EE 211 - Homework 2 Solutions

1.
Problem 2.21
(a)
Circuit C3: everything in series; combine two resistors into single 7.1 k$\Omega$ resistor, then $i_x = 10{\rm V}/7.1 {\rm k}\Omega = 1.4 {\rm mA}$. $v_x = 5.6 {\rm k}\Omega \times i_x = 7.9 {\rm V}$.

(b)
Circuit C4: Combine 3.3k and 9.1k resistor to make single 12.4k resistor, then use current divider to find ix:


\begin{displaymath}i_x = \frac{\frac{1}{12.4{\rm {k}}}}{\frac{1}{220{\rm {k}}} + \frac{1}{12.4{\rm {k}}}} 2{\rm {mA}} = 1.89 {\rm mA}\end{displaymath}

Then $v_x = 9.1{\rm k} \times i_x = 17.2 {\rm V}$.

2.
Problem 2.26
(a)
ix = 0.2A. Voltage drop across each 25$\Omega$ resistor is 0.2A $\times$ 25$\Omega$ = 5V. By KVL, vx = 25V - 5V - 5V = 15V. Voltage drop across each 2$\Omega$ resistor is 0.2A $\times$ 2$\Omega$ = 0.4V. By KVL, vy = vx - 0.4V - 0.4V = 14.2V.
(b)
Power delivered at line input is $p_{IN} = v_x \times i_x = $3 W. Everything is in series so ix goes through load. Power delivered to load is $p_{OUT} = v_y \times i_x = $2.84 W.
(c)
$\eta = (p_{OUT}/p_{IN}) \times 100 = (2.84/3) \times 100 = $95%.

3.
Problem 2.33
(a)
Do source transformation: 10V voltage source in series with 50$\Omega$resistor goes to 200mA current source in parallel with 50$\Omega$ resistor, all in parallel with 200$\Omega$ resistor. Combine 50$\Omega$ and 200$\Omega$ in parallel to get 40$\Omega$ in parallel with 200mA current source. Another source transformation gives 8V voltage source in series with 40$\Omega$ resistor.

\epsfig{file=hw2_soln_2_33_a.eps, width=4.5in}

(b)
Do source transformation: 200mA current source in parallel with 100$\Omega$ resistor goes to 20V voltage source in series with 100$\Omega$resistor, all in series with 33$\Omega$ resistor. Combine 100$\Omega$ and 33$\Omega$ into single 133$\Omega$ resistor in series with 20V voltage source. Another source transformation gives 150mA current source in parallel with 133$\Omega$ resistor.

\epsfig{file=hw2_soln_2_33_b.eps, width=4.5in}

(c)
Do a source transformation on the voltage source with resistor in series to get a current source with resistor in parallel: 8V voltage source in series with 40$\Omega$ resistor gives 200mA current source in parallel with a 40$\Omega$ resistor. Now you have a 200mA current source in parallel with a 150mA current source - these two add to give a 350mA current source. You also have a 40$\Omega$ resistor in parallel with a 133$\Omega$ resistor - these add to give a 31$\Omega$ resistor.

\epsfig{file=hw2_soln_2_33_c.eps, width=5in}

4.
Problem 2.34
(a)
Circuit C1: Any resistor in series with 2A current source is equivalent to a 2A current source, and the problem is done: 5$\Omega$ resistor parallel to 2A current source.
(b)
Circuit C2: Do source transformation on 2A current source parallel to 5$\Omega$resistor. Get 10V current source in series with 5$\Omega$ resistor, all in series with the 10$\Omega$ resistor. Add the 5$\Omega$ resistor and 10$\Omega$resistor in series; final circuit is 10V voltage source in series with 15$\Omega$ resistor.
(c)
Circuit C3: 10V voltage source in parallel with any resistor is equivalent to the 10V voltage source, and the problem is done: a 10V voltage source in series with a 50$\Omega$ resistor.
5.
Problem 2-41
(a)
Circuit C1: Simple current division:

\begin{displaymath}i_x = \frac{\frac{1}{R_1}}{\frac{1}{R_1} + \frac{1}{R_2}} i_s = \frac{R_2}{R_1 + R_2}i_s \end{displaymath}

(b)
Circuit C2: Simple voltage division:

\begin{displaymath}v_x = \frac{R2}{R1+R2+R3} v_S\end{displaymath}

(c)
Circuit C3: Can ignore R3 - resistor in series with current source is equivalent to current source alone. Then have simple current divider, but ix going in opposite direction to is:

\begin{displaymath}i_x = - \frac{\frac{1}{R_2}}{\frac{1}{R_1} + \frac{1}{R_2}} i_s = - \frac{R_1}{R_1 + R_2}i_s \end{displaymath}

(d)
Circuit C4: Combine R2 and R4 into single resistor Req = R2 R4 / ( R2 + R4). vx is voltage across Req. vx can be found with simple voltage divider:

\begin{displaymath}v_x = \frac{R_{eq}}{R1+R_{eq}+R3} v_S = \frac{R_2 R_4} {R_1 R_2 + R_1 R_4 + R_2 R_4 + R_2 R_3 + R_3 R_4} v_S \end{displaymath}

