s}\)
\(I(s) ( {2s+ {2\over s} }) = {10\over s}\)
\(I(s)= {5 \over s^2+1}\)
i(t) = 5 sint
\(\therefore v_{c1}(t) = 10 \space + {1\over C_1}\int_{0}^{\pi}5sin\space t\space dt\)
\( v_{c1}(t) = 10 \space + (5cost)_{o}^{\pi}\)
\( v_{c1}(t) = 0\space V\)
\(\therefore v_{c2}(t) = {1\over C_2}\int_{0}^{\pi}5sin\space t\space dt\)
\( v_{c2}(t) = (5cost)_{o}^{\pi}\)
\( v_{c2}(t) = 10\space V\)
Transient Analysis in SDomain Question 5:
Which one of the following techniques can be used to analyze and design the effect of loop gain upon the system's transient response and stability?
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 5 Detailed Solution
The root locus graphically displays both transient response and stability information.
The locus can be sketched quickly to get a general idea of the changes in the transient response generated by changes in gain K.
The root locus typically allows us to choose the proper loop gain to meet a transient response specification.
The root locus technique can be used to analyze and design the effect of loop gain upon the system's transient response and stability.
Top Transient Analysis in SDomain MCQ Objective Questions
In the sdomain circuit analysis, the circuit natural response is contributed by
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 6 Detailed Solution
Download Solution PDFForced response:
 The response of a network with a source present in it is called forced response (F.R) and it gives a steadystate response.
 This response is independent of the nature of the passive element, but it depends entirely on the type of input.
 This part of the solution is found out by finding the particular integral part (P.I) of the differential equation
Natural response:
 The response of a network without a source in it is called a natural response (NR) and it gives the transient response
 This response depends on the nature of the passive element and it is independent of the type of input.
 This part of the response is found and by solving the complementary function in the differential equation.
Total Response (T.R) = 
Natural Response (N.R) + 

Forced Response (F.R) 

↓ 

↓ 

Zero input Response 

Zero state Response 

↓ 

↓ 

Transient Response 

Steadystate Response 

↓ 

↓ 

Complementary function (C.F) 

