Equivalent resistance with possible short circuit

[rhemmin]

Homework Statement

Find the current flowing through the voltage source in mA. Assume Vs = 3.9 V and R = 8.3 kohm. Round off your answer to two decimal points.

Homework Equations

1/R(equivalent) = 1/R1 + 1/R2 + ...

I=V/R

Kirchoff's voltage law

The Attempt at a Solution

I considered the 3 resistors in the upper left of diagram (10k, R=8.3k, 10k) to be in parallel and calculated their equivalent resistance to be 3.12 kOhms. I basically ignored the two 10kOhm resistors in the bottom right because a) they didn't seem to be in series, or parallel so I didn't know what to do with them and b) it seemed that the current would avoid them by taking the path of least resistance after going through the 3 resistors in parallel in the upper left. I then wrote the KVL equation: Vs - I(3.12*10^3) = 0 and solved for I determining the current to be 1.25mA.

 

[phinds]

Uh, dude, you REALLY need to rethink that analysis. How can the current through R possibly NOT flow through on or both of the resistors on the right & bottom?

 

[NascentOxygen]

Try redrawing the diagram, this time repositioning enough of the elements so that you can draw it so there are no sloping lines. I.e., arrange it so all elements and connecting wires are either vertical or horizontal, and with no ugly cross-overs.

You may need to make a few attempts before you manage it. Just keep trying.

 

[rhemmin]

Wow. Thank you both very much for the help. My first big problem was that I assumed there was a node where the wires crossed. For sure, I'll pay much closer attention to nodes from now on. Also, redrawing the diagram with the R=8.3k resistor going around the other circuit elements made the diagram much easier to understand.

Thanks again!

 

[asteriatic]

I would like to point out that the assumption of a short circuit in this scenario is not physically possible. A short circuit occurs when there is a low resistance path that bypasses the load, causing a large current to flow through it. In this case, the two 10kOhm resistors in the bottom right are not in series or parallel with the other resistors, so they cannot create a short circuit. Additionally, the assumption of a short circuit would result in an infinite current, not a finite value of 1.25mA.

To accurately solve this problem, all resistors in the circuit must be taken into account. The two 10kOhm resistors in the bottom right can be combined into a single equivalent resistance of 5kOhms, which is in series with the 8.3kOhm resistor. This results in a total resistance of 13.3kOhms. Using Ohm's Law, the current can be calculated as I=Vs/R, which gives a value of 0.29mA. This is significantly lower than the value calculated under the assumption of a short circuit.

In conclusion, it is important to consider all components in a circuit when calculating the equivalent resistance and current. Assumptions of short circuits can lead to incorrect solutions and should be avoided.