Solved Problems

First-Order Circuits

Master Transient Analysis

Dr. Mithun Mondal

Demonstrative Video

Concepts of First-Order Circuits

Problem-1

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Solution-1

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\[\begin{aligned} R_{Th} & = R || (2R) = \dfrac{2}{3} R \\ \tau & = R_{Th} \cdot C = \dfrac{2}{3} RC~ \mathrm{sec.} \end{aligned}\]

Problem-2

  • In the circuit shown, the initial capacitor voltage is 4 V. Switch \(S_1\) is closed at \(t=0\). Find the charge lost by the capacitor from \(t=25~\mu s\) to \(t=100~\mu s\) ?

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Solution-2

  • \[\begin{aligned} \tau & = R\cdot C = 25\times 10^{-6}~\text{sec} \\ i(t) & = \dfrac{V}{R} e^{-t/\tau} = \dfrac{4}{5} e^{-t/\tau} \end{aligned}\]
    Current
  • \[\begin{aligned} Q & = \int_{25~\mu s}^{100~\mu s} i(t) dt = 6.99\times 10^{-6}~\mathrm{C} \end{aligned}\]
    Charge lost by the capacitor:

Problem-3

  • \(S_1\) was closed and \(S_2\) was open for a very long time. At \(t=0\), \(S_1\) is opened and \(S_2\) is closed. Determine the voltage across the capacitor at \(t=5~\mu s\) ?

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Solution-3

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Step-1
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Step-2
\[\begin{aligned} \text{At}~t&=\infty\\ V_c(\infty)&=3\dfrac{2}{2+1}=2~\mathrm{V} \end{aligned}\]
\[\begin{aligned} & \tau = RC = \dfrac{2}{3}\times 10 \times 10^{-6}\\ & V_c(t) =[V_c(0^{+})-V_c(\infty)]e^{-t/\tau}+V_c(\infty)\\ &=[1-2]e^{-t/\tau}+2 \end{aligned}\]
Step-3

Problem-4

  • The switch in the figure below has been closed for a long time ,and it opens at \(t=0\) .Find \(V(t)\) for \(t \geq 0\)

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Solution-4

\[\begin{aligned} &\mathrm{v}_{\mathrm{o}}=\mathrm{v}(0)=\frac{2}{10+2}(24)=4 \mathrm{~V} \\ &\mathrm{v}(\mathrm{t})=\mathrm{v}_{\mathrm{o}} \mathrm{e}^{-\mathrm{t} / \tau}\\ &\tau=\mathrm{RC}=40 \times 10^{-6} \times 2 \times 10^{3}=\frac{2}{25} \\ &\mathrm{v}(\mathrm{t})=4 \mathrm{e}^{-12.5 \mathrm{t}} \mathrm{V} \end{aligned}\]

Problem-5

  • For the circuit in figure below, find \(v_0(t)\) for \(t > 0\) . Determine the time necessary for the capacitor voltage to decay to one-third of its value at \(t = 0\).

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Solution-5

  • For \(\mathrm{t}<0\), \(v\left(0^{-}\right)=\frac{3}{3+9}(36 V)=9 V\)

  • \[\begin{aligned} \tau=R C&=3 \times 10^{3} \times 20 \times 10^{-6} =0.06 s \\ v_{0}(t) &=9 \mathbf{e}^{-16.667 t} \mathbf{V} \end{aligned}\]
    circuit , we have a source-free For
  • \[\begin{aligned} 3&=9 \mathrm{e}^{-16.667t_0}\\ \mathrm{e}^{16.667 t_0}&=9 / 3=3\\ t_0&=\ln (3) / 16.667=\mathbf{6 5 . 9 2} \mathbf{~ m s} \end{aligned}\]
    . Let the time be

Problem-6

  • Determine the current \(i(t)\) through the capacitor?