Particular solution (P.S) 
This clearly indicates,
The natural response clearly indicates System function poles,
The forced response indicates forcing function poles,
The total response of the System indicates system poles, forcing function poles, and zeros respectively.
A coil has a resistance of 10 Ω and an inductance of 1 H, what will be the value of current 0.1 second after switching on to a 50 volts d.c. supply?
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 7 Detailed Solution
Download Solution PDFConcept:
 When a series RL network is connected across a DC supply for t>0, it is a charging circuit or source network.
 Whenever there is the 1^{st} order charging or discharging circuit, the transient equation is applicable.
 1st order means no. of the energystoring element in the circuit must be one i.e either capacitor or inductor must be present.
Transient equation for 1st order RL circuit is as follows:
_{IL}(t) = I_{L}(∞) + { IL(0^{+})  IL(∞) }e^{t/τ }
 where, I_{L}(∞) = steady state current across inductor
 _{IL}(0^{+}) = I_{L}(0^{}) = initial current across inductor
 t = time period
 τ = time constant
 The time constant of the RL circuit (τ) is given by:
τ = \( {L \over R}\)
 where L and R are inductance and resistance respectively
Calculation:
Step 1: When the switch is not closed at t = 0^{}
At t = 0^{} the switch is not closed, therefore the 50 V DC source can not flow the current in the circuit
So, the initial current across the inductor is zero.
IL(0+) = IL(0) = 0 A
Step 2: When the switch gets closed at t > 0
At the steadystate, the inductor always acts as a short circuit.
IL(∞) = \( {V \over R}\)
IL(∞) = \( {50 \over 10}\)
IL(∞) = 5 A
Step 3: Calculation of time constant
τ = \( {L \over R}\)τ = \( {1 \over 10}\)
τ = 0.1 sec
IL(t) = IL(∞) + { IL(0+)  IL(∞) }et/τ
IL(t) = 5 + {0  5}e^{t/τ}
IL(t) = 5(1  e^{t/τ})
IL(t) at t= 0.1 sec is:
IL(t) = 5(1  e^{0.1/0.1})
I_{L}(t) = 5(1  e^{1})
∵ e^{1} = 0.367
IL(t) = 5(1  0.367)
IL(t) = 3.16 A
Key Points
 For the 2^{nd} order network, the transient equations are not valid.
 In such cases, the Laplace Transform approach is used for the evaluation of current and voltage across the inductor and capacitor.
In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated. The switch is closed at time t = 0. The current i(t) at a time t after the switch is closed is
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 8 Detailed Solution
Download Solution PDFQ = 2.5 mC
\(V_{initial} = \frac{2.5 × 10^{3}C}{50 × 10^{6}F}\)
V_{initial} = 50 V
Thus the net voltage = 100 + 50 = 150 V
The initial current at t = 0^{+} will be:
I = 150/10 = 15 A
The current at any time 't' will now be:
\(i(t) = \frac{150}{50}\) exp(2 × 10^{3}t) A = 15 exp(2 × 10^{3}t)A
The switch in the circuit in the figure is in position P for long time and then moved to position Q at time t = 0.
The value of \(\frac{{dv\left( t \right)}}{{dt}}\) at t = 0^{+} is
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 9 Detailed Solution
Download Solution PDFConcept:
Under steadystate:
 When a capacitor, is present with a D.C supply, it behaves as an open circuit.
 When an inductor is present with D.C supply, it behaves as a short circuit
At t = 0+ :
Inductor replace with current source IL (0+) = IL (0)
The inductor does NOT allow the sudden change in current”
The capacitor is replaced with the voltage source
Vc (0+) = Vc (0)
The capacitor does NOT allow the sudden change in voltage.
Calculation:
At t = 0^{} → switch is at P position network is in steadystate
