Voltage Divider Rule Example # 2: Three Resistors of 22 Ω, 44 Ω, and 66 Ω

We previously studied the application of voltage divider rule with various examples. In this post, you'll learn the method to apply voltage divider on three series resistors .

Statement: Three resistors of 22 Ω, 44 Ω, and 66 Ω are connected in a series configuration. The input voltage source is a 12 V battery. Apply the voltage divider rule to find the current across three resistors. Mention the observations.
Solution: Let's rewrite the general voltage divider formula:

In the present case we have three resistors. So

V _x . = [R _x / {R ₁ + R ₂ + R ₃ } ] *V _in
asdas
For R ₁ > V ₁ = [R ₁ / {R ₁ + R ₂ + R ₃ } ] *V _in = [22 Ω/ {22 Ω + 44 Ω + 66 Ω} ] * 12 V = 2 V
For R ₂ > V ₂ = [R ₂ / {R ₁ + R ₂ + R ₃ } ] *V _in = [44 Ω/ {22 Ω + 44 Ω + 66 Ω} ] * 12 V = 4 V
For R ₃ > V ₃ = [R ₃ / {R ₁ + R ₂ + R ₃ } ] *V _in = [66 Ω/ {22 Ω + 44 Ω + 66 Ω} ] * 12 V = 6 V

After solving this problem we observe that the higher the resistance, the more voltage is dropped across it.