Example 16
1 kΩ
+
_
v1
1.2 kΩ
20 V
6 kΩ
4 kΩ
+
_
3 kΩ
6.6 kΩ
+
+
v2
v3
_
_
2 kΩ
Use voltage division to find the voltages v1 , v 2 , v 3 .
In this problem, we want to use the voltage division principle to solve the voltages v1 , v 2 ,
v 3 . For N series-connected resistors, if we know the total voltage across all the resistors,
the voltage across one of resistors should be the total voltage multiplied by the ratio of its
resistance to the total resistance. A voltage divider circuit consists of series-connected
resistors.
In this circuit, we know the total voltage cross two terminal is 20V. 1 kΩ is in series with
the combination of these 5 resistors and then in series with the 6.6 kΩ resistor. If we can
find the equivalent resistance across terminals a-b. We can use the voltage division to
find the voltage v1 . The 1.2 kΩ resistor is parallel connected with the 6 kΩ resistor
because they share the same pair of nodes.
1. 2  6
 1k .
1 .2  6
The 1 kΩ resistor is series connected with the 3 kΩ resistor since the same current flows
through both of the resistors. To combine the series capacitors, we simply add together
the resistors 1  3  4k .
Similarly, the 4 kΩ resistor and the 2 kΩ resistor are in series and the equivalent
resistance is 4  2  6k
We can identify that the 4 kΩ resistor and the 6 kΩ resistor are parallel connected they
have the same voltages across the two terminals. The equivalent resistance is
46
 2.4k .
46
At this point, we can recognize that this is a voltage divider circuit. The total voltage is
divided among the 3 resistors. The voltage across the 1 kΩ should be the total voltage
20V multiplied by the ratio of the resistance 1k to the total resistance.
v1  20 
1k
1k
 20 
 2V
1k  2.4k  6.6k
10 k
.
The voltage across the combination of the resistors should be
v ab  20 
2.4k
2.4k
 20 
 4.8V
1k  2.4k  6.6k
10 k
From the original circuit, we know the two resistors can be combined to a 1 kΩ resistor
and it is in series with 3 kΩ. So the 4.8V voltage is divided between the two resistors in
direct proportion to their resistances. It is also a voltage divider circuit.
So the voltage across the 3 kΩ resistor should be
v 2  4.8 
3k
 3.6V
1k  3k
Similarly, we can use the voltage division to get v 3 because the 4 kΩ resistor is seriesconnected with the 3 kΩ resistor.
v 3  4. 8 
2k
 1.6V
4k  2k