# Find the Equivalent Resistance of a Network

## Homework Statement

A network consists of a combination of parallel and series connections. What is the equivalent resistance of the network? (Hint: Use the series and parallel resistance formulas).

The network looks like this:

## Homework Equations

When a circuit is in series, R = R1 + R2 + R3...

When a circuit is in parallel, R = 1/R1 + 1/R2 + 1/R3...

## The Attempt at a Solution

First, c and b are in series, so I just add them up. 5 + 3 = 8.

Now, a, d, and this new resistor of value 8 are in series. So add them up again.
8 + 4 + 1 = 13.

Did I do that correctly?

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LowlyPion
Homework Helper
Sorry. B and C are ||.

Not in series.

They look like they're in series to me... how can you tell?

That changes my answer to 1/5 + 1/3 = 0.533. And 1/0.533 = 1.875.

Then 1.875 + 4 + 1 = 6.875

LowlyPion
Homework Helper
They look like they're in series to me... how can you tell?

That changes my answer to 1/5 + 1/3 = 0.533. And 1/0.533 = 1.875.

Then 1.875 + 4 + 1 = 6.875
That looks more like it.

When a circuit is in parallel, R = 1/R1 + 1/R2 + 1/R3...
This is wrong. I think you meant to write:
1/Req = 1/R1 + 1/R2 + 1/R3...

They look like they're in series to me... how can you tell?
Notice that the upper terminals of C and B share a common node. Similarly, the bottom terminals of C and B share a common node.

Yeah, I meant to write 1/Req = 1/R1 + 1/R2...

So a is in series with c because it only shares one common node? And b and c are in parallel because they share two nodes?

So a is in series with c because it only shares one common node? And b and c are in parallel because they share two nodes?
Well, a better way to say it is this: A is in series with the combination C||B.

If the same current flows through two resistive elements, the elements are in series.
If the same voltage potential is across two elements, the elements are in parallel.

Does that help?

That helps a lot - I understand now. Thank you!