Voltage/Current division

I. Voltage Division
When you know the amount of voltage that will divide across resistors, it is easy to apply voltage
division to determine the value of the drop across each resistor because it is proportional to the
value of resistance. An example is given below:
Determine V R1 and V R2
+ V R1 -
+
V R2
-
KVL: -V S + V R1 + V R2 = 0
V S = V R1 + V R2
Apply Ohms Law: V S = I S R 1 + I S R 2
Using this general expression, find the voltage across each resistor.
V R1 = I S R 1 = (V S ) * R 1 = V S * R 1
(R 1 +R 2 ) (R 1 +R 2 )
V R1 = V R2 =
Notice that the voltages are proportional to their resistance values (if R 2 >R 1 then V R2 >V R1 ).

The voltage across series resistors divides in direct proportion to the R value (Can be stated:
Voltage drop across the resistor of interest is equal to the source voltage times the resistance of
interest divided by the sum of resistances).
More of the voltage drop goes to the larger resistance. If R 1 = R 2 , the division is equal.
Same example with numerical values…
II.
Current Division
When you know the current that is supplying parallel branches, you can apply current division to
help find currents in the resistive branches. Unlike voltage division, the current flowing through
a branch depends on the resistance of the opposite branch.
Example:
V = I S R eq = I S (R 1 *R 2 )
R 1 +R 2
What is I 1 I 1 = V/R 1
= =
So to determine the current in a branch containing R 1 , we will multiply the supply current by the
resistance of the opposite branch and then divide by the sum of the two resistors.

It follows that:
Same example with numerical values…
Notice the current favors the path of least resistance.
Example: Find V o
First reduce the circuit:
→
30 kΩ and 60 kΩ are in parallel yielding 20 kΩ. V 1 = 6 V because the 2 resistors are equal

or by voltage division: V 1 = 12 (20k/40k) = 12 (1/2) = 6 V.
Go back to the original circuit and label V 1
Now apply voltage division again:
V o = V 1 (20k/60k) = 6(1/3) = 2 V.
You could also use current division to solve this problem:
From the circuit below: I S = 12/40 kΩ = 3/10 mA
So by current division:
I o = 3/10 mA (30k/(30k + 60k)) = 3/10 (1/3)
= 1/10 mA = 0.1 mA
V o = 20k*I o = 20k (0.1 mA) = 2 V.
Same answer as above.

More examples of current division.
Find I S
You are given the voltage across the 30 kΩ
resistor.
So, I 3 = 3/30k = 0.1 mA
Current division:
If I 3 = I S /3 then I S = 3I 3 = 3(0.1 mA) so I S = 0.3 mA
Check your work.
Voltage across 90 kΩ (V 90 ) is
I 3 (90k) = 0.1 mA(90 kΩ) = 9 V.
Voltage across 60 kΩ (V 60 ) is 9 + 3 = 12 V
(KVL; -V 60 + 9 + 3 = 0)
V 60 = I S *R eq
Checks!

Another example illustrating both voltage and current division
Find I o
Label nodes (a, b, c) and define currents in each resistor.
Notice 4 kΩ and 12 kΩ are in parallel.
That equivalent resistance is 3 kΩ. Redraw the circuit.
You can redraw again noting that 3 kΩ is in series with 3 kΩ
giving 6 kΩ.
Because the branches are equal in resistance,
I 1 = I 2 = 6 mA.
Or by current division,
I 1 = 12 mA (6k/12k) = 12 mA (1/2) = 6 mA.
So now that you know both branch currents, you are closer to finding I o although you still need
more information. Recognizing that all branches are in parallel means they have the same
voltage across them. You could either combine resistors again (6 kΩ in parallel with 6 kΩ = 3
kΩ) and use Ohm’s Law V = (12 mA) * (3 kΩ) = 36 V OR you could look at a single resistive
branch and use Ohm’s Law: V = 6 mA (6 kΩ) = 36 V.

Now go back to an earlier circuit. If 36 V is the voltage across each branch, by voltage division
the voltage across each of the series 3 kΩ resistors is 18 V. (V ab = 36 (3k/6k) = 36 (1/2) = 18 V.)
Go back to the original circuit.
V ab is the voltage across the 4 kΩ resistor so …
I o = 18/4k = 4.5 mA.
To verify, write KCL at node b:
I o + I 3 = I 1
I o = I 1 - I 3 = 6 mA – I 3 and I 3 = 18/12k = 1.5 mA
I o = (6-1.5) mA = 4.5 mA