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Solution-6

Applying Source Transformation

\[\begin{aligned} Z &=\sqrt{R^{2}+\frac{1}{(C \omega)^{2}}}\\ &=\sqrt{4+\frac{1}{(0.1 \times 5)^{2}}} =2 \sqrt{2} \\ i &=\frac{2 \sqrt{2} \sin \left(5 t-45^{\circ}\right)}{2 \sqrt{2} \angle(-45)}=\sin 5 t \mathrm{~A} \end{aligned}\]

Problem-7

  • The capacitor shown in the figure is initially charged to 10 V. The switch closes at time \(t=0\) . Find the value of \(V_c(t)\) in volts at time \(t=10\) ms ?

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Solution-7

\[\begin{aligned} v_{c}(t) &=V_{o} e^{-t / \tau} \\ &=10 e^{-100 t}\left[\because \tau=R C=\frac{1}{100}\right] \\ \therefore v_{c}(\text { at } 10 ~& \text{msec})=10 e^{\left(-100 \times 10 \times 10^{-3}\right)} \\ &=3.678 \mathrm{~V} \end{aligned}\]

Problem-8

  • In Figure, the capacitor initially has a charge of 10 coloumb. The current in the circuit 1 second after the switch ‘S’ is closed will be

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Solution-8

\[\begin{aligned} V_{c}\left(0^{-}\right) &=V_{c}\left(0^{+}\right)=\frac{Q}{C} = \frac{10}{0.5}=20 \mathrm{~V} \\ V_{c}(\infty)=& 100 \mathrm{~V} \\ \end{aligned}\]
switch is closed At
\[\begin{aligned} V_{c}(t)=& V_{c}(\infty)+\left\{V_{c}\left(0^{+}\right)-V_{c}(\infty)\right\} e^{-t/ R C} \\ V_{c}(t)=& 100+(20-100) e^{--t/ 1} \\ \therefore \quad i_{c}(t)=& C \frac{d V_{c}(t)}{d t} \\ =& 0.5 \times-80 \times\left(-e^{-t}\right)=40 e^{-t} \\ \left.i_{c}(t)\right|_{t=1}=& 40 e^{-1} = 14.71 \mathrm{~A} \end{aligned}\]

Problem-9

  • The initial charge in the 1F capacitor present in the circuit shown is zero. Find the energy in joules transformed from the DC source until steady state condition is reached?

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Solution-9

DELTA \(\rightarrow\) STAR

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\[i(t)=\frac{V}{R} e^{-t / \tau}=\frac{10}{3} e^{-t / 5}=2 e^{-0.2 t}\]
\[E=\int_{0}^{\infty} 10 \times 2 e^{-0.2 t} d t=100 \mathrm{~J}\]

. Now current, Here,

Problem-10

  • Find the energy absorbed by the 4 \(\Omega\) in time interval \(( 0, \infty)\) ? . Given \(V_c(0) = 6\) V

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Solution-10

\[\begin{aligned} &i_{C}(t)=i_{C}(\infty)+\left[i_{C}\left(0^{+}\right)-i_{C}(\infty)\right] e^{-t / \tau}\\ &V_{C}\left(0^{-}\right)=V_{C}\left(0^{+}\right)=6 \mathrm{~V} \quad \text { [Given] } \end{aligned}\]
We know:
\[\tau = RC = 8~\text{sec}\]
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At \(t>0\)Transient
\[\begin{aligned} i_{C}(t) &=e^{-t / 8} \mathrm{~A} \end{aligned}\]

Substituting we get,

\[\begin{aligned} P &=i_{C}^{2}(t) R \\ E &=\int_{0}^{\infty} P d t=\int_{0}^{\infty} i_{C}^{2}(t) R d t=\int_{0}^{\infty}\left(e^{-t / 8}\right)^{2} \times 4 d t \\ E &=\int_{0}^{\infty} 4 e^{-t / 4} d t=4 \times(-4)\left[e^{-t / 4}\right]_{t=0}^{\infty} \\ E &=-16[0-1]=16 \text { Joules } \end{aligned}\]

RL Circuits

Problem-11

  • Assuming that \(i(0)\)=10 A, Calculate \(i(t)\) and \(i_x(t)\)

