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Problem 1.
The input to this circuit is the voltage of the
independent voltage source. The output is the
voltage measured by the meter. Show that the
output is proportional to the input. Determine the
value of the constant of proportionality.
Problem 2.
The input to this circuit is the voltage of the
independent voltage source. The output is the
current measured by the meter. Show that the
output is proportional to the input. Determine the
value of the constant of proportionality.
Problem 3
Determine the value of the voltage measured by
the meter.
1
Problem 4
Determine the value of the current measured by
the meter.
Problem 5
Determine the value of the resistance R required
to cause the value of the voltage measured by the
voltmeter to be 4 V.
Problem 6
The input to this circuit is the voltage of the
independent voltage source, v s . The output is the
current measured by the meter, i m .
(a) Suppose v s = 15 V . Determine the value of the
resistance R that causes the value of the current
measured by the ammeter to be i m = 5 A.
(b) Suppose v s = 15 V and R = 24 Ω. Determine
the value of the current measured by the ammeter.
(c) Suppose R = 24 Ω. Determine the value of the
input voltage, v s , that causes the value of the
current measured by the ammeter to be i m = 3 A.
2
Solution 1
va =
20 & 20
1
vs = vs
20 + ( 20 & 20 )
3
3
1
 12 
vo = 
 (10v a ) = × 10 × v s = 2v s
5
3
 12 + 8 
So vo is proportional to vs and the constant of
V
proportionality is 2
.
V
Solution 2
ib =
vs
2 + ( 40 & 10 )
=
vs
10
 40 
 4   vs  4
ia = 
 ib =     = vs
 40 + 10 
 5   10  50
100  4 
8
 40 
io = − 
 ( 50i a ) = −
  vs = − vs
3  50 
3
 20 + 40 
The output is proportional to the input and the
8 A
constant of proportionality is −
.
3 V
Solution 3
Replace the voltmeter with the equivalent open
circuit and label the voltage measured by the
voltmeter as v m . Then
ib =
24
24
=
=2A
4 + ( 40 & 10 ) 4 + 8
 40 
4
ia = 
 i b =   2 = 1.6 A
 40 + 10 
5
v m = 8 i a = 8 (1.6 ) = 12.8 V
3
Solution 4
Replace the ammeter with the equivalent short
circuit and label the current measured by the
ammeter as i m . Then
v b = 3 || ( 2 + 4 )  3 = ( 2 ) 3 = 6 V
 4 
2
va = 
 vb =   6 = 4 V
 4+2
3
i m = 8 v a = 8 ( 4 ) = 32 A
Solution 5
Replace the voltmeter with the equivalent open circuit and label the
voltage measured by the voltmeter as v m . Then use current division
in the top part of the circuit to get
 40 
ia = − 
 ( 3) = −2.4 A
 40 + 10 
Next, use voltage division in the bottom part of the circuit to get
 R 
 −5 R 
vm = −
 (5 ia ) = 
 ia
 18 + R 
 18 + R 
Combining these equations gives:
12 R
 −5 R 
vm = 
 ( −2.4 ) =
18 + R
 18 + R 
When vm = 4 V,
4=
12 R
18 + R
⇒ R=
4 ×18
=9 Ω
12 − 4
4
Solution 6
Replace the ammeter with the equivalent short circuit and label the
current measured by the ammeter as i m . Then use voltage division in
the top part of the circuit to get
2
 12 
va = − 
 (vs ) = − vs
5
 12 + 18 
Next, use current division in the bottom part of the circuit to get
80 
 16 

im = − 
 (5 v a ) =  −
 va
 16 + R 
 16 + R 
Combining these equations gives:
80   2   32 

im =  −
  − vs  = 
 vs
 16 + R   5   16 + R 
a. When vs = 15 V and im = 5 A
400
 32 
5=
= 80 Ω
 15 ⇒ 80 + 5 R = 480 ⇒ R =
5
 16 + R 
b. When vs = 15 V and R = 24 Ω
 32 
im = 
 15 = 12 A
 16 + 24 
c. When im = 3 A and R = 24 Ω
4
 32 
3=
 vs = vs
5
 16 + 24 
⇒ vs =
15
= 3.75 V
4
5