Series Resistors with Different Areas
Part A = 6/25 * V0
- Wire 1 has resistance R1 and cross-sectional area A.
- Wire 2 has resistance R2 and cross-sectional area 2A.
- Wire 3 has resistance R3 and cross-sectional area 3A.
- Wire 4 has resistance R4 and cross-sectional area 4A.
A voltage V0 is applied across the series, as shown in the figure.
Give your answer in terms of V0, the voltage of the battery.
The resistance of a resistor is given by R = ρL/A. Since the lengths and resistivities (ρ) for each resistor are the same, just set ρ * L = 1. So the overall resistance is:
R = 1/A + 1/(2A) + (1/3A) + (1/4A)
Set A = 1, for simplification purposes (the area of each resistor is just a multiple of A, so it’s ok to do this):
R = 1/1 + 1/2 + 1/3 + 1/4
R = 12/12 + 6/12 + 4/12 + 3/12
R = 25/12
Next, find the current. The total voltage, V0, is given by V0 = IR:
V0 = IR
I = V0 / R
I = V0 / (25 / 12)
I = 12/25 * V0
And the voltage drop across the second resistor (using ohm’s law) is just I * R2. R2 was found to be 1/2, above, so:
VR2 = I * R2
VR2 = (12/25 * V0) * 1/2
VR2 = 6/25 * V0
6/25 * V0