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Solution-11

\[\begin{aligned} &2\left(i_{1}-i_{2}\right)+1=0 \\ &6 i_{2}-2 i_{1}-3 i_{1}=0 \\ &i_{1}=-3 \mathrm{~A}, \quad i_{o}=-i_{1}=3 \mathrm{~A} \\ &R_{\mathrm{eq}}=R_{\mathrm{Th}}=\frac{v_{o}}{i_{o}}=\frac{1}{3} \Omega \\ &\tau=\frac{L}{R_{\mathrm{eq}}}=\frac{\frac{1}{2}}{\frac{1}{3}}=\frac{3}{2} \mathrm{~s} \\ &i(t)=i(0) e^{-t / \tau}\\ &=10 e^{-(2 / 3) t} \mathrm{~A}, \quad t>0 \end{aligned}\]
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Method-1:
\[\begin{aligned} &\frac{1}{2} \frac{d i_{1}}{d t}+2\left(i_{1}-i_{2}\right)=0 \Rightarrow \frac{d i_{1}}{d t}+4 i_{1}-4 i_{2}=0 \\ &6 i_{2}-2 i_{1}-3 i_{1}=0 \quad \Rightarrow \quad i_{2}-\frac{5}{6} i_{1} \\ &\frac{d i_{1}}{d t}+\frac{2}{3} i_{1}=0 \quad \Rightarrow \quad \frac{d i_{1}}{i_{1}}=-\frac{2}{3} d t \\ &\left.\ln i\right|_{i(0)} ^{i(0)}=-\left.\frac{2}{3} t\right|_{0} ^{t} \quad \Rightarrow \ln \frac{i(t)}{i(0)}=-\frac{2}{3} t \\ &i(t)=i(0) e^{-(2 / 3) t}=10 e^{-(2 / 3) t} \mathrm{~A}, \quad t>0 \end{aligned}\]
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Method-2:
\[\begin{aligned} &v=L \frac{d i}{d t}=0.5(10)\left(-\frac{2}{3}\right) e^{-(2 / 3) t}=-\frac{10}{3} e^{-(2 / 3) t} \mathrm{~V} \\ &i_{x}(t)=\frac{v}{2}=-1.6667 e^{-(2 / 3) t} \mathrm{~A}, \quad t>0 \end{aligned}\]

Problem-12

  • The switch has been closed for a long time. At \(t=0\), the switch is opened. Calculate \(i(t)\) for \(t>0\).

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Solution-12

  • \(t<0~\Rightarrow~S~\text{closed}~\Rightarrow ~L\) short \(\Rightarrow~16\Omega\) short

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\[\begin{aligned} R_{eq} & = \dfrac{4\times 12}{4+12} = 3~\Omega \\ i_1 & = \dfrac{40}{2+3} = 8~\mathrm{A} \\ i(t) & = \dfrac{12}{12+4}i_1 = 6~\mathrm{A}, \quad t<0 \\ \end{aligned}\]
\[\begin{aligned} i(0^+) & = i(0^{-}) = 6~\mathrm{A} \\ R_{eq} & = (12+4)||16 = 8~\Omega \\ \tau & = \dfrac{L}{R_{eq}} =\dfrac{1}{4}~\mathrm{s}\\ i(t) & = i(0)e^{-t/\tau}= 6e^{-4t}~\mathrm{A} \end{aligned}\]

Problem-13

  • Find \(i_0\), \(v_0\), and \(i\) for all time , assuming that the switch was open for a long time.

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Solution-13

  • \(t<0~\Rightarrow~S~\text{open}~\Rightarrow ~L\) short \(\Rightarrow~6\Omega\) short

\[\begin{aligned} i_0&=0 \\ i(t) & = \dfrac{10}{2+3} = 2~\mathrm{A}, \quad t<0 \\ v_0(t)&=3i(t)=6~\mathrm{V},\quad t<0\\ i(0) & = 2 \end{aligned}\]
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\[\begin{aligned} R_{Th} & = 3||6=2~\Omega \\ \tau & = \dfrac{L}{R_{Th}}=1~\mathrm{s} \\ i(t) & = i(0)e^{-t/\tau}=2e^{-t}~\mathrm{A}, \quad t>0\\ v_0(t)&=-v_L=-L\dfrac{di}{dt} =4e^{-t}~\mathrm{V},\quad t>0 \end{aligned}\]
\[i_{o}(t)=\frac{v_{L}}{6}=-\frac{2}{3} e^{-t} \mathrm{~A}, \quad t>0\]
Thus, for all time,
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Problem-14

