Example 18
Use voltage division to find the current i through the 30 kΩ resistor and the voltage v
across the 6 kΩ resistor.
20 kΩ
1 kΩ
50 V
5 kΩ
4 kΩ
30 kΩ
15 kΩ
+
_
+
v 6 kΩ
i
_
Solution
In this problem, we want to use the voltage division principle to solve the voltages v and
the current i.
A voltage divider circuit consists of series-connected resistors. The voltage across one of
resistors should be the total voltage multiplied by the ratio of its resistance to the total
resistance.
In this circuit, we know the total voltage across all the resistors is 50V. The voltage
division only works for series-connected resistors. The 20kΩ and 5kΩ resistors could be
combined to a single resistor and let’s label the equivalent resistance as Rab. We can also
find the equivalent resistance across b and c and it is labeled as Rbc. After the circuit is
reduced to 3 resistors in series. Then voltage division can used to solve the voltage across
any of the 3 resistors, say Rbc. If we know the voltage across Rbc, the current can be
solved by taking the voltage divided by the resistance. The voltage division can be
applied again to find the voltage across the 6 kΩ resistor.
The 20 kΩ and 5 kΩ resistors share the same nodes a and b, hence they are parallel
connected. The equivalent resistance of two parallel resistors is the product of the two
resistances divided by the sum of the two resistances.
R ab 
20  5
 4k .
20  5
Note that the same current flows through the 4 kΩ and 6kΩ resistors so
that they are in series. Series resistances simply add together.
4 +6=10 kΩ.
The 3 resistors are parallel connected because they have the same voltages across them.
The equivalent resistance is
Rbc 
1
30

 5k
1
1
1
2 1 3


15 30 10
Notice that for 3 resistors in parallel, we cannot take the product of them and divided by
the sum of them to find the equivalent resistance.
For the simplified circuit, the voltage 50V is divided among the 3 resistors. The voltage
across Rbc should be the total voltage multiplied by the ratio of its resistance to the total
resistance.
v bc  50
5
 25V
1 4  5
.
If we consider a the original circuit, the current i can be calculated as v bc divided by the
resistance 20kΩ, that is
i
25
 0.83 mA
30 k
v bc is divided between the two series-connected resistors. Apply voltage division again,
v  25 
6
 15V
46