E-84 Electric & Magnetic Circuits & Devices
Solution Key for PS #2
Sam Tanenbaum
Problems from Chapter 2
2. a. Given the circuit shown below, find the value for RL which maximizes the power to
the load.
PL is a maximum when RL = RTh, the Thevenin resistance for the rest of the
circuit. RTh is the resistance from A to ground with the voltage sources off and
RL removed. Hence
RTh = 2||6 = 12/8 = 1.5 Ω
b. For RL = 1.5 , use KCL to determine VA, and find the power drawn from (or given to)
each of the voltage sources and the power to the load.
Using KCL at A, we have:
(12 – VA)/2 + (8 – VA)/6 = VA/1.5
so that we find
VA = 5.5 V
Hence PL = VA2/RL = 20.17 W
P12V = 12I12V = 12•(12 - VA)/2 = 39 W (drawn from source)
P8V = 8•(8 - VA)/6 = 3.33 W (drawn from source)
3. A 120 V heater has two resistive elements. Switches allow them to be attached to the
voltage source in four configurations:
 R1 alone
 R2 alone
 R1 in parallel with R2
 R1 in series with R2.
If the “high” setting draws 1500 W, and the “low” setting draws 360 W:
a. Find R1 and R2.
Note first that for all configurations, P = V2/REqivalent, where V = 120 V, and there
are four values for REq:
 R1,. R2, R1 + R2, and R1•R2/(R1 + R2)
The highest power will come from the resistors in parallel (lowest REq) and the
lowest power will come from the resistors in series (highest REq).
Hence:
(120) 2 • (R1  R2 )
(120) 2
 1500W and
= 360W or
R1 • R2
R1  R2

Sam Tanenbaum
1
E-84 fall, 2010
(1) (R1 + R2)/R1R2= 1/9.6 and
(2) R1 + R2 = 40 so that R1 = 40 – R2
Substituting (2) into (1) we find that R22 – 40R2 + 384 = 0, so that
R2 = 24 or 16 and R1 = 40 – R2. Hence
one resistor has R = 24 Ω and the other has R = 16 Ω
b. Find the power in the two intermediate settings.
The two intermediate power levels are (using just one resistor)
V2/16 = 900 W and V2/24 = 600 W
6. a. Find the Thevenin equivalent circuit for the circuit shown below. Show that VTh =
2100 V and RTh = 700 .
VTh is the open circuit voltage when RL is removed. Thus it is the voltage at
the node above the 100 V source, which we call A. We can find this voltage
immediately by using KCL at A. Since both current sources must flow into A
we have:
At A: 2 + 3 = (VA – 100)/400 or VTh = VA = 2100 V
To find RTh we want the resistance looking left from RL with open circuits
replacing both current sources, and a short circuit replacing the voltage source.
We see that the 100 Ω and 200 Ω resistors are left dangling, so that
RTh = 300 + 400 = 700 Ω
b. Working with the original circuit (not the Thevenin equivalent) show that IN = 3 A.
IN is the short circuit current through RL.. To find it we once again use KCL at
A. Once again both current sources flow into A, but with RL = 0 we have
At A: 5 + (100 – VA)/400 = VA/300 or 2100 – VA = (4/3)VA
Solving for VA we find VA = 900 V, so that IN = 900/300 = 3 A.
This agrees with the requirement that IN = VTh/RTh = 2100/700 = 3 A
c. Show that the maximum power that can be delivered to the load is 1,575 W.
Using the Thevenin Equivalent Circuit, peak power to the load occurs when RL =
RTh. For this case,
IL = VTh/2RTh = 2100/1400 = 1.5 A. Hence the max power to the load is
PL = IL2RL = (1.5)2700 = 1575 W
8. Consider the circuit shown below and answer the following questions:.
Sam Tanenbaum
21
E-84 fall, 2010
a. How much power is produced by the 100 V source?
P = VI = 1,000 W
b. What is the value for RL?
V = IR, so that R = V/I = 10 Ω. But R = 50||RL = 50RL/(50 + RL) = 10
Thus (dividing by 10 and collecting terms)
4RL = 50, so that RL = 12.5 Ω
c. Show that power to RL is: PL = 800 W.
PL = V2/RL = 10,000/12.5 = 800 W
9. Find IL in the circuit shown below by three different methods:



Use Thevenin’s Equivalent Circuit
Use Norton’s Equivalent Circuit
Use Node Analysis
Using Thevenin Equivalent Circuit
VTh = VA when IL = 0. Since VC = 10 V, we use the voltage divider formula to
find VA = 10•(2/8) = 2.5 Volts = VTh
RTh = resistance from A to ground with short circuit in place of voltage source
and load resistance removed. Hence RTh = 2||6 = 12/8 = 1.5 Ω
Thus IL = VTh/(Rth + 8) = 2.5/(9.5) = .263 A
Using Norton’s Equivalent Circuit
IN = IL when RL = 0 (so that no current flows through the 2 Ω resistor)
Hence IN = VC/6 = 10/6 = 5/3 A
From the current divider formula, we have for the Norton circuit:
IL = IN•RTh(RL + RTh) = (5/3)•(1.5)/(8 + 1.5) = 2.5/9.5 = .263 A
Sam Tanenbaum
31
E-84 fall, 2010
Using Node Analysis (Note that VC = 10 V, so the only unknown is the voltage at A)
At A: (10 – VA)/6 = VA/2 + VA/8
Multiplying through by 24 and collecting terms we find
40 = VA(4 + 12 + 3), so that VA = 40/19 V
Hence IL = VA/8 = 5/19 = .263 A
EXTRA PROBLEM
As shown in the figure below
 An ammeter can be modeled as an ideal measurement instrument and a
series resistor, ra, and
 A voltmeter can be modeled as an ideal measurement instrument and a
parallel resistance rv.
In the circuit below, if these instruments are ideal (ra = 0 and rv = ∞), when V0 = 10 V:
 The voltmeter reads V = 5 Volts and
 The ammeter reads I = 5/R Amps
However, these readings are significantly altered unless R satisfies the condition:
ra << R << rv
Use KCL to find V and I for ra = 10 Ω and rv = 104 Ω, when:
a. R = 5 Ω so that IIdeal = 1 A
b. R = 500 Ω so that IIdeal = 10 mA
c. R = 5•104 Ω so that IIdeal = 0.1 mA
(Fortunately, real ammeters have smaller values for ra than this and real voltmeters have larger
values for rv!)
10  V
V
V
a. At V: Current In = I =
 Current Out =  4
15
5 10
Solving for V, we find V = 2.5 volts, so that I = 0.5 A
10  V
V
V
b. At V: Current In = I =
= Current Out =
 4
510
500 10
 Solving for V, we find V = 4.83 volts, so that I = .0101A = 10.1 mA

Sam Tanenbaum
41
E-84 fall, 2010
10  V
V
V
= Current Out =
 4
4
50,010
5 • 10
10
-4
Solving for V, we find V = 1.43 volts, so that I = 1.71•10 A = .171 mA
c. At V: Current In = I =

Sam Tanenbaum
51
E-84 fall, 2010