Electric Circuits 1st Edition by Kang - Test Bank

Electric Circuits 1st Edition by Kang – Test Bank

Instant Download – Complete Test Bank With Answers

Sample Questions Are Posted Below

1
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5
Operational Amplifier Circuits
Test Bank
5.1 In the circuit shown below, let
R1 = 1 k, R2 = 2 k, R3 = 6 k, R4 = 500 , R5 = 5 k, R6 = 3 k, R7 = 1.5 k, v s = 1 V.
(a) write a node equation at node 1 by summing the currents leaving node 1. Express va as a
function of vo .
(b) write a node equation at node 2 by summing the currents leaving node 2. Express vn as a
function of vo .
(c) write a node equation at node 3 by summing the currents leaving node 3.
(d) find the numerical value of vo , vp , vn , va.
5.2 The circuit shown below generates output signal vo given by
vo = 10v 1  5v 2
where v1 and v2 are two input signals. Find R1 , R2 , R3 , and R4 .
U1
OPAMP
+
–
OUT
U2
OPAMP
+
–
OUT
R1
R2
R7
R4
R5
R6
R3
v s
0
0
0
0
v o
v a
3
v n
1
2
v p
2
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
5.3 The circuit shown below generates output signal vo given by
v o = 2v 1 + 3v 2
where v 1 , and v2 are input signals. Find R1 and R2 if R 3 = R4 = 6 k and R5 = 30 k.
5.4 Let
R1 = R3 = 2 k, R2 = 1 k, R4 = R5 = 1 k, R6 = R7 = 5 k, v 2 = 0.05 V, v 1 = 0.05 V
in the circuit shown below. Find the numerical values of v o1 , v o2 , and v o3 .
U1
OPAMP
+
–
OUT
0
v 2
U2
OPAMP
+
–
OUT Vo
Va1
R4
2
R3
0
v 1
R1
0
R2
3
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
5.5. In the circuit shown below, let
R1 = 1 k, R2 = 6 k, Ri = 4 k, Ro = 500 , A = 2000, v s = 1V, v p = 0.
(a) Write a node equation at node 1 by summing currents leaving node 1.
(b) Write a node equation at node 2 by summing currents leaving node 2
(c) Solve the two node equations to find numerical values of v n and v o .
(d) Find the value of is through R1 . Find the input impedance Rin = v s /i s .
0
+
–
0
v n
0
A(v p – v n)
v o
1 2
R1
R2
Ro
Ri
is
v s
v p
4
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
5.6 In the circuit shown below, let
R1 = 1 k, R2 = 4 k, Ri = 3 k, Ro = 2 k, A = 1000, v s = 1 V, v p = v s .
(a) Write a node equation at node 1 by summing currents away from node 1.
(b) Write a node equation at node 2 by summing currents away from node 2.
(c) Solve the two node equations to find numerical values of v n and v o .
(d) Find the value of is through Ri . Find the input impedance Rin = v s /i s .
5.7 In the circuit shown below, let
R1 = 1 k, R2 = 3.6 k, Ri = 10 k, Ro = 600 , A = 5000, v t = 1 V.
Find the value of it and the output resistance Rout = v t /i t .
0
+
0
0
v n
–
A (v p-v n)
1
v o
R1
2
R2
Ro
Ri
v s
v p is

1
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Solutions to Test Bank
Chapter 5
5.1
(a)
6 7
0 0 0o av v
R R
 
 
7
6
a o
R
v v
R
 
(b)
4 5
0 0n n ov v v
R R
 
 
4 5 5
1 1 o
n
v
v
R R R
 
  
 
5 4
4 5
4 5
1
1 1n o o
R R
v v v
R R
R R
  
(c)
1 3 2
0p s p p av v v v v
R R R
 
  
p nv v
1 3 2 2 1
1 1 1 1 s
p a
v
v v
R R R R R
 
    
 
74
1 3 2 4 5 2 6 1
1 1 1 s
o o
R vR v v
R R R R R R R R
 
      
2
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1
74
1 3 2 4 5 2 6
1
1 1 1
o s
R
v v
RR
R R R R R R R
  
     
vn = 0.2264 V, vp = 0.2264 V, va = – 1.2453 V, vo = 2.4906 V
%Chapter 5, Test Bank 5.1
clear all;format long;
R1=1000;R2=2000;R3=6000;R4=500;R5=5000;R6=3000;R7=1500;vs=1;
syms vn vp vo va
[vn,vp,vo,va]=solve((0-va)/R7+(0-vo)/R6,…
(vn-vo)/R5+vn/R4,…
(vp-vs)/R1+vp/R3+(vp-va)/R2,vp==vn,…
vn,vp,vo,va)
vn=vpa(vn,7)
vp=vpa(vp,7)
va=vpa(va,7)
vo=vpa(vo,7)
Answers:
vn =
0.2264151
vp =
0.2264151
va =
-1.245283
vo =
2.490566
5.2
One solution:
R1 = 4 k, R2 = 5 k, R3 = 1 k, R4 = 9 k,
There are many solutions. Superposition principle can be used to show that the answer is correct.
5.3
1
6 3
2
k
R k

