Chapter 5 Solution P5.2- 2, 3, 6 P5.3-3, 5, 8, 15 P5.4-3, 6, 8, 16 P5.5-2, 4, 6, 11 P5.6-2, 4, 9 P 5.2-2 Consider the circuit of Figure P 5.2-2. Find ia by simplifying the circuit (using source transformations) to a single-loop circuit so that you need to write only one KVL equation to find ia.

Figure P 5.2-2

Solution:

Finally, apply KVL: 1610 3 4 0 2.19 A3a a ai i i− + + − = ∴ =

(checked using LNAP 8/15/02)

P 5.2-3 Find vo using source transformations if i = 5/2 A in the circuit shown in Figure P 5.2-3. Hint: Reduce the circuit to a single mesh that contains the voltage source labeled vo. Answer: vo = 28 V

Figure P 5.2-3

Solution:

Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors:

Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top:

Source transformation at left; series resistors at right:

Parallel resistors, then source transformation at left:

Finally, apply KVL to loop

o6 (9 19) 36 0i v− + + − − =

o5 / 2 42 28 (5 / 2) 28 Vi v= ⇒ = − + =

(checked using LNAP 8/15/02) P 5.2-6 Use source transformations to find the value of the voltage va in Figure P 5.2-6. Answer: va = 7 V

Figure P 5.2-6

Solution: A source transformation on the right side of the circuit, followed by replacing series resistors with an equivalent resistor:

Source transformations on both the right side and the left side of the circuit:

Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources with an equivalent current source:

Finally, ( ) ( ) ( )50 100 1000.21 0.21 7 V50 100 3av = = =

+

(checked using LNAP 8/15/02)

P5.3-3 The circuit shown in Figure P5.3-3 has two inputs, vs and is, and one output io. The output is related to the inputs by the equation

o s si a i bv= + Given the following two facts:

The output is io = 0.45 A when the inputs are is = 0.25 A and vs = 15 V. and

The output is io = 0.30 A when the inputs are is = 0.50 A and vs = 0 V.

Determine the values of the constants a and b and the values of the resistances are R 1 and R 2. Answers: a = 0.6 A/A, b = 0.02 A/V, R 1 = 30 Ω and R 2 = 20 Ω.

Figure P5.3-3

Solution: From the 1st fact:

( ) ( )0.45 0.25 15a b= + From the 2nd fact:

( ) ( ) 0.300.30 0.50 0 0.600.50

a b a= + ⇒ = =

Substituting gives ( )( ) ( ) ( )( )0.45 0.60 0.250.45 0.60 0.25 15 0.02

15b b

−= + ⇒ = =

Next, consider the circuit:

s

1

1 2s o so1 0v

Ra i i i i

R R=

= = = +

so 11 2

1 2

0.60 2 3R

R RR R

= ⇒ =+

and s 1 2

ss oo2 0i

vb v i i

R R== = =

+

so 1 21 2

1 10.02 50 0.02

R RR R

= ⇒ + = = Ω+

Solving these equations gives R 1 = 30 Ω and R 2 = 20 Ω.

Figure P5.3-5

P5.3-5 Determine v(t), the voltage across the vertical resistor in the circuit in Figure P5.3-5. Solution; We’ll use superposition. Let v1(t) the be the part of v(t) due to the voltage source acting alone. Similarly, let v2(t) the be the part of v(t) due to the voltage source acting alone. We can use these circuits to calculate v1(t) and v2(t).

Notice that v1(t) is the voltage across parallel resistors. Using equivalent resistance, we calculate 40||10 = 8 Ω. Next, using voltage division we calculate

( ) ( )18 12 2 V

8 40v t = =

+

Similarly v2(t) is the voltage across parallel resistors Using equivalent resistance we first determine 40||40 = 20 Ω and then calculate

( ) ( )( ) ( )220 12cos 5 8cos 5 V

10 20v t t t= =

+

Using superposition ( ) ( ) ( ) ( )1 2 2 8cos 5 Vv t v t v t t= + = +

P 5.3-8 Use superposition to find the value of the current ix in Figure P 5.3-8. Answer: i = 3.5 mA

Figure P5.3-8

Solution: Consider 8 V source only (open the 2 A source)

Let i1 be the part of ix due to the 8 V voltage source. Apply KVL to the supermesh:

( ) ( ) ( )1 1 16 3 3 8 0i i i+ + − =

18 2 A

12 3i = =

Consider 2 A source only (short the 8 V source)

Let i2 be the part of ix due to the 2 A current source. Apply KVL to the supermesh:

( ) ( )2 2 26 3 2 3 0i i i+ + + =

26 1 A

12 2i −= = −

Finally, 1 22 1 1 A3 2 6xi i i= + = − =

P 5.3-15 The circuit shown in Figure P 5.3-15 has three inputs: v1, i2, and v3. The output of the circuit is the current io. The output of the circuit is related to the inputs by

i1 = avo + bv2 + ci3 where a, b, and c are constants. Determine the values of a, b, and c.

