Download Homework 1 Solution

Homework 1
1)
On your first day on your new job as an engineer, your boss comes to you with a problem. She
says your company needs to model what is inside a small black box. It has two terminals. She
tells you it contains 12 resistors in some unknown circuit and you cannot open the box.
In the lab, you apply a series of voltages to the terminals and measure the current as shown:
I -->
V
You get the following values:
V (volts)
0
1
2
3
I (amps)
0
0.25
0.5
0.75
Graph the I vs. V relationship. What would you use to model the black box? Show the values of
any elements you use.
Solution:
Black Box
0.8
0.7
Amps
0.6
0.5
0.4
0.3
Black Box
0.2
0.1
0
0
1
2
Volts
3
4
R
1
3 1

 4
slope 0.75  0.25
The block box circuit can be modeled as a single 4 ohm resistor across the terminals. There is no need to
divide the circuit into 12 resistors since:
1. We do not know the actual form of the circuit
2. We are only interested in the I/V curve and relationship
This shows how we can simplify complex circuits into easily understood models.
2)
In this circuit:
4V
R1
10
R2
20
AC =
DC = V
TRAN =
a.
b.
c.
d.
e.
f.
How many nodes are there?
How many meshes are there?
Find the current in each resistor.
Find the total current delivered by the source.
Find the power being delivered or absorbed by each element, including the source.
(Watch your +/- signs)
What is the sum of these powers?
Solution:
a. 2
b. 2
c.
4Volts
 0.4 A
10
4Volts

 0.2 A
20
I R1 
IR2
d. IS = IR1+ IR2 = 0.6A
e. Source: PS = 4V(-0.6A)=-2.4W R1: PR1 = 4V(0.4A)=1.6W R2: PR2 = 4V(0.2A) = 0.8W
f. Total Power = PS + PR1 + PR2 = -2.4W + 1.6W + 0.8W = 0W
3) In this circuit:
R1
10
I1
4Adc
a.
b.
c.
d.
e.
f.
R2
20
How many nodes are there?
How many meshes are there?
Find the voltage across each resistor.
Find the voltage across the source.
Find the power being delivered or absorbed by each element, including the source.
(Watch your +/- signs)
What is the sum of the powers?
Solution:
g. 3
h. 1
i.
VR1  4 A(10)  40V
VR 2  4 A(20)  80V
j. VS = VR1+ VR2 = 120V
k. Source: PS = 120V(-4A)=-480W R1: PR1 = 40V(4A)=160W R2: PR2 = 80V(4A) = 320W
l. Total Power = PS + PR1 + PR2 = -480W + 160W + 320W = 0W
4) Given the following circuit, answer the questions below. Show all your work and indicate which
of Kirchoff’s Laws or Ohm’s Law you are using.
R1
R3
5
10V
1
+
V=?
+
R2
-
1A
2V
4
a. The voltage measured across R2 is 2V as shown. What is the voltage across the 1A
current source?
Ans: Ohm’s Law: IR2=2V/4Ω = 0.5A
KCL: Current into node between R1 and R2 = Current out of node.
IR1=IR2=0.5A
Ohm’s Law: VR1=0.5A(1Ω) =0.5V
KVL: Voltage across the current source = VR1+VR2 = 2V+0.5V=2.5V
b. What is the voltage across R1?
Ans: From above: Ohm’s Law: VR1=0.5A(1Ω)=0.5V
c. Is the 1A current source generating or absorbing power? How much?
Ans: Absorbing, current is flowing INTO the positive voltage terminal.
P=IV=1A(2.5V)=2.5W
d. Is the 10V source generating or absorbing power? How much?
Ans: KCL: Current out of center node = 0.5A + 1A = 1.5A. Therefore current into node
from R3 is 1.5A. This is the same as the current FROM the source.
PS = 10V(-1.5A) = -15W
5) Given the circuit below:
V
a.
b.
c.
d.
R1
R2
2
4
What is the current through R1 in terms of V?
What is the current through R2 in terms of V?
What is the total current being supplied by the voltage source?
Given the answer from c., what is the equivalent resistance seen by the voltage source?
In other words, what single resistance value would have to be put across the voltage
source to draw the same total current?
e. What is the general equation of this resistance in terms of R1 and R2?
Solution:
V V

