E84 Midterm Exam 2

Instructions

• Take-home, open notes, feel free to use a calculator, but not any software package.
• PRINT your name and mark question numbers clearly on top of each page. Indicate the total number of pages submitted.
• When solving a problem, list all the steps. In each step, indicate concisely what you are doing in English, then show the calculation and the result of for the step. Box the final answer.
• A final answer, even if correct, without evidence of the steps leading to it will receive ZERO credit.

• Problem 1. (50 pts)

  Find the current $i(t)$ through the $5\Omega$ resistor in the circuit shown in the figure below. Here the voltage source is $v_0(t)=10\sqrt{2}\cos 2t$ and the current source is $i_0(t)=2\sqrt{2}\cos(3t+45^\circ)$.

  

  Solution:

  • Phasor representation of the voltage source: $\dot{V}=10\angle 0^\circ$ V.
  • Find current through $R$ due to voltage source:
    
    $$\dot{I}_1=\frac{\dot{V}}{j2/3+(j2\vert\vert-j0.5)+5}=\frac{10}{j2/3-j2/3+5}=2\;A$$
    
    

  • Phasor representation of the voltage source: $\dot{I}=2\angle 45^\circ$ A.
  • Find current through $R$ due to current source:
    
    $$\dot{I}_2=\dot{I}\frac{j1+j3\vert\vert(-j1/3)}{5+j1+j3\vert... ...5^\circ \times \frac{j5/8}{5+j5/8}=0.25\angle 127.87^\circ\;A$$
    
    

  • The total phasor current is
    
    $$I(t)=I_1(t)+I_2(t)=2+0.25\angle 127.87^\circ$$
    
    

  • Convert to time function:
    
    $$i(t)=2\sqrt{2}\cos 2t+0.25\sqrt{2}\cos(3t+127.87^\circ)\;A$$
    
    

• Problem 2. (50 pts)

  In the circuit shown in the figure below, $R_1=R_2=15\,k\Omega$, $R_3=30\,k\Omega$, $C=50\,\mu F$, and the input voltage is $V=40$ V. The switch is closed at $t=0$. Assume the circuit was at steady state before $t=0$. Find currents $i_2(t)$ and $i_3(t)$, together with $i_c(t)$ as the complete responses of the circuit for $t>0$. Make sure the signs of these currents are consistent with the assumed directions as shown in the figure.

  

  Solution:

  • Find voltage across $C$ at $t=0^-$: (0 pts)
    
    $$v_c(0^-)=v_c(0^+)=40\frac{30}{15+15+30}=20\;V$$
    
    

  • Find initial current $i_c(0^+)$ through $C$: (5 pts)
    
    $$i_c(0^+)=-\frac{v_c(0^+)}{R_2 \vert\vert R_3}=-\frac{20}{15 \vert\vert 30}=-2\; mA$$
    
    

  • The steady state current through $C$ is $i_c(\infty)=0$. (5 pts)
  • Find time constant: (5 pts)
    
    $$\tau=R_{eq}C=(R_2\vert\vert R_3)C=(15\vert\vert 30)\times 10^3\times 50\times 10^{-6}=0.5\;s$$
    
    

  • The current through $C$ as the complete response: (5 pts)
    
    $$i_c(t)=i_c(\infty)+[i_c(0^+)-i_c(\infty)] e^{-t/tau}=-2 e^{-2t}\;mA$$
    
    

  • Find current through $R_1$ for $t>0$: $i_1(t)=40/15=8/3=2.67\;mA$. (5 pts)

  • Find initial current $i_2(0^+)$ through $R_2$: (5 pts)
    
    $$i_2(0^+)=-\frac{v_c(0^+)}{R_2}=\frac{20}{15}=-\frac{4}{3}\;mA$$
    
    

  • The steady state current through $R_2$ is $i_2(\infty)=0$. (5 pts)
  • The current through $R_2$ as the complete response: (5 pts)
    
    $$i_2(t)=i_2(\infty)+[i_2(0^+)-i_2(\infty)] e^{-t/tau}=-4/3 \; e^{-2t}\;mA$$
    
    

  • Find initial current $i_3(0^+)$ through $R_3$: (5 pts)
    
    $$i_3(0^+)=\frac{v_c(0^+)}{R_3}=\frac{20}{30}=\frac{2}{3}\;mA$$
    
    

  • The steady state current through $R_3$ is $i_3(\infty)=0$. (3 pts)
  • The current through $R_3$ as the complete response: (5 pts)
    
    $$i_3(t)=i_3(\infty)+[i_3(0^+)-i_3(\infty)] e^{-t/tau}=2/3 \; e^{-2t}\;mA$$