Figure P 5.3-15

Solution:

( ) ( )

( ) ( )

1o 2

3

10 10 2010 40 20 12 40 10 10 40 20 12 40 10

20 1240 20 12 10 40 20 12

vi i

v

= − + − + + + + + + +

+ − + + + +

o 1 2 31 1 1

200 10 62.5i v i v = − + − + −

So 0.05, 0.1 and 0.016a b c= − = − = −

(checked: LNAP 6/19/04) P 5.4-3 The circuit shown in Figure P 5.4-3b is the Thévenin equivalent circuit of the circuit shown in Figure P 5.4-3a. Find the value of the open-circuit voltage, voc, and Thévenin resistance, Rt. Answer: voc = 2 V and Rt = 4 Ω

Figure P 5.4-3

Solution: The circuit from Figure P5.4-3a can be reduced to its Thevenin equivalent circuit in five steps:

(a)

(b)

(c)

(d)

(e) Comparing (e) to Figure P5.4-3b shows that the Thevenin resistance is Rt = 4 Ω and the open circuit voltage, voc = 2 V.

(checked using LNAP 8/15/02)

P 5.4-6 Find the Thévenin equivalent circuit for the circuit shown in Figure P 5.4-6.

Figure P 5.4-6

Solution: Find voc:

Apply KCL at the top, middle node: 2 3 0 18 V3 6

a a aa

v v v v−= + + ⇒ =

The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V Find isc:

Apply KCL at the top, middle node: 2 3 18 V3 6 3

a a a aa

v v v v v−= + + ⇒ = −

Apply Ohm’s law to the right-hand 3 Ω resistor : 18 6 V3 3a

scvi −

= = = −

Finally: 18 36

oct

sc

vRi

= = = − Ω−

(checked using LNAP 8/15/02)

P 5.4-8 A resistor, R, was connected to a circuit box as shown in Figure P 5.4-8. The voltage, v, was measured. The resistance was changed, and the voltage was measured again. The results are shown in the table. Determine the Thévenin equivalent of the circuit within the box and predict the voltage, v, when R = 8 kΩ.

Figure P 5.4-8

Solution:

oct

Rv vR R

=+

From the given data:

200062000 1.2 V

1600400024000

oct oc

toc

t

vR v

Rv

R

= + = ⇒ = − Ω=+

When R = 8000 Ω,

( )8000 1.2 1.5 V1600 8000

v = =− +

P 5.4-16 An ideal voltmeter is modeled as an open circuit. A more realistic model of a voltmeter is a large resistance. Figure P 5.4-16a shows a circuit with a voltmeter that measures the voltage vm. In Figure P 5.4-16b the voltmeter is replaced by the model of an ideal voltmeter, an open circuit. The voltmeter measures vmi, the ideal value of vm. As Rm → ∞, the voltmeter becomes an ideal voltmeter and vm → vmi. When Rm < ∞, the voltmeter is not ideal and vm > vmi. The difference between vm and vmi is a measurement error caused by the fact that the voltmeter is not ideal. (a) Determine the value of vmi. (b) Express the measurement error that

occurs when Rm = 1000 Ω as a percentage of vmi.

(c) Determine the minimum value of Rm required to ensure that the measurement error is smaller than 2 percent of vmi.

Figure P 5.4-16

Solution: Replace the circuit by its Thevenin equivalent circuit:

mm

m

550

Rv

R

= +

(a) mi mmlim 5 V

Rv v

→∞= =

(b) When m m1000 , 4.763 VR v= Ω = so

% error = 5 4.762 100 4.76%5

−× =

(c)

m

m mm

m

5 550

0.02 0.98 2450 5 50

RR R

RR

− + ≥ ⇒ ≥ ⇒ ≥ Ω

+

(checked: LNAP 6/16/04)

P 5.5-2 Two black boxes are shown in Figure P 5.5-2. Box A contains the Thévenin equivalent of some linear circuit, and box B contains the Norton equivalent of the same circuit. With access to just the outsides of the boxes and their terminals, how can you determine which is which, using only one shorting wire?

Figure P 5.5-2

Solution: When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will cool off. P 5.5-4 Find the Norton equivalent circuit for the circuit shown in Figure P 5.5-4.

