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Example 20
R4
+
v4
_
+
+
is
vs
i1
R1
i2
_
R2
i3
v3
R3
_
Solution
Given that i s  6mA ,
v s  6V , 2i1  3i2 , i2  2i3 , v4 : v3  2 : 1 , we need to specify the
resistors to meet the following specification.
The total current of all the branches is given. Based on KCL, the sum of the currents
which enter the node is equal to the sum of the currents which leaves the node.
(1)
i S  i1  i 2  i3
Given that 2i1  3i2 , it can be written as
i1  1.5i 2 (2)
The equation i2  2i3 is divided by 2 on both sides,
i3  0.5i 2 (3)
Substitute (2) and (3) into (1), we get
i S  1.5i 2  i 2  0.5i 2  3i 2  6mA
We can solve for i2  2mA (4)
Substitute (4) into (2) and (3),
i1  1.5i 2  1.5  2  3mA . i3  0.5i 2  0.5  2  1mA
Using Ohm’s law,
R1 
vs 6
v
6
  6k , R2  s   3k
i1 1
i2 2
For the resistors R3 and R 4 , they are in series, the voltage across the two resistors are
v S  v3  v 4  v3  2v3  6V
We can solve for v 3  2V
v 4  6  2  4V .
Apply ohms’ law,
R3 
v3 2
v
4
  2k and R4  4   4k .
i3 1
i3 1
We have solved the problem completely. Here we can verify our result easily by voltage
and current division.
R3 and R 4 are series connected so that they can be considered as a voltage divider.
The voltage across the resistor R3 should be the total voltage across the two resistors
multiplied by the ratio of its resistance divided by the total resistance.
v3  6 
2k
 2V
2k  4k
Similarly, v 4  6 
.
4k
 4V
2k  4k
.
So v4 : v3  2 : 1 , which is consistent with the specification given before.
Also we can verify the result using the current divider, which is a parallel-connected
resistance circuit. The two resistors can be combine into a since they are in series. For the
parallel circuit, the current through one of the resistor is proportional to the total current.
The proportionality is the equivalent resistance of the parallel resistors divided by its
resistance.
The equivalent resistance is
1
 1k .
1 1 1
 
2 3 6
R
1k
The current i1  6  eq  6   3mA . It is also consistent with the result we got before.
2k
2k
Req 
Similarly we can calculate the currents for the other two branches. So the values of the
currents should also be consistent with the specification in the problem.