  • At \(t=0\), switch-1 is closed, and switch-2 is closed 4 s later. Find \(i(t)\) for \(t>0\). Calculate \(i\) for \(t=2\) s and \(t=5\) s.

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Solution-14

  • For \(0 \leq t \leq 4, ~S_{1}\) closed \(\Rightarrow 4-\Omega\) and \(6-\Omega\) are in series. (Remember, at this time, \(S_{2}\) is still open.)

  • \[\begin{gathered} i(\infty)=\frac{40}{4+6}=4 \mathrm{~A}, \quad R_{\mathrm{Th}}=4+6=10 \Omega \\ \tau=\frac{L}{R_{\mathrm{Th}}}=\frac{5}{10}=\frac{1}{2} \mathrm{~s} \end{gathered}\]
    is closed forever, Hence, assuming for now that
  • \[\begin{aligned} i(t) &=i(\infty)+[i(0)-i(\infty)] e^{-t / \tau} \\ &=4+(0-4) e^{-2 t}=4\left(1-e^{-2 t}\right) \mathrm{A}, \quad 0 \leq t \leq 4 \end{aligned}\]
    Thus,

  • Intervals \(t \leq 0,0 \leq t \leq 4\), and \(t \geq 4\) separately.

  • For \(t<0\), \(S_{1}\) and \(S_{2}~\Rightarrow\) open \(\Rightarrow i=0\).

  • \(i\left(0^{-}\right)=i(0)=i\left(0^{+}\right)=0\)

  • For \(t \geq 4, ~S_{2}\) closed \(\Rightarrow\)10 - V connected.

  • \[i(4)=i\left(4^{-}\right)=4\left(1-e^{-8}\right) \simeq 4 \mathrm{~A}\]
    This sudden change does not affect the inductor current because the current cannot change abruptly. Thus, the initial current is
  • \[\begin{gathered} \frac{40-v}{4}+\frac{10-v}{2}=\frac{v}{6} \quad \Rightarrow \quad v=\frac{180}{11} \mathrm{~V} \\ i(\infty)=\frac{v}{6}=\frac{30}{11}=2.727 \mathrm{~A} \end{gathered}\]
    . Using KCL, be the voltage at node , let To find
  • \[R_{\mathrm{Th}}=4 \| 2+6=\frac{4 \times 2}{6}+6=\frac{22}{3} \Omega\]
    \[\tau=\frac{L}{R_{\mathrm{Th}}}=\frac{5}{\frac{22}{3}}=\frac{15}{22} \mathrm{~s}\]
    The Thevenin resistance at the inductor terminals is
  • \[i(t)=i(\infty)+[i(4)-i(\infty)] e^{-(t-4) / \tau}, \quad t \geq 4\]
    Hence,
  • \[\begin{array}{rlr} i(t) & =2.727+(4-2.727) e^{-(t-4) / \tau}, & \tau=\frac{15}{22} \\ & =2.727+1.273 e^{-1.4667(t-4)}, & t \geq 4 \end{array}\]
    in the exponential because of the time delay. Thus, We need
  • \[i(t)= \begin{cases}0, & t \leq 0 \\ 4\left(1-e^{-2 t}\right), & 0 \leq t \leq 4 \\ 2.727+1.273 e^{-1.4667(t-4)}, & t \geq 4\end{cases}\]
    \[i(5)=2.727+1.273 e^{-1.4667}=3.02 \mathrm{~A}\]
    \(t=5\)
    \[i(2)=4\left(1-e^{-4}\right)=3.93 \mathrm{~A}\]
    \(t=2\)Putting all this together,
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