Figure P 5.5-4

Solution:

P 5.5-6 The circuit shown in Figure P 5.5-6b is the Norton equivalent circuit of the circuit shown in Figure P 5.5-6a. Find the value of the short-circuit current, isc, and Thévenin resistance, Rt. Answer: isc = –24 A and Rt = –3 Ω

Figure P 5.5-6

Solution: To determine the value of the short circuit current, Isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit current. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage. Consequently,

1 24 Vv = − . The voltage at node 3 is equal to the voltage across a short, 3 0v = . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. 2av v= . The voltage at node 3 is equal to the voltage across a short, i.e. 3 0v = . Apply KCL at node 2 to get

1 2 2 31 3 22 3 48 3 16 V

3 6 a a

v v v vv v v v v

− −= ⇒ + = ⇒ − = ⇒ = −

Apply KCL at node 3 to get

( )2 32

4 9 9 16 24 A6 3 6 6sc a sc sc

v vv i v i i

−+ = ⇒ = ⇒ = − = −

Figure (a) Calculating the short circuit current, Isc, using mesh equations.

To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as

Tth

T

vRi

=

Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. 1 0v = . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. 2av v= . The voltage at node 3 is equal to the voltage across the current source, i.e. 3 Tv v= . Apply KCL at node 2 to get

1 2 2 31 3 22 3 3

3 6 T a

v v v vv v v v v

− −= ⇒ + = ⇒ =

Apply KCL at node 3 to get

2 32 2 3

4 0 9 6 06 3

9 6 03 6 0 2 6

T T

a T T

T T T T T

v vv i v v i

v v iv v i v i

−+ + = ⇒ − + =

⇒ − + =

⇒ − + = ⇒ = −

Finally,

3Tt

T

vRi

= = − Ω

Figure (b) Calculating the Thevenin resistance, T

thT

vRi

= , using mesh equations.

To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage. Consequently,

1 24 Vv = − . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. 2av v= . The voltage at node 3 is equal to the open circuit voltage, i.e. 3 ocv v= . Apply KCL at node 2 to get

1 2 2 31 3 22 3 48 3

3 6 oc a

v v v vv v v v v

− −= ⇒ + = ⇒ − + =

Apply KCL at node 3 to get

2 32 2 3

4 0 9 0 96 3 a oc

v vv v v v v

−+ = ⇒ − = ⇒ =

Combining these equations gives

( )3 48 9 72 Voc a oc ocv v v v− + = = ⇒ =

Figure (c) Calculating the open circuit voltage, voc, using node equations. As a check, notice that

( )( )3 24 72th sc ocR I V= − − = =

(checked using LNAP 8/16/02)

P 5.5-11 Determine values of Rt and isc that cause the circuit shown in Figure P 5.5-11b to be the Norton equivalent circuit of the circuit in Figure P 5.5-11a. Answer: Rt = 3 Ω and isc = –2 A

Figure P 5.5-11

Solution:

2 12

3 A6

2 6 V

aa a

oc a

ii i

v i

−= ⇒ = −

= = −

( )

12 6 2 3 A23 2 3 2 A3

a a a

sc a sc

i i i

i i i

+ = ⇒ = −

= ⇒ = − = −

6 32tR −

= = Ω−

P 5.6-2 For the circuit in Figure P 5.6-2, (a) find R such that maximum power is dissipated in R and (b) calculate the value of maximum power. Answer: R = 60 Ω and Pmax = 54 mW

Figure P 5.6-2

Solution: a) For maximum power transfer, set RL equal to the Thevenin resistance:

100 1 101L tR R= = + = Ω

b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin equivalent circuit:

The voltage across RL is ( )101 100 50 V101 101Lv = =

+

Then 2 2

max50 24.75 W101

L

L

vpR

= = =

P 5.6-4 Find the maximum power to the load RL if the maximum power transfer condition is met for the circuit of Figure P 5.6-4. Answer: max pL = 0.75 W

Figure P 5.6-4

Solution:

LL S

S L

2 2L S L

L 2L S L

( )

Rv v

R R

v v Rp

R R R

=

+

∴ = =+

By inspection, pL is max when you reduce RS to get the smallest denominator.

∴ set RS = 0 P 5.6-9 A resistive circuit was connected to a variable resistor, and the power delivered to the resistor was measured as shown in Figure P 5.6-9. Determine the Thévenin equivalent circuit. Answer: Rt = 20 Ω and voc = 20 V

Figure P 5.6-9

Solution: From the plot, the maximum power is 5 W when R = 20 Ω. Therefore:

Rt = 20 Ω and

( )2

max max 4 5 4 20 20 V4

ococ t

t

vp v p RR

= ⇒ = = =