Question

Part A: Determine the equivalent resistance of the "ladder" of equal 115-Ω resistors shown in (Figure 1) . In other words, what resistance would an ohmmeter read if connected between points A and B? Part B : What is the current through resistor a if a 65.0-V battery is connected between points A and B? Part C: What is the current through resistor b? Part D: What is the current through resistor c?

Answer 1

A) . Resistors in series simply add. Rt = R1 + R2 + R3 .......
Resistors in parallel combine such that 1/Rt = 1/R1 + 1/R2 + 1/R3.........
or for two resistors, Rt = (R1*R2)/(R1 + R2).
Let's start at the far right. We have three 115 ohm resistors in series. Total resistance 345 ohms.
Then we have that 345 ohms in parallel with a 115 ohm resistor. That is equivalent to 86.25 ohms.
There are two 115 ohm resistors in series with this combination for a total resistance of 316.25 ohms.
Now there is 316.25 ohms in parallel with another 115 ohm resistor, for an equivalent of 84.333 ohms.
We are left with two 115 ohm resistors in series with this for an total resistance measure between A and B of 314.333 ohms.

B) current in a = 65/314.333 = 0.206787 A

C) current in b = 0.206787 x 316.25/115 = 0.568664 A

D) current in c = 0.206787 A