  
2
6 2
3
k
R k

  
5.4
2
21
1 R
vv
i 
 = – 0.1 mA
1 1 1 1vo i R v  = – 0.25 V
3
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2 1 3 2vo i R v   = 0.25 V
 61
3 2 1
2 4
2
1 RR
vo v v
R R
 
   
 
= 2.5 V
%Ch5 Test Bank 5.4
%Instrumentation Amp
clear all;format long;
v2=0.05;v1=-0.05;
R1=2000;R2=1000;R3=2000;R4=1000;R5=1000;R6=5000;R7=5000;
i1=(v1-v2)/R2
vo1=i1*R1+v1
vo2=-i1*R3+v2
vo3=(1+2*R1/R2)*R6/R4*(v2-v1)
Answers:
i1 =
-1.000000000000000e-04
vo1 =
-0.250000000000000
vo2 =
0.250000000000000
vo3 =
2.500000000000000
5.5
1 2
0n s n o n
i
v v v v v
R R R
 
  
 
2
0 0o no n
o
v A vv v
R R
   
 2
1 2 1 2 1 1
i o s
n
o i o i i i
R R R v
v R R R R R R R R R R AR R

      = 0.003235 V
 2
1 2 1 2 1 1
i o s
o
o i o i i i
R R R A v
v R R R R R R R R R R AR R
  
      = – 5.9725 V
1
s n
s
v v
i R

 = 0.0009968 A
s
in
s
v
R i
 = 1003.2457 
% Chapter 5, Test Bank 5.5
clear all;
R1=1000;R2=6000;Ri=4000;Ro=500;vs=1;A=2000;
syms vn vo
[vn,vo]=solve((vn-vs)/R1+vn/Ri+(vn-vo)/R2,…
(vo-vn)/R2+(vo-A*(0-vn))/Ro,vn,vo);
4
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
is=(vs-vn)/R1;
Rin=vs/is;
vn=vpa(vn,11)
vo=vpa(vo,11)
is=vpa(is,11)
Rin=vpa(Rin,11)
Answers:
vn =
0.0032352392211
vo =
-5.9725004666
is =
0.00099676476078
Rin =
1003.24574
5.6
2 1
0n s n o n
i
v v v v v
R R R
 
  
 
2
0o s no n
o
v A v vv v
R R
   
 1 2
1 2 1 2 1 1
o i s
n
o i o i i i
R R R AR v
v R R R R R R R R R R AR R
 
      = 0.99306244 V
 1 1 2
1 2 1 2 1 1
o i i s
o
o i o i i i
R R R R A R R A v
v R R R R R R R R R R AR R
 
      = 4.9561 V
s n
s
i
v v
i R

 = 0.00000231252 A
s
in
s
v
R i
 = 432,428.5714 
% Chapter 5, Test Bank 5.6
clear all;
R1=1000;R2=4000;Ri=3000;Ro=2000;vs=1;A=1000;
vp=vs;
syms vn vo
[vn,vo]=solve(vn/R1+(vn-vs)/Ri+(vn-vo)/R2,…
(vo-vn)/R2+(vo-A*(vp-vn))/Ro,vn,vo);
is=(vs-vn)/Ri
Rin=vs/is
vn=vpa(vn,11)
vo=vpa(vo,11)
is=vpa(is,11)
Rin=vpa(Rin,11)
Answers:
vn =
0.99306243806
vo =
5
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
4.9560621077
is =
0.0000023125206475
Rin =
432428.57143
5.7
1
1 1
1 1 2 2 1
2
1
i
i i
n t t
i i i
i
R R
R R R R
v v v
R R R R R R R RR R R

    
= 0.2016 V
  1 1
1 2 2 1 1 2 2 1
2 2
0
i i
t t t t
t nt n i i i i
t
o o
R R R R
v v v A v
v A vv v R R R R R R R R R R R R
i R R R R
 
     
    = 1.682 A
1 1
1 2 2 1 1 2 2 1
2
1
1 1
t
out
i it
i i i i
o
v
R R R R Ri A
R R R R R R R R R R R R
R R
 
 
   

= 0.5945 
% Chapter 5, Test Bank 5.7
clear all;
R1=1000;R2=3600;Ri=10000;Ro=600;vt=1;A=5000;
syms vn
vn=solve(vn/R1+vn/Ri+(vn-vt)/R2,vn)
it=(vt-vn)/R2+(vt-A*(0-vn))/Ro
Rout=vt/it;
vn=vpa(vn,11)
it=vpa(it,11)
Rout=vpa(Rout,11)
Answers:
vn =
0.20161290323
it =
1.6819959677
Rout =
0.59453174632