Capacitor → open circuit
Inductor → S.C.
i_{L}(0^{}) = 20/20 = 1 mA
i_{L}(0^{+}) = iL(0) = 1 mA
V_{c} (0^{}) = 10 × 1 = 10 V
V_{c} (0^{+}) = Vc (0) = 10 V
At t = 0^{+} → switch is at Q position.
Applying KCL at node A
10/5 + i_{c}(0^{+}) + 1 = 0
i_{c} (0^{+}) = 3 mA
\(C{\left. {\frac{{d{V_c}\left( t \right)}}{{dt}}} \right_{t = {0^ + }}} =  3mA\)
∵ \({i_c} = C\frac{{d{V_c}}}{{dt}}\)
\({\left. {\frac{{d{V_c}\left( t \right)}}{{dt}}} \right_{t = {0^ + }}} =  3V/sec\)
The circuit in the figure contains a current source driving a load having an inductor and a resistor in series, with a shunt capacitor across the load. The ammeter is assumed to have zero resistance. The switch is closed at time t = 0.
Initially, when the switch is open, the capacitor is discharged and the ammeter reads zero ampere. After the switch is closed, the ammeter reading keeps fluctuating for some time till it settles to a final steady value. The maximum ammeter reading that one will observe after the switch is closed (rounded off to two decimal places) is _______ A.
Answer (Detailed Solution Below) 1.40  1.50
Transient Analysis in SDomain Question 10 Detailed Solution
Download Solution PDFConcept:
Capacitor do not allow sudden change in voltage i.e. V_{c}(0^{}) = V_{c}(0^{+})
Inductors do not allow a sudden change in current.
i.e. i_{L}(0^{}) = i_{L}(0^{+})
Calculation:
Since at t < 0 … switch is open
∴ i_{L} (0^{}) = 0 Amp
V_{L} (0^{}) = 0 V
As we know, when Inductor and capacitor both present, we use the Laplace domain concept:
Now, the ckt for t > 0 in the Laplace domain
Apply nodal concept:
\(\frac{1}{s} = \frac{{{V_c}\left( s \right)}}{{\frac{1}{{sC}}}} + \frac{{{V_c}\left( s \right)}}{{R + sL}} = 0\)
\({V_c}\left( s \right) = \frac{1}{s} \cdot \frac{{R + sL}}{{{s^2}LC + RCs + 1}}\)
∵ we need, \({I_L}\left( s \right) = \frac{{{V_c}\left( s \right)}}{{R + sL}}\)
\({I_L}\left( s \right) = \frac{{1/LC}}{{s\left( {{s^2} + \frac{R}{L}s + \frac{1}{{LC}}} \right)}}\)
From denominator:
\({s^2} + \frac{R}{L}s + \frac{1}{{LC}} = {s^2} + 2\xi {\omega _n}s + \omega _n^2\)
By comparing with std. 2nd order equation
\({\omega _n} = \frac{1}{{\sqrt {LC} }}\;\;\& \;2\xi {\omega _n} = \frac{R}{L}\)
∴ ξ = 0.25
Now
First maxima or peak overshoot \( = {e^{\frac{{  \xi \pi }}{{\sqrt {1  {\xi ^2}} }}}}\)
(m_{p}) = 0.4443
Steadystate value of \({I_L}\left( s \right) = \mathop {\lim }\limits_{s \to 0} s \cdot \frac{{\frac{1}{{LC}}}}{{s\left( {{s^2} + \frac{R}{L}s + \frac{1}{{LC}}} \right)}}\)
= 1
∴ The maximum ammeter reading just after the switch closed is
= 1.444 Amp
In the circuit shown in the figure, the switch is closed at time t = 0, while the capacitor is initially charged to  5 V (i.e. vc(0) = 5V)
The time after which the voltage across the capacitor becomes zero (rounded off to three decimal places) is ______ ms.
Answer (Detailed Solution Below) 0.132  0.146
Transient Analysis in SDomain Question 11 Detailed Solution
Download Solution PDFConcept:
Voltage across capacitor is given by
V_{c}(t) = V_{c}(∞) + [V_{c} (0^{+}) – V_{c} (∞)] e^{t/τ }
Analysis:
Given V_{c}(0^{}) = 5 V
V_{c}(0^{}) = V_{c} (0^{+}) = 5 V
At t → ∞ capacitor will act as open circuit
Applying KCL at node A:
\(\frac{{{V_c}\left( \infty \right)  5}}{{250}} + \frac{{V_R}}{{500}} + \frac{{{V_c}\left( \infty \right)}}{{250}} = 0\)
2 V_{c} (∞) – 10 + V_{R} + 2 V_{c} (∞) = 0 …i)
\(V_R = \left\{ {\left( {\frac{{{V_c}\left( \infty \right)  5}}{{250}}} \right) \times 250} \right\}\times  1\)
V_{R} = V_{c} (∞) + 5 …ii)
Putting V_{R} from equation ii) to equation i)