R1 2
V V


R2 4
a.
I R1 
b.
IR2
V V
1 1 3
 V     V
2 4
2 4 4
V
V
4
d. Req 

 
Is 3 V 3
4
c.
I S  I R1  I R 2 
e.
1 
 1
IS  V  

 R1 R 2 
V
V
1
Req  

1   1
1 
IS
 1
V 
  

 R1 R 2   R1 R 2 
6) Given the circuit below:
R1
100V
100
101V
a. What is the voltage across R1?
Ans: Using KVL(assuming VR1 has + on left side): -100V+VR1+101V=0 Therefore VR1=-1V
b. If you removed the resistor, R1 and touched the plus terminals of both sources with
your hands, would you get a shock? Why or why not?
Ans: DO NOT TRY THIS AT HOME! You would probably not get a shock. The voltage
difference is only 1 volt. However, in the real world there are other things that may
make this dangerous. The point of the question is that, despite the high voltages, the
difference is only 1V.
c. Calculate the power being absorbed or generated by each source (with R1 back in
place).
Ans: There is 1V across the resistor. IR1= 1V/100Ω = .01A This current is flowing right to
left through the resistor.
P101V=101V(-.01A)= -1.01W (generating power)
P100V=100V(.01A)=+1W (absorbing power)
7) Your flashlight needs two new batteries. If you put them in correctly with the plus terminal of
one touching the minus terminal of the other, the flashlight works. However, if you put the
batteries in with their minus terminals touching as shown below, the flashlight does not light.
Why? What law prevents the light from turning on? (Hint: the lamp behaves like a resistor.)
1.5V
1.5V
1.5V
1.5V
CORRECT
WRONG
Solution: Using KVL and specifying the voltage across the lamp as VL, in the correct arrangement:
-1.5V-1.5V +VL=0, therefore VL= 3.0V
In the wrong arrangement:
1.5V -1.5V+VL = 0, therefore VL=0V
8) In the circuit below:
R1
R2
10
1
I1
1A
5V
R3
4
a. What is the voltage across the 1A current source?
Ans:-VI1+VR1+5V=0 VR1=(10Ω)1A=10V, therefore VI1=15V
b. Is it generating or absorbing power? How much?
Ans: PI1=15V(-1A)=-15W generating
c. How much power is being absorbed by R1?
Ans:PR1=I2(10)=10W
d. How much power is being absorbed by R3?
Ans:KVL around the right mesh: -5V+VR2+VR3=0 Therefore, VR2+VR3=5V
Using voltage division:
 R3 
4
VR 3  5 
  5    4V
5
 R3  R2 
VR 32 16
PR 3 
  4W
R3
4
e. Is power being generated or absorbed by the 5V voltage source? How much?
Ans: Using KCL at the node on the positive terminal of the 5V source, we see that 1A is
entering the node from the 10 ohm resistor and also 1A is leaving the terminal through
the 1 ohm resistor:
5V
 1A .Therefore, NO CURRENT can be flowing in or out of
1  4 
the 5V source. It is neither generating or absorbing power, even though we used it in the
KVL calculation above.
f. If the current source were to suddenly fail (go to zero), what would be the new power
being absorbed by R3?
Ans: It would not change. The voltage across the resistor would still be determined by
the 5V voltage source and voltage division.
g. If the current source fails, is power being generated or absorbed by the 5V voltage
source? How much?
Ans: Since the 1A from the current source is gone, the 5V would have to supply this
current to maintain KVL in the right mesh. The current would flow OUT from the source:
P5v Source=5V(-1A)= -5W, generating.
9) In the circuit below, the voltage source, V1, will burn up if it has to generate more than 10 watts.
However, R3 will burn up if it has to absorb more than 9 watts. If we are able to slowly increase
V1 from zero, which will fail first: the voltage source or R3? At what voltage on V1 will this
occur? (Hint: Work backwards.)
R2
1
V1
R1
R3
5
4
Solution: This is a typical engineering problem: two constraints. One way to solve this is to
divide the problem into two parts and work backwards on each from power back to V1: One, a
relation between V1 and the power it must generate and two, a relation between V1 and the
power absorbed by R3.
One: Combining R2 and R3 in series: R=R2+R3 = 1Ω+4Ω=5Ω. This in parallel with R1:
RT 
5  5 
 5  5 
 2.5
This is the total load seen byV1. Therefore the power is:
PV 1 
V12
solving for V1: V1  PV 1 RT
RT
V1 will burn up if it generates 10W. This will occur when V1=5V
Two: Using voltage division, the voltage across R3 is:
 R3  4
VR 3  V1 
  V1
 R3  R2  5
Therefore, the power is:
16 2
V
VR23 25 1
PR 3 

 .16V12
R3
4
solving for V1 :
V1 
PR 3
.16
Therefore, for 9 watts in R3, V1=7.5V
This is greater than the 5V at which V1 will burn out and therefore V1 will burn out first.