1/21/14
1
Current Divider Circuit
I
s
i
1
i
2
R
1
R
2
V
Current source and two resistors connected in parallel
KCL:
I
s
=
i
1
+
i
2
V
=
I
s
R
1
R
2
R
1
+
R
2
⎧
⎨
⎩
⎫
⎬
⎭
I
s
=
V
R
1
+
V
R
2
I
s
=
V
R
2
+
R
1
R
1
R
2
⎧
⎨
⎩
⎫
⎬
⎭
i
1
=
?
i
2
=
?
i
1
=
V
R
1
i
1
=
I
s
R
2
R
1
+
R
2
⎧
⎨
⎩
⎫
⎬
⎭
i
2
=
V
R
2
i
2
=
I
s
R
1
R
1
+
R
2
⎧
⎨
⎩
⎫
⎬
⎭
i
1
i
2
=
R
2
R
1
Example (2)
10V
2
k
Ω
10
k
Ω
R
x
v
o
In the circuit shown, select a value for the resistor R
x
so that v
o
=8V

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
1/21/14
2
Solution
KVL (abca):
10
=
v
+
8
v
=
2
V
i
=
2
2
k
=
1
mA
Ohm
’
s Law:
i
o
=
8
10
k
=
0.8
mA
KCL node b:
i
=
i
x
+
i
o
10V
2
k
Ω
10
k
Ω
R
x
v
o
i
v
a
b
c
c
c
i
o
i
x
1
m
=
i
x
+
0.8
m
i
x
=
0.2
mA
Current division rule:
i
o
i
x
=
R
x
10
k
0.8
0.2
=
R
x
10
k
R
x
=
40
k
Ω
Given:
v
o
=
8
V
Example (3)
V
S
R
2
R
1
R
3
v
o
Use the voltage division rule to find the output
voltage
v
o

1/21/14
3
Solution:
V
S
R
2
R
1
R
3
v
o
•
current through R
3
is zero !
a
b
c
b
v
o
Hint:
•
The voltage source and R
1
and R
2
connected in
series !
Voltage division rule:
v
o
=
V
s
R
2
R
1
+
R
2
⎛
⎝
⎜
⎞
⎠
⎟
5A
20
Ω
20
Ω
5
Ω
5
Ω
i
x
i
y
i
z
Example (4)
•
Find the current
i
x
•
Find the current
i
y
•
Find the current
i
z

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*