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:
general-voltage-divider-formula
In the present case we have three resistors. So

Vx. = [Rx/ {R1 + R2 + R3} ] *Vin
asdas
For R> V1 = [R1/ {R1 + R2 + R3} ] *Vin = [22 Ω/ {22 Ω + 44 Ω + 66 Ω} ] * 12 V = 2 V
For R> V2 = [R2/ {R1 + R2 + R3} ] *Vin = [44 Ω/ {22 Ω + 44 Ω + 66 Ω} ] * 12 V = 4 V
For R> V3 = [R3/ {R1 + R2 + R3} ] *Vin = [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.

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