2 V_{c} (∞) – 10 – V_{c} (∞) + 5 + 2 V_{c} (∞) = 0
3 V_{c} (∞) = 5
\({V_c}\left( \infty \right) = \frac{5}{3}\;V\)
Time constant τ = R_{eQ}.C
R_{eQ} = R_{th} across capacitor C
Applying KCL at node A
\(I = \frac{V}{{250}} + \frac{{V_R}}{{500}} + \frac{V}{{250}}\) …iii)
\(V_R =  \frac{V}{{250}} \times 250 =  V\)…iv)
Put V_{R} = V into equation iii)
\(I = \frac{V}{{250}}  \frac{V}{{500}} + \frac{V}{{250}}\)
\(I = V\left\{ {\frac{3}{{500}}} \right\}\)
\(\frac{V}{I} = \frac{{500}}{3}{\rm{\Omega }} = {R_{th}} = {R_{eQ}}\)
\(τ = {R_{eQ}}.C = \frac{{500}}{3} \times 6 \times {10^{  6}} = {10^{  4}}\;sec\)
\({V_c}\left( t \right) = {V_c}\left( \infty \right) + \left\{ {{V_c}\left( {{0^ + }} \right)  {V_c}\left( \infty \right)} \right\}{e^{  \frac{t}{τ }}}\)
\({V_c}\left( t \right) = \frac{5}{3} + \left\{ {  5  \frac{5}{3}} \right\}{e^{  \frac{t}{{{{10}^{  4}}}}}}\)
Let V_{c}(t) = 0 at t = t_{1}
\(0 = \frac{5}{3}  \frac{{20}}{3}{e^{  \frac{{{t_1}}}{{{{10}^{  4}}}}}}\)
\(\frac{{{e^{  {t_1}}}}}{{{{10}^{  4}}}} = \frac{1}{4}\)
t_{1} = 10^{– 4 }ln 4 = 0.1386 msec
In the circuit shown above, the switch is closed after a long time. The current i_{S} (0^{+}) through the switch is
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 12 Detailed Solution
Download Solution PDFA t = 0, steady state is reached
\({I_2}\left( {{0^  }} \right) = \frac{{12}}{{8 + 4}}\)
= 1 A
\({V_1}\left( {{0^  }} \right) = \left( {\frac{6}{{6 + 3}}} \right){V_L}\)
\( = \left( {\frac{6}{9}} \right) \times 4\)
\( = \frac{8}{3}V\)
\({V_2}\left( {{0^  }} \right) = \frac{3}{{6 + 3}} \times 4\)
\( = \frac{4}{3}V\)
At t = 0^{+}, we can draw the circuit as follows
\({I_1} = \frac{{\frac{8}{3}}}{4} = \frac{2}{3}A\)
I_{1} + I_{s} = 1 A
\({I_s} = \left( {1  \frac{2}{3}} \right)A\)
\( = \frac{1}{3}A\)
The switch S in the circuit shown has been closed for a long time.
It is opened at time t = 0 and remains open after that. Assume that the diode has zero reverse current and zero forward voltage drop.
The steadystate magnitude of the capacitor voltage V_{c} (in Volts) is ________.
Answer (Detailed Solution Below) 100
Transient Analysis in SDomain Question 13 Detailed Solution
Download Solution PDFConcept:
Capacitor and inductor in sdomain
Calculation:
at t = 0^{} : switch is closed
\({i_L}\left( {{0^  }} \right) = \frac{{10}}{1} = 10\;Amp\)
The diode is in R.B. since V_{p} < V_{N}
V_{C} (0^{}) = 0 V
when t ≥ 0 drawing circuit in sdomain
Applying KVL in the loop:
\(\frac{1}{{{{10}^{  5}}s}}I\left( s \right) + {10^{  3}}\;sI\left( s \right)  {10^{  2}} = 0\)
\(I\left( s \right) = \frac{{{{10}^{  2}}\;}}{{{{10}^{  3}}s\; + \;\frac{{{{10}^5}}}{s}}}\)
\(I\left( s \right) = \frac{{10s}}{{{s^2}\; + \;{{10}^8}}}\frac{s}{{{s^2}\; + \;ω _0^2}}\mathop \leftrightarrow \limits^{ILT} \cos {ω _0}t\)
\(i\left( t \right) = 10\cos \underbrace {10^4}_{wt}t\)
When i(t) is positive then only diode will be in forward bias \(\left( {0 < ω t < \frac{π }{2}} \right)\)
so diode → S.C.
When i(t) is negative then the diode will be in Reverse Bias \(\left( {\frac{π }{2} < ω t < π } \right)\)
so diode → O.C.
\(0 < ω t < \frac{π }{2}\)
\({V_c}\left( s \right) = I\left( s \right) \cdot \frac{1}{{sc}} = \frac{{10s}}{{{s^{2\;}} + \;{{10}^8}}} \times \frac{1}{{{{10}^{  5}}s}}\)
\( \Rightarrow \frac{{{{10}^6}\;}}{{{s^2}\; + \;{{10}^8}}} = 100 \times \left\{ {\frac{{{{10}^4}}}{{{s^2}\; + \;{{\left( {{{10}^4}} \right)}^2}}}} \right\}\)
\({V_c}\left( t \right) = 100\sin \underbrace {10^4}_{wt}t\)
\(at\;ω t = \frac{π }{2}\)
\({V_c}\left( t \right) = 100\sin \frac{π }{2} = 100\;V\)
