ECE35_SP17_HW04_Solution.pdf

ECE 35

Spring 2017

Homework #4 Solution

All homework problems come from the textbook, “Introduction to Electric Circuits”, by Svoboda & Dorf, 9th _{Edition. Question numbers in the 8}th _{edition is listed for reference.}

Question Number Svoboda & Dorf, 8th Edition Svoboda & Dorf, 9th Edition

1 P 5.2-1 P 5.2-1

2 P 5.2-3 P 5.2-3

3 P 5.2-6 P 5.2-6

4 P 5.3-6 P 5.3-6

5 P 5.3-10 P 5.3-9

6 P 5.3-15 P 5.3-13

7 P 5.4-5 P 5.4-5

8 P 5.4-13 P 5.4-12

9 P 5.4-15 P 5.4-14

10 P 5.4-17 P 5.4-16

11 P 5.5-7 P 5.5-7

12 P 5.5-9 P 5.5-8

13 P 5.5-12 P 5.5-11

P 5.2-1 The circuit shown in Figure P 5.2-1a has been divided into two parts. The circuit shown in Figure P 5.2-1b was obtained by simplifying the part to the right of the terminals using source transformations. The part of the circuit to the left of the terminals was not changed. (a) Determine the values of Rt and vt

in Figure P 5.2-1b.

(b) Determine the values of the current i and the voltage v in Figure P 5.2-1b. The circuit in Figure P 5.2-1b is equivalent to the circuit in Figure P 5.2-1a. Consequently, the current i and the

voltage v in Figure P 5.2-1a have the same values as do the

current i and the voltage v in Figure P 5.2-1b.

(c) Determine the value of the current ia in Figure P 5.2-1a.

Figure P 5.2-1

Solution:

= 2 = 0.5 V

t t R

v

 



(b) 9 4 2 ( 0.5) 0

9 ( 0.5)

1.58 A

4 2 i i

i

     

  

  



9 4 9 4( 1.58) 2.67 V

v  i   

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  resistors:

Source transformation at left; series resistors at right:

Parallel resistors, then source transformation at left:

Finally, apply KVL to loop

o

6 (9 19) 36i v 0

     

o

5 / 2 42 28 (5 / 2) 28 V

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

   





50 100 3

a

v   

P 5.3-6 Use superposition to find the value of the current i in Figure P 5.3-6.

Answer: i = 3.5 mA

Figure P5.3-6

Solution:

Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to the 30

mA current source.

1

2 6

30 6 mA 2 mA

2 8 6 12

a a

i  _ _   i  i _ _ 

 

   

Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i due to the

15 mA current source.

2

4 6

15 6 mA 2 mA

4 6 6 12

b b

i  _ _   i  i _ _ 

 

   





3

6 || 6 3

2.5 10 0.5 mA

6 || 6 12 3 12

i   _ _   _ _  

   

 

Finally, i   i1 i2 i3    2 2 0.5  3.5 mA

P 5.3-9 The input to the circuit shown in Figure P 5.3-9 is the voltage source voltage, vs. The output is the voltage vo. The current source current, ia, is used to adjust the relationship between the input and output. Design the circuit so that input and output are related by the equation vo = 2vs + 9.

Hint: Determine the required values of A and ia.

Figure P 5.3-9 Soluton:

s a

x

1

v v

i

R

 

s a

a o x

1

v v

v v A i A

R



  

1 o s

a

1

R v A v

v

R A

 



a s a o a

1 2 3

0

v v v v

i

R R R



   

1 2 s o

a a

1 2 1 3

0

R R v v

v i

R R R R



   

1 2 1 o s s o

a

1 2 1 1 3

0

R R R v A v v v

i

R R R A R R

           _   













3 1

2 1 1 2 1

1 1

0

R R A

R R

v v i

R R

R R A R R R A

 _   _                   

















3 1 2 2 1 2

o s a

2 3 1 2 1

0

R R R R R A A R

v v i

R R R A R R A

   

  

 

























o s a

3 1 2 2 1 3 1 2 2 1

R R A R R R A

v v i

R R R R R A R R R R R A

 

