Circuit Analysis Question (Electrical Engineering majors holla)

[optimus_prime]

Alright so finished up my exam in electricity and magnetism and was kind of thrown off by a question.

I'm 90% certain it was a trick question of sorts but was hoping some of the 2p2 brains could confirm/deny.

Here's the question as I remember it:

http://img104.imageshack.us/img104/9437/eeqoo2.jpg (http://imageshack.us)

The question asked for the potential difference across R1 and R4, the current magnitude and direction across R3, and the source current.

My buddy found the equivalent resistance:

(1/R1+1/R2)^-1+R3+(1/R4+1/R5)

but I'm damn near positive the whole "path of least resitance" thing means that the current wouldn't even pass through R1, R2, R4 and R5 and would instead take this path:

http://img338.imageshack.us/img338/9455/eeq2ye3.jpg (http://imageshack.us)

and that I1 and I2 would be 32/R3 and V1 and V2 would both be 32V.

Any help appreciated.

[The Legend]

the five resistors are in parallel. It's just drawn a little funky. So the voltage across all of them is 32. Currents are 32/R.

[The Legend]

Currents across each resistor is Ix=32/Rx I2 is the sum of all the currents across all the resistors.

[optimus_prime]

N/M

[Matt R.]

Your current can split at any junction. You can't just ignore a possible pathway because a resistance is lower through another path. Unless it's some theoretical question where your resistance is infinite at one point in the circuit, you need to apply Kirchoff's rules (like your friend did, I think).

Your buddy is on the right track, although I didn't check his derivation. You need to find the current passing through R1 and R4, then just use Ohm's Law to get the voltage drop. Same for R3. Find the equivalent resistance across the entire circuit, and use V=IR again to get the source current.

Everything follows from Kirchoff's rules and V=IR, so just do the analysis from there.

[Matt R.]

I type slow.

As Legend noted, an easy way to do this is to use your first "loop" as the outer-most loop of the circuit. Since sum(V) = 0 you know immediately that V2=32V and I2=32V/R1. This circuit is really simple once you break it down into the loops, but it's hard to see at first since it's drawn kind of weird.

[carlo]

Been awhile(with trepidation)-isn't node @R1 the same as R4? therefore isn't the voltage= 0 from R1 to R4?

[BruceZ]

Each resistor is connected directly across the battery, so all 5 resistors are in parallel. V1 = 32, V2 = 32, I1 = 32/R3. The equivalent resistance is actually Req = R1 || R2 || R3 || R4 || R5 = 1/(1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5). I2 = 32/Req.

Electricity doesn't only take the "path of least resistance". It takes all possible paths to ground, with a current in each path in inverse proportion to the resistance of that path.

EDIT: If the voltage source is actually a 32 volt battery with some internal resistance Rint, then compute I2 as 32/(Req+Rint) and scale the voltage across each resistor as 32*Req/(Req + Rint). This would not apply to an idealized 32 volt source, or if the 32 volts refers to the actual voltage across the battery in the circuit.

[carlo]

Ok, try again.Aren't R1 and R2 in parellel and in series with R4 and R5 which are also in parellel? Also, isnt R3 in parallel with the series circuit just referred to?

Thereforethe voltage across R1 to R4 is 32 volts and the voltage across R3 is 32 volts. My drawing doesn't hve them all in parallel but R1/R2 which series with R4/R5 and this resultant resistance in parallel with R3. The voltage is across the resulting circuit.

[optimus_prime]

In case I wasn't clear, my buddy was arguing that R1 and R2 were in parallel, R4 and R5 were in parallel, and that these were both in series with R3.

I started to look at R1, R2 and R3 being in parallel but couldn't put it all together. That, and my prof is a bit of a douche who loves tricky [censored].

Anyways I'm pretty sure I see how they're all in parallel now and am sobbing gently.

Edit to add: The resistors had values on the exam, but I couldn't remember them and they weren't important for my purposes here.