When ωt > π/2, then the diode will be opencircuited and there will be no path available for capacitor discharging/charging so V_{c}(t) will remain constant at 100 V.
So steady state value of V_{c}(t) = 100 V.
The value of the current i(t) in amperes in the above circuit is
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 14 Detailed Solution
Download Solution PDFSince at t = 0, the switch is closed.
So at t > 0, the circuit will be.
Redrawing above circuit in sdomain –
\({Z_{eQ}} = \left( {s + 1} \right)\left( {1 + \frac{1}{s}} \right)\) \(\because {Z_1}\parallel {Z_2} = \frac{{{Z_1}{Z_2}}}{{{z_1} + {Z_2}}}\)
\({Z_{eQ}} = \left( {s + 1} \right)\left( {1 + \frac{1}{s}} \right) \Rightarrow \frac{{{{\left( {s + 1} \right)}^2}}}{{{s^2} + 2s + 1}}\)
\({Z_{eQ}} = \frac{{{{\left( {s + 1} \right)}^2}}}{{{{\left( {s + 1} \right)}^2}}} = 1\)
\(I\left( s \right) = \frac{{V\left( s \right)}}{{{Z_{eQ}}}} = \frac{{10/s}}{1} = \frac{{10}}{s}\)
i(t) = 10 u(t) Amp
\(u\left( t \right)\;\mathop \leftrightarrow \limits^{\;\;L.T.\;\;} \;\frac{1}{s}\)
NOTE:
Since the switch is closed t ≥ 0
So 10 V will work only after t ≥ 0 that’s why we take voltage source as 10 u(t). In Laplace or sdomain \(V\left( s \right) = \frac{{10}}{s}\)
The switch SW shown in the circuit is kept at position ‘1’ for a long duration. At t = 0+, the switch is moved to position ‘2’ Assuming V_{02} > V_{01}, the voltage V_{C}(t) across capacitor is
Answer (Detailed Solution Below)
Transient Analysis in SDomain Question 15 Detailed Solution
Download Solution PDFConcept:
I) Inductor do NOT allow sudden change in current i.e. iL(0) = iL(0+)
II) Capacitor do NOT allow sudden change in voltage i.e. VL(0) = VL(0+)
Considering the initial conditions, Laplace transform of capacitor and inductor will be replaced as follows
Calculation:
Given circuit diagram
When the switch is at position 1 for a long time (time t < 0):
The capacitor will be completely charged to V_{01} and it will become an open circuit
When the position of switch is changed from position 1 to position 2 at time t = 0^{+}
Apply KVL in the loop and let I(s) be the current flowing in the circuit
\(\frac{{  {V_{02}}}}{S} + 2I\left( s \right)R + \left[ {\frac{1}{{SC}}} \right]\left[ {I\left( s \right)} \right] + \frac{{{V_{01}}}}{S} = 0\)
\( I\left( s \right) = \frac{{\frac{{{V_{01}}}}{S}  \frac{{{V_{02}}}}{S}}}{{2R + \frac{1}{{SC}}}} = \frac{{{V_{01}}  {V_{02}}}}{{S\left( {2R + \frac{1}{{SC}}} \right)}}\\ \)
The voltage across the capacitor considering the initial conditions will be V_{c}(s)
\({V_c} = \frac{{  {V_{01}}}}{S} + \frac{1}{{SC}}I\left( s \right) \)
Substitute the value of I(s) which is obtained above
\( = \frac{{  {V_{01}}}}{S} + \frac{1}{{SC}}\left[ {\frac{{{V_{01}}  {V_{02}}}}{{SR\left( {2 + \frac{1}{{SRC}}} \right)}}} \right]\)
\(\begin{array}{l} \frac{{  {V_{01}}}}{S} + \frac{1}{{SC}}\left[ {\frac{{\left( {{V_{01}}  {V_{02}}} \right)}}{{SR\frac{{\left( {2RCS + 1} \right)}}{{SRC}}}}} \right]\\ = \frac{{  {V_{01}}}}{S} + \frac{1}{{SC}}\left[ {\frac{{{V_{01}}  {V_{02}}}}{{2RCS + 1}}} \right] \end{array}\)
\(\begin{array}{l} = \frac{{  {V_{01}}}}{S} + \frac{{{V_{01}}  {V_{02}}}}{{2RC}}\left[ {\frac{1}{{S\left( {S + \frac{1}{{RC}}} \right)}}} \right]\\ = \frac{{  {V_{01}}}}{S} + \frac{{{V_{01}}  {V_{02}}}}{{2RC}}\left[ {\frac{1}{S}  \frac{1}{{S + \frac{1}{{2RC}}}}} \right]2RC\\ =  {v_{01}} + \left( {{V_{01}}  {V_{02}}} \right)\left[ {1  {e^{  \frac{t}{{2RC}}}}} \right] \end{array}\)
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