E84 Midterm Exam 1

Instructions

• Take home, open everything except discussion. Due Wednesday in class.
• Mark your start and end times. Don't spend more than 2 hours.
• Compare your print-out of the exam with the online version to make sure your hard copy is complete.
• Mark your name and question number clearly on top of each page. Indicate the total number of pages submitted.
• When solving a problem, list all the steps. In each step, describe concisely what you are doing in English, then show the calculation and the result of the step. A final answer, even if correct, without evidence of the steps leading to the answer will not receive credit.

1. Problem 1. (30 pts)

   Find the two currents labeled as $I_a$ and $I_b$ in the figure.

   

   Solution:

   By KCL (observation), $I_b=6+12=18A$.

   Use loop current method to around the loop and get:
   

   $$18I_a+12(I_a+12)+6(I_a+12+6)=0,\;\;\;\;\mbox{i.e.,}\;\;\;\; 36I_a+252=0$$
   
   
   Solving this we get $I_a=-7$.

   Use branch current method. Assume currents going through 1$\Omega$, 2$\Omega$ and 3$\Omega$ resistors are, respectively, $I_1$, $I_2$ and $I_3$. Using KCL and KVL, we get
   

   $$\left\{ \begin{array}{l} I_1+I_2=6 I_2+I_3=-12 I_1-2 I_2+3I_3=0 \end{array} \right.$$
   
   
   Solving this we get: $I_1=11$, $I_2=-5$, $I_3=I_a=-7$, $I_b=I_1-I_3=18$.

   Alternatively, one can use superposition theorem. First, turn the 6A current off and get
   

   $$I'_a=-12 \frac{6+12}{6+12+18}=-6 A$$
   
   
   Second, turn the 12A off and get
   
   $$I''_a=-6 \frac{6}{6+12+18}=-1 A$$
   
   
   Now we have $I_a=I'_a+I''_a=-7 A$. $I_b=18A$ is simply the sum of 6A and 12A.

2. Problem 2. (30 points)

   Find the voltage $V$ across the 2$\Omega$ resistor (with assumed polarity shown), and the current $I$ through the 1$V$ voltage source (with assumed direction shown).

   

   Solution:

   Assume lower-left node is grounded as 0V. upper-left node is 5V, upper-right node is 1V, and lower-right node is 3V. Voltage across $2\Omega$ resistor is $1V-3V=-2V$. Current through $1\Omega$ resistor (right-ward) is 4A, current through $2\Omega$ resistor (up-ward) is 1A, Applying KCL to upper-rigt node, we get $I=5A$.

   Or, use superposition theorem:

   • short the 1V and 2V sources, $V_1=-3V$, $I_1=3A+1.5A=4.5A$
   • short the 1V and 3V sources, $V_1= 0V$, $I_1= 2A$
   • short the 2V and 3V sources, $V_1= 1V$, $I_1=-(1+0.5)A=-1.5A$
   Total: $V=-2V$, $I=5A$.

3. Problem 3. (40 pts)

   The following two parts of the problem are independent, each worth 20 pts.

   • Find the resistance $R$ so that the voltage across the resistor is 1V.
   • Find the resistance $R$ so that the current through the resistor is 0.2A.

   

   Solution:

   The problems can be most easily solved by assuming the desired voltage/current and then figuring out the resistance $R$ needed.

   Fiirst, if $V=1$ as desired, then the current (rightward) through the $6\Omega$ resistor is $(5+3-1)V/6\Omega=7/6A$, and the current through $R$ (downward) is $I=7/6-1=1/6\;A$, and we get $R=V/I=6\Omega$.

   Second, if $I=0.2A$ as desired, then the current (rightward) through $6\Omega$ resistor has to be $1.2A$ and the voltage drop across it is $7.2V$ (positive on the left). Now the voltage to the right of the $6\Omega$ resistor is $V=5+3-7.2=0.8V$, and $R=V/I=0.8/0.2=4\Omega$.

   Superposition method could also be used.

   • Only turn 3V voltage source on (5V shorted together with $3\Omega$ resistor, 1A current source is open), we get
     
     $$I'=\frac{3}{6+R},\;\;\;\;\; V'=\frac{3R}{6+R}$$
     
     

   • Only turn 5V voltage source on (3V shorted, 1A current source is open), we get
     
     $$I''=\frac{5}{6+R},\;\;\;\;\; V''=\frac{5R}{6+R}$$
     
     

   • Only turn 1A current source on (3V and 5V are both shorted), we get
     
     $$I'''=\frac{-6}{6+R},\;\;\;\;\; V'''=\frac{-6R}{6+R}$$
     
     

   Now we have:
   
   $$I=I'+I''+I'''=\frac{2}{6+R}=0.2A;\;\;\;\; V=V'+V''+V'''=\frac{2R}{6+R}=2V$$
   
   
   Solving these for $R$, we get $R=4\Omega$ and $R=6\Omega$.

   Alternatively, we can use Thevenin's theorem or Norton's theorem by treating the $R$ as a load. In either case, we need to find the internal resistance of the rest of the circuit excluding R. When all energy sources are turned off, the internal resistance is simply $R_0=6\Omega$.

   • Find open circuit voltage $V_{oc}$:

     Use superposition, when current source is open, $V'_{oc}=5V+3V=8V$; when voltage sources are short, $V''_{oc}=-1A\times 6\Omega=-6V$. Therefore $V_T=V_{oc}=V'_{oc}+V''_{oc}=2V$

   • Find short circuit current $I_{sc}$:

     Use superposition, when current source is open, $I'_{sc}=8V/6\Omega=4/3\;A$; when voltage sources are short, $I''_{sc}=-1A$. Therefore $I_N=I_{sc}=I'_{sc}+I''_{sc}=1/3\;A$, which is consistent with the previous result: $V_T=V_{oc}=I_{sc} R_0=6/3 = 2V$.

   The rest of the circuit is equivalent to a voltage source $V_T=2V$ in series with a resistor $R=6\Omega$, for $V=1V$, we need $R=6\omega$. For $I=0.2A$, we need $R=4\Omega$.