Solved Problems on First-Order Circuits
Demonstrative Video
Concepts of First-Order Circuits
Problem-1
Determine the Time-constant of the network?
Solution-1
\[\begin{aligned} R_{Th} & = R || (2R) = \dfrac{2}{3} R \\ \tau & = R_{Th} \cdot C = \dfrac{2}{3} RC~ \mathrm{sec.} \end{aligned}\]
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 \(t=100~\mu s\) ?
Solution-2
Current \[\begin{aligned} \tau & = R\cdot C = 25\times 10^{-6}~\text{sec} \\ i(t) & = \dfrac{V}{R} e^{-t/\tau} = \dfrac{4}{5} e^{-t/\tau} \end{aligned}\]
Charge lost by the capacitor: \[\begin{aligned} Q & = \int_{25~\mu s}^{100~\mu s} i(t) dt = 6.99\times 10^{-6}~\mathrm{C} \end{aligned}\]
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 \(t=5~\mu s\) ?
Solution-3
\[\begin{aligned} \text{At}~t&=0^{-}\\ V_c(0^{-})&=1 \end{aligned}\]
\[\begin{aligned} \text{At}~t&=\infty\\ V_c(\infty)&=3\dfrac{2}{2+1}=2~\mathrm{V} \end{aligned}\]
Step-3 \[\begin{aligned} & \tau = RC = \dfrac{2}{3}\times 10 \times 10^{-6}\\ & V_c(t) =[V_c(0^{+})-V_c(\infty)]e^{-t/\tau}+V_c(\infty)\\ &=[1-2]e^{-t/\tau}+2 \end{aligned}\] \[\begin{aligned} &\text{At}~t=5~\mu s \Rightarrow V_c = 1.52~\mathrm{V} \end{aligned}\]
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 \(t \geq 0\)
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 \(t = 0\).
Solution-5
For \(\mathrm{t}<0\), \(v\left(0^{-}\right)=\frac{3}{3+9}(36 V)=9 V\)
For \(t>0\), we have a source-free \(R C\) circuit \[\begin{aligned} \tau=R C&=3 \times 10^{3} \times 20 \times 10^{-6} =0.06 s \\ v_{0}(t) &=9 \mathbf{e}^{-16.667 t} \mathbf{V} \end{aligned}\]
Let the time be \({t}_{0}\). \[\begin{aligned} 3&=9 \mathrm{e}^{-16.667t_0}\\ \mathrm{e}^{16.667 t_0}&=9 / 3=3\\ t_0&=\ln (3) / 16.667=\mathbf{6 5 . 9 2} \mathbf{~ m s} \end{aligned}\]
Problem-6
Determine the current \(i(t)\) through the capacitor?
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 ?
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
Solution-8
At \((t=0)\) switch is closed \[\begin{aligned} V_{c}\left(0^{-}\right) &=V_{c}\left(0^{+}\right)=\frac{Q}{C} = \frac{10}{0.5}=20 \mathrm{~V} \\ V_{c}(\infty)=& 100 \mathrm{~V} \\ \end{aligned}\]
\[\begin{aligned} V_{c}(t)=& V_{c}(\infty)+\left\{V_{c}\left(0^{+}\right)-V_{c}(\infty)\right\} e^{-t/ R C} \\ V_{c}(t)=& 100+(20-100) e^{--t/ 1} \\ \therefore \quad i_{c}(t)=& C \frac{d V_{c}(t)}{d t} \\ =& 0.5 \times-80 \times\left(-e^{-t}\right)=40 e^{-t} \\ \left.i_{c}(t)\right|_{t=1}=& 40 e^{-1} = 14.71 \mathrm{~A} \end{aligned}\]
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?