 

     

When R1 6 , R212 and  R312 





o s a

12 6 12

24 24

A A

v v i

A A

 

 

 

Comparing this equation to v_o 2v_s9, we requires

12 V 2 12 24 A A A A  _{   } 

Then 2vs 9 vo 2vs6iaso we require

a a

P 5.3-13 Theinput to the circuit shown in P 5.3-13 is the current i1, the output is

is the voltage vo. The current i2 is used to

adjust the relationship between the input and output. Determine values of the current i2 and the resistance, R, that

cause the output to be related to the input by the equation: v_o  0.5i₁4

Figure P 5.3-13

Solution:

Using superposition









o 1 2

|| 4 4

2 2

6 || 4 2 || 4 4

R

v i i

R R

   

  _{ }  _{ }

  

   

Comparing to vo  0.5i14, we require













2 0.5 4 || 4 6 || 4 || 4 2 4

6 || 4

R

R R R R

R

 

 _ _         



 

and









4 4

2 4 2 4 4 A

2 R|| 4 4 i 2 4 || 4 4 i i

   

    

   

 _{ }   _{ } 

P 5.4-5 Find the Thévenin equivalent circuit for the circuit shown in Figure P 5.4-5.

Answer: voc = –2 V and Rt = –8/3 Ω

Figure P 5.4-4

Solution:

Find voc:

Notice that voc is the node voltage at node a. Express

the controlling voltage of the dependent source as a function of the node voltage:

va = voc

Apply KCL at node a:

6 3

0

8 4 4

oc oc oc v v v      _{ }  _{ }    

6 v_oc 2v_oc 6v_oc 0 v_oc 2 V

       

Find Rt:

We’ll find isc and use it to calculate Rt. Notice that the

short circuit forces

va = 0

Apply KCL at node a:

6 0 0 3

0 0

8 4 4 isc

     _{ }  _{ }      6 3 A 8 4 sc

i  

2 8

3 4 3

P 5.4-12 The circuit shown in Figure P 5.4-12 contains an adjustable resistor. The resistance R

can be set to any value in the range 0 ≤ R ≤ 100 kΩ.

(a) Determine the maximum value of the current ia that can be obtained by adjusting R. Determine the corresponding value of R.

(b) Determine the maximum value of the voltage va that can be obtained by adjusting R. Determine the corresponding value of R.

(c) Determine the maximum value of the power supplied to the adjustable resistor that can be obtained by adjusting R. Determine the corresponding value of R.

Figure P 5.4-12

Solution: Replace the part of the circuit that is connected to the variable resistor by its Thevenin equivalent circuit:





a

36 12000 i

R



 and a 36

12000 R v

R

 

2

a a

36 12000

p i v R

R

 

 _{  }



 

(a) _a 36 3 mA

0 12000

i  

 when R = 0  (a short circuit).

(b)

5

a ₅

10

36 32.14 V 10 12000

v  

 when R is as large as possible, i.e. R = 100 k.

(c) Maximum power is delivered to the adjustable resistor when RR_t 12 k. Then

2

a a

36

12000 0.027 27 mW 12000 12000

pi v _ _  



P 5.4-14 The circuit shown in

Figure P 5.4-14 contains an unspecified resistance, R. Determine the value of R in each of the following two ways.

(a) Write and solve mesh equations. (b) Replace the part of the circuit

connected to the resistor R by a Thévenin equivalent circuit. Analyze the resulting circuit.

