Series Resistors with Different Areas

Part A = 6/25 * V_{0}

**Solution Below:**

- Wire 1 has resistance R
_{1}and cross-sectional area A. - Wire 2 has resistance R
_{2}and cross-sectional area 2A. - Wire 3 has resistance R
_{3}and cross-sectional area 3A. - Wire 4 has resistance R
_{4}and cross-sectional area 4A.

A voltage V_{0} is applied across the series, as shown in the figure.

Part A

_{2}across wire 2.

Give your answer in terms of V

_{0}, 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, V_{0}, is given by V_{0} = IR:

V_{0} = IR

I = V_{0} / R

I = V_{0} / (25 / 12)

I = 12/25 * V_{0}

And the voltage drop across the second resistor (using ohm’s law) is just I * R_{2}. R_{2} was found to be 1/2, above, so:

V_{R2} = I * R_{2}

V_{R2} = (12/25 * V_{0}) * 1/2

V_{R2} = 6/25 * V_{0}

6/25 * V_{0}