Equivalent resistance with a short circuit

[sun18]

Homework Statement

Find the equivalent resistance of the circuit shown below.

Homework Equations

R=$\Sigma_{i}$R$_{i}$
1/R=$\Sigma_{i}$1/R$_{i}$

The Attempt at a Solution

I'm having a lot of trouble understanding how this circuit can be simplified. All I see is a big short circuit where the only element that matters is R$_{4}$. What I tried was considering R$_{2}$ and R$_{3}$ as being in parallel, but I still see a short circuit happening. I don't think I understand how short-circuits behave, because I don't think the equivalent resistance is simply R$_{4}$. Any guidance would be greatly appreciated (also sorry for the terrible drawing)

 

[gneill]

Homework Statement

Find the equivalent resistance of the circuit shown below.

Homework Equations

R=$\Sigma_{i}$R$_{i}$
1/R=$\Sigma_{i}$1/R$_{i}$

The Attempt at a Solution

I'm having a lot of trouble understanding how this circuit can be simplified. All I see is a big short circuit where the only element that matters is R$_{4}$. What I tried was considering R$_{2}$ and R$_{3}$ as being in parallel, but I still see a short circuit happening. I don't think I understand how short-circuits behave, because I don't think the equivalent resistance is simply R$_{4}$. Any guidance would be greatly appreciated (also sorry for the terrible drawing)

Hi sun18, Welcome to Physics Forums.

Your intuition is correct; The subnetwork consisting of R1 through R3 is bypassed by the wire running from the top terminal to R4. A short circuit is equivalent to a resistance of zero Ohms, so anything in parallel with it is effectively bypassed (A zero Ohm resistance in parallel with any other resistor value is zero).

 

[sun18]

Thanks so much for the response gneill. I guess I was overthinking it instead of concluding the obvious.