Figure P 5.4-14

Solution: (a)

3 2 0.25 A

i  i Apply KVL to mesh 1 to get









20 i i 20 i i 400

Apply KVL to the supermesh corresponding to the unspecified resistance to get









2 3 1 3 1 2

40i 10i 20 i i 20 i i 0

1 1

2 2

3 3

0 1 1 0.25 1.875

40 20 20 40 0.750

40 60 30 0 1.000

i i

i i

i i

   



     

   

 _{ }  _  _{ } 

   

     

   

     

 _{ }   _{ }  

Apply KVL to mesh 2 to get













2 2 3 1 2

2 3

20 40

40i R i i 20 i i 0 R i i i 30

i i

 

        



(b)

oc

20 40

40 40 12 V

20 20 10 40

v _ _ _ _  

 

   

t 18

R  

12

0.25 30

18 R R

   

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:

m m

m

5 50 R v

R

 

 _ _



(a) _{mi m} m

lim 5 V

R

v v



 

(b) When Rm 1000 ,  vm4.763 V so

% error = 5 4.762 100 4.76% 5

 _{ }

(c)

m

m m

m m

5 5

50

0.02 0.98 2450

5 50

R

R R

R R

 

 _ _



 

     



P 5.5-7 Determine the value of the resistance R in the circuit shown in Figure P 5.5-7 by each of the following methods:

(a) Replace the part of the circuit to the left of terminals a–b by its the Norton equivalent circuit. Use current division to determine the value of R.

(b) Analyze the circuit shown Figure P 5.5-6 using mesh equations. Solve the mesh equations to determine the value of R.

Figure P 5.5-7

Solution: (a) Replace the part of the circuit that is connected to the left of terminals a-b by its Norton equivalent circuit:

 

oc 25 5000 b 25 V

v   i 

Apply KVL to the supermesh corresponding to the dependent source to get

 

b b b

5000i 10000 3i 25 0 i 1 mA

     

Apply KCL to get

sc 3 b 3 mA

i  i 

Then oc t sc 8.33 k v R i   

Current division gives 8333

0.5 3 41.67 k

8333 R

R

   



(b)

Notice that i_band 0.5 mA are the mesh currents. Apply KCL at the top node of the dependent source to get

3

b b b

1

0.5 10 4 mA

6 i     i  i 

Apply KVL to the supermesh corresponding to the dependent source to get









b

5000i 10000 R 0.5 10 25 0

     









3 3

1

5000 10 10000 0.5 10 25

6 R      _  _      3 125

6 _{41.67 k}

0.5 10

R _  

P 5.5-8 Find the Norton equivalent circuit ofthis circuit:

Solution

Simplify the circuit using a source transformation:

Identify the open circuit voltage and short circuit current.

Apply KVL to the mesh to get:





Then

oc 3 x 3 V

Express the controlling current of the CCVS in terms of the mesh currents:

x 1 sc

i  i i

The mesh equations are



 



1 1 sc 1 sc 1 sc

10i 2 i i 3 i i 150  15i 5i 15 and





sc 1 sc 1 sc

4

3 0

3 i  i i   i  i

so

sc sc sc

4

15 5 15 1 A

3i i i

 _{    }

 

 

The Thevenin resistance is

t

3 3 1

R   

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 A 6

2 6 V

a

a a

oc a

i

i i

v i



   

  

 

12 6 2 3 A

2

3 2 3 2 A

3

a a a

sc a sc

i i i

i i i

    

     

6 3 2

t

R    

P 5.6-1 The circuit shown in Figure P 5.6-1 consists of two parts separated by a pair of terminals. Consider the part of the circuit to the left of the terminals. The open circuit voltage is voc = 8V, and the short circuit current is isc = 2A. Determine the values of

(a) The voltage source voltage, vs, and the resistance R2. (b) The resistance R that maximizes the power delivered to

the resistor to the right of the terminals, and the corresponding maximum power

Figure P 5.6-1

Solution:

(a) The value of the current in R2 is 0 A so voc 4ia. Then

KVL gives

   

a a s s a a oc

8i 4i V 0  V 12i 3 4i 3 v 24 V

Next, KVL gives,

a a a

8i 4i 240  i 2 A

and

 

 

2 2 2

a sc

4i R i  4 2 R 2  R  4

(b) The powerdelivered to the resistor to the right of the terminals is maximized by setting R equal to the Thevenin resistance of the part of the circuit to the left of the terminals:

oc t

sc

8 4 2 v R R

i

    

Then

 

2 ₂

oc max

t

8

4 W

4 4 4

v p

R