Solution-9
DELTA \(\rightarrow\) STAR
Here, \(\quad \tau=R C=5 \mathrm{sec}\). Now current, \[i(t)=\frac{V}{R} e^{-t / \tau}=\frac{10}{3} e^{-t / 5}=2 e^{-0.2 t}\] Energy supplied by the source, \[E=\int_{0}^{\infty} 10 \times 2 e^{-0.2 t} d t=100 \mathrm{~J}\]
Problem-10
Find the energy absorbed by the 4 \(\Omega\) in time interval \(( 0, \infty)\) ? . Given \(V_c(0) = 6\) V
Solution-10
We know: \[\begin{aligned} &i_{C}(t)=i_{C}(\infty)+\left[i_{C}\left(0^{+}\right)-i_{C}(\infty)\right] e^{-t / \tau}\\ &V_{C}\left(0^{-}\right)=V_{C}\left(0^{+}\right)=6 \mathrm{~V} \quad \text { [Given] } \end{aligned}\]
At \(t>0\)Transient
\[\tau = RC =
8~\text{sec}\] At \(t>0^{+}\)
\[i_c(0^{+}) =
\dfrac{10-6}{4}=1~\text{A}\] At \(t =
\infty\) Steady state
\[i_c(\infty) = 0\]
Substituting we get, \[\begin{aligned} i_{C}(t) &=e^{-t / 8} \mathrm{~A} \end{aligned}\]
\[\begin{aligned} P &=i_{C}^{2}(t) R \\ E &=\int_{0}^{\infty} P d t=\int_{0}^{\infty} i_{C}^{2}(t) R d t=\int_{0}^{\infty}\left(e^{-t / 8}\right)^{2} \times 4 d t \\ E &=\int_{0}^{\infty} 4 e^{-t / 4} d t=4 \times(-4)\left[e^{-t / 4}\right]_{t=0}^{\infty} \\ E &=-16[0-1]=16 \text { Joules } \end{aligned}\]
RL Circuits
Problem-11
Assuming that \(i(0)\)=10 A, Calculate \(i(t)\) and \(i_x(t)\)
Solution-11
Method-1:
\[\begin{aligned}
&2\left(i_{1}-i_{2}\right)+1=0 \\
&6 i_{2}-2 i_{1}-3 i_{1}=0 \\
&i_{1}=-3 \mathrm{~A}, \quad i_{o}=-i_{1}=3
\mathrm{~A} \\
&R_{\mathrm{eq}}=R_{\mathrm{Th}}=\frac{v_{o}}{i_{o}}=\frac{1}{3}
\Omega \\
&\tau=\frac{L}{R_{\mathrm{eq}}}=\frac{\frac{1}{2}}{\frac{1}{3}}=\frac{3}{2}
\mathrm{~s} \\
&i(t)=i(0) e^{-t / \tau}\\
&=10 e^{-(2 / 3) t} \mathrm{~A}, \quad t>0
\end{aligned}\]
Method-2:
\[\begin{aligned}
&\frac{1}{2} \frac{d i_{1}}{d t}+2\left(i_{1}-i_{2}\right)=0
\Rightarrow \frac{d i_{1}}{d t}+4 i_{1}-4 i_{2}=0 \\
&6 i_{2}-2 i_{1}-3 i_{1}=0 \quad \Rightarrow \quad
i_{2}-\frac{5}{6} i_{1} \\
&\frac{d i_{1}}{d t}+\frac{2}{3} i_{1}=0 \quad \Rightarrow
\quad \frac{d i_{1}}{i_{1}}=-\frac{2}{3} d t \\
&\left.\ln i\right|_{i(0)} ^{i(0)}=-\left.\frac{2}{3}
t\right|_{0} ^{t} \quad \Rightarrow \ln \frac{i(t)}{i(0)}=-\frac{2}{3} t
\\
&i(t)=i(0) e^{-(2 / 3) t}=10 e^{-(2 / 3) t} \mathrm{~A},
\quad t>0
\end{aligned}\]
\[\begin{aligned}
&v=L \frac{d i}{d t}=0.5(10)\left(-\frac{2}{3}\right) e^{-(2
/ 3) t}=-\frac{10}{3} e^{-(2 / 3) t} \mathrm{~V} \\
&i_{x}(t)=\frac{v}{2}=-1.6667 e^{-(2 / 3) t} \mathrm{~A},
\quad t>0
\end{aligned}\]
Problem-12
The switch has been closed for a long time. At \(t=0\), the switch is opened. Calculate \(i(t)\) for \(t>0\).
Solution-12
\(t<0~\Rightarrow~S~\text{closed}~\Rightarrow ~L\) short \(\Rightarrow~16\Omega\) short
\[\begin{aligned}
R_{eq} & = \dfrac{4\times 12}{4+12} = 3~\Omega
\\
i_1 & = \dfrac{40}{2+3} = 8~\mathrm{A} \\
i(t) & = \dfrac{12}{12+4}i_1 = 6~\mathrm{A},
\quad t<0 \\
\end{aligned}\]
\[\begin{aligned} i(0^+) & = i(0^{-}) = 6~\mathrm{A} \\ R_{eq} & = (12+4)||16 = 8~\Omega \\ \tau & = \dfrac{L}{R_{eq}} =\dfrac{1}{4}~\mathrm{s}\\ i(t) & = i(0)e^{-t/\tau}= 6e^{-4t}~\mathrm{A} \end{aligned}\]
Problem-13
Find \(i_0\), \(v_0\), and \(i\) for all time , assuming that the switch was open for a long time.