6.
Problem 2-44:
(a)
Circuit C3: Combine two 1$\Omega$ resistors in parallel into single 0.5$\Omega$resistor. Do source transformation on 5A current source and 1$\Omega$resistor in parallel to get 5V voltage source in series with 1$\Omega$resistor. You then have: 5V voltage source, 1$\Omega$ resistor, 0.5$\Omega$resistor and 1.5$\Omega$ resistor all in series. Use a voltage divider to find vx across 1.5$\Omega$ resistor:

\begin{displaymath}v_x = \frac{1.5}{1 + 0.5 + 1.5} 5{\rm V} = 2.5 {\rm V}\end{displaymath}

(b)
Circuit C4: Combine 60$\Omega$ resistor and 28$\Omega$ resistor in series to form 88$\Omega$ resistor in series with 100V voltage source. To source transformation to get 100V/88$\Omega$ = 1.14A current source in parallel with 88$\Omega$ resistor. Now have 1.14mA current source in parallel with 88$\Omega$, 30$\Omega$ and 20$\Omega$ resistors. Use current division to find ix, current through 20$\Omega$ resistor:

\begin{displaymath}i_x = \frac{\frac{1}{20}}{\frac{1}{88} + \frac{1}{30} + \frac{1}{20}} 1.14{\rm mA} = 0.6{\rm V}\end{displaymath}

7.
Problem 2-43
(a)
Circuit C1: Combine two 20$\Omega$ reistors in parallel to get singe 10$\Omega$ resistor. Combine the equivalent 10$\Omega$ resistor with the other 10$\Omega$ resistor to get a 20$\Omega$ resistor. Now have two 20$\Omega$resistors in parallel, and ix is current through right 20$\Omega$ resistor. Use current division to find ix:

\begin{displaymath}i_x = \frac{\frac{1}{20}}{\frac{1}{20} + \frac{1}{20}} 5{\rm A} = 2.5{\rm V}\end{displaymath}

(b)
Combine 10$\Omega$ and 5$\Omega$ resistors in parallel to get single 3.3$\Omega$ resistor. vx is voltage across this 3.3$\Omega$ resistor. Use voltage divider to find vx:

\begin{displaymath}v_x = \frac{3.3}{15 + 3.3 + 6.7} 5{\rm V} = 0.66 {\rm V}\end{displaymath}

8.
Problem 2-49: A simple interface circuit is shown below, with unknown R in series. You could also do the problem with unkown R in parallel.

\epsfig{file=hw2_soln_2_49.eps, width=3.5in}

(a)
To find iL, combine all resistors in series, then $i_L = 50/(1{\rm {k}} + R + 2{\rm {k}})$. You want iL = 1mA; this give R = 47k$\Omega$.
(b)
To find vL use a voltage divider:

\begin{displaymath}v_L = \frac{2{\rm {k}}}{1{\rm {k}}+R+2{\rm {k}}} 50\end{displaymath}

To get 10V across load, set vL to 10V, solve for R, get R = 7k$\Omega$.
(c)
From Part (b), use seven 1k$\Omega$ resistors in series to get R.
9.
Problem 2-50: Call bottom part of pot resistance Rx; then upper part is 1M$\Omega$ - Rx. Combine Rx in parallel with 1M$\Omega$ resistor: $R_{eq} = R_x 1{\rm {M}}/(R_x + 1{\rm {M}})$. 5V is voltage across Req. Then use voltage divider:


\begin{displaymath}5 = \frac{R_{eq}}{(1 - R_x) + R_{eq}} 10 = \frac{R_x 1{\rm {M}}} {R_x + 1{\rm {M}} - R_x^2}\end{displaymath}

This gives Rx = 0.62M$\Omega$ or Rx = -1.62$\Omega$. Negative resistance is impossible, so the corrrect answer is 0.62M$\Omega$. This means the wiper is 62% of the way up from the bottom.

10.
Problem 2-29: The device can have either 0 current through it or 1V across it. To do the problem, assume iD = 0, then solve (using KVL and KCL). Then assume vD = 1V, and solve again. Only one of these solutions will work - this is the correct solution.
(a)
vs = 2V. If iD = 0, find vD = vS = 2V. (Can't be true, because vD cannot be higher than 1V.) If vD = 1V, then $i_D = (2V-1V)/10\Omega = 0.1$A. This solution is okay: if vD = 1V, then iD can be any positive current. Thus, vD = 1V, iD = 0.1A.

vs = 0.5V: If iD = 0 then vD = vS = 0.5V. (Can be true). If vD= 1V then $i_D = (0.5V - 1V)/10\Omega = -50$mA. This cannot be true: iDcannot be negative. Thus, iD = 0, vD = 0.5V.

vs = -2V: If iD = 0 then vD = vS = -2V. (Can be true). If vD= 1V then $i_D = (-2V - 1V)/10\Omega = -300$mA. This cannot be true: iDcannot be negative. Thus, iD = 0, vD = -2V.

(b)
Device in nonlinear, nonbilateral, and passive ( $p_D = v_D i_D \ge 0$).


Bill Rison
1998-09-11