Mastering Physics Solutions: Series Resistors with Different Areas

Series Resistors with Different Areas

Part A = 6/25 * V0

Solution Below:

Series Resistors with Different Areas: Four wires are made of the same highly resistive material, cut to the same length, and connected in series.

  1. Wire 1 has resistance R1 and cross-sectional area A.
  2. Wire 2 has resistance R2 and cross-sectional area 2A.
  3. Wire 3 has resistance R3 and cross-sectional area 3A.
  4. Wire 4 has resistance R4 and cross-sectional area 4A.

A voltage V0 is applied across the series, as shown in the figure.

Mastering Physics Solutions: Series Resistors with Different Areas

Part A

Find the voltage V2 across wire 2.
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

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