Solution-13
\(t<0~\Rightarrow~S~\text{open}~\Rightarrow ~L\) short \(\Rightarrow~6\Omega\) short
\[\begin{aligned}
i_0&=0 \\
i(t) & = \dfrac{10}{2+3} = 2~\mathrm{A}, \quad
t<0 \\
v_0(t)&=3i(t)=6~\mathrm{V},\quad t<0\\
i(0) & = 2
\end{aligned}\]
\[\begin{aligned} R_{Th} & = 3||6=2~\Omega \\ \tau & = \dfrac{L}{R_{Th}}=1~\mathrm{s} \\ i(t) & = i(0)e^{-t/\tau}=2e^{-t}~\mathrm{A}, \quad t>0\\ v_0(t)&=-v_L=-L\dfrac{di}{dt} =4e^{-t}~\mathrm{V},\quad t>0 \end{aligned}\]
\[i_{o}(t)=\frac{v_{L}}{6}=-\frac{2}{3} e^{-t} \mathrm{~A}, \quad t>0\] Thus, for all time, \[\begin{gathered} i_{o}(t)=\left\{\begin{array}{ll} 0 \mathrm{~A}, & t<0 \\ -\frac{2}{3} e^{-t} \mathrm{~A}, & t>0 \end{array}, \quad v_{o}(t)= \begin{cases}6 \mathrm{~V}, & t<0 \\ 4 e^{-t} \mathrm{~V}, & t>0\end{cases} \right. \\ i(t)= \begin{cases}2 \mathrm{~A}, & t<0 \\ 2 e^{-t} \mathrm{~A}, & t \geq 0\end{cases} \end{gathered}\]
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.
Solution-14
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 \(0 \leq t \leq 4, ~S_{1}\) closed \(\Rightarrow 4-\Omega\) and \(6-\Omega\) are in series. (Remember, at this time, \(S_{2}\) is still open.)
Hence, assuming for now that \(S_{1}\) is closed forever, \[\begin{gathered} i(\infty)=\frac{40}{4+6}=4 \mathrm{~A}, \quad R_{\mathrm{Th}}=4+6=10 \Omega \\ \tau=\frac{L}{R_{\mathrm{Th}}}=\frac{5}{10}=\frac{1}{2} \mathrm{~s} \end{gathered}\]
Thus, \[\begin{aligned} i(t) &=i(\infty)+[i(0)-i(\infty)] e^{-t / \tau} \\ &=4+(0-4) e^{-2 t}=4\left(1-e^{-2 t}\right) \mathrm{A}, \quad 0 \leq t \leq 4 \end{aligned}\]
For \(t \geq 4, ~S_{2}\) closed \(\Rightarrow\)10 - V connected.
This sudden change does not affect the inductor current because the current cannot change abruptly. Thus, the initial current is \[i(4)=i\left(4^{-}\right)=4\left(1-e^{-8}\right) \simeq 4 \mathrm{~A}\]
To find \(i(\infty)\), let \(v\) be the voltage at node \(P\). Using KCL, \[\begin{gathered} \frac{40-v}{4}+\frac{10-v}{2}=\frac{v}{6} \quad \Rightarrow \quad v=\frac{180}{11} \mathrm{~V} \\ i(\infty)=\frac{v}{6}=\frac{30}{11}=2.727 \mathrm{~A} \end{gathered}\]
The Thevenin resistance at the inductor terminals is \[R_{\mathrm{Th}}=4 \| 2+6=\frac{4 \times 2}{6}+6=\frac{22}{3} \Omega\] and \[\tau=\frac{L}{R_{\mathrm{Th}}}=\frac{5}{\frac{22}{3}}=\frac{15}{22} \mathrm{~s}\]
Hence, \[i(t)=i(\infty)+[i(4)-i(\infty)] e^{-(t-4) / \tau}, \quad t \geq 4\]
We need \((t-4)\) in the exponential because of the time delay. Thus, \[\begin{array}{rlr} i(t) & =2.727+(4-2.727) e^{-(t-4) / \tau}, & \tau=\frac{15}{22} \\ & =2.727+1.273 e^{-1.4667(t-4)}, & t \geq 4 \end{array}\]
Putting all this together, \[i(t)= \begin{cases}0, & t \leq 0 \\ 4\left(1-e^{-2 t}\right), & 0 \leq t \leq 4 \\ 2.727+1.273 e^{-1.4667(t-4)}, & t \geq 4\end{cases}\] At \(t=2\), \[i(2)=4\left(1-e^{-4}\right)=3.93 \mathrm{~A}\] At \(t=5\), \[i(5)=2.727+1.273 e^{-1.4667}=3.02 \mathrm{~A}\]