EE 211 - Homework 2 Solutions

1.
Problem 2.21
(a)
Circuit C3: everything in series; combine two resistors into single 7.1 k resistor, then . .

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

Then .

2.
Problem 2.26
(a)
ix = 0.2A. Voltage drop across each 25 resistor is 0.2A 25 = 5V. By KVL, vx = 25V - 5V - 5V = 15V. Voltage drop across each 2 resistor is 0.2A 2 = 0.4V. By KVL, vy = vx - 0.4V - 0.4V = 14.2V.
(b)
Power delivered at line input is 3 W. Everything is in series so ix goes through load. Power delivered to load is 2.84 W.
(c)
95%.

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

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

(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 resistor gives 200mA current source in parallel with a 40 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 resistor in parallel with a 133 resistor - these add to give a 31 resistor.

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 resistor parallel to 2A current source.
(b)
Circuit C2: Do source transformation on 2A current source parallel to 5resistor. Get 10V current source in series with 5 resistor, all in series with the 10 resistor. Add the 5 resistor and 10resistor in series; final circuit is 10V voltage source in series with 15 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 resistor.
5.
Problem 2-41
(a)
Circuit C1: Simple current division:

(b)
Circuit C2: Simple voltage division:

(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:

(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:

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

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

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

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

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.

(a)
To find iL, combine all resistors in series, then . You want iL = 1mA; this give R = 47k.
(b)
To find vL use a voltage divider:

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

This gives Rx = 0.62M or Rx = -1.62. Negative resistance is impossible, so the corrrect answer is 0.62M. 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 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 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 mA. This cannot be true: iDcannot be negative. Thus, iD = 0, vD = -2V.

(b)
Device in nonlinear, nonbilateral, and passive ( ).

Bill Rison
1998-09-11