[BruceZ]

Actually, even if you don't recognize immediately that all 5 resistors are in parallel, this should become clear once you try to solve the circuit. Since there are fewer nodes than loops, you would pick node analysis (kirchoff's current law - sum of currents leaving a node is zero). This involves writing an equation for each node at which you don't already know the voltage. At that point you would notice that there are actually ZERO nodes for which you don't already know the voltage since there are only 2 nodes all together, and both connect directly to the battery, so one has a voltage of 32 volts, and the other is at ground potential (0 volts). Zero nodes implies that all of the components have the same voltage across them, i.e. they are in parallel. So there is nothing to do as you already have all the voltages, and this means that you can solve for the current through each resistor. Since the only 2 voltages are 32 and ground, it becomes clear that each resistor is connected across the battery.

[AquaSwing]

I hate this crap. I had a couple professors who pulled this [censored] on exams to humor us. Now that I'm in the real world no one would ever present a circuit in that form so what's the point?

My last question on my last final my last semester was about being Gilligan Islanded with your pet pigeon. You were given a specific list of parts and had to come up with a simple transmitter to save yourself. Of course, I was drunk 45 minutes after the thing ended and came back to help proctor a junior level exam by the same professor. I ended up giving a whole bunch of people the answers to questions. Sweet revenge.

By the way, this really scares me:
[ QUOTE ]
but I'm damn near positive the whole "path of least resitance" thing means that the current wouldn't even pass through R1, R2, R4 and R5 and would instead take this path:


[/ QUOTE ]

[BruceZ]

[ QUOTE ]
I hate this crap. I had a couple professors who pulled this [censored] on exams to humor us. Now that I'm in the real world no one would ever present a circuit in that form so what's the point?

[/ QUOTE ]

If you ever have to fix a circuit without a schematic, or even check a printed circuit board against a schematic, you would run into exactly this kind of problem.

If you know your stuff, you should prefer this problem to one where you actually have to do a lot of work.


[ QUOTE ]
My last question on my last final my last semester was about being Gilligan Islanded with your pet pigeon. You were given a specific list of parts and had to come up with a simple transmitter to save yourself. Of course, I was drunk 45 minutes after the thing ended and came back to help proctor a junior level exam by the same professor. I ended up giving a whole bunch of people the answers to questions. Sweet revenge.


[/ QUOTE ]

How dare he try to separate the students who could think for themselves from the ones that just regurgitate information. /images/graemlins/smile.gif

A simple transmitter is just an oscillator and an antenna, maybe an amplifier. You can make a simple oscillator from an inverter with feeback through a crystal. Two or three components to match the antenna would be nice. What were the parts?

[bigalt]

[ QUOTE ]

By the way, this really scares me:
[ QUOTE ]
but I'm damn near positive the whole "path of least resitance" thing means that the current wouldn't even pass through R1, R2, R4 and R5 and would instead take this path:


[/ QUOTE ]

[/ QUOTE ]

haha, just don't hire him to hook up your bathroom wiring and you should be fine.

[AquaSwing]

[ QUOTE ]
If you ever have to fix a circuit without a schematic, or even check a printed circuit board against a schematic, you would run into exactly this kind of problem.

If you know your stuff, you should prefer this problem to one where you actually have to do a lot of work.

[/ QUOTE ]
True. Fortunately all I've ever had to debug were my own designs.

[ QUOTE ]
My last question on my last final my last semester was about being Gilligan Islanded with your pet pigeon. You were given a specific list of parts and had to come up with a simple transmitter to save yourself. Of course, I was drunk 45 minutes after the thing ended and came back to help proctor a junior level exam by the same professor. I ended up giving a whole bunch of people the answers to questions. Sweet revenge.


[/ QUOTE ]

How dare he try to separate the students who could think for themselves from the ones that just regurgitate information. /images/graemlins/smile.gif

A simple transmitter is just an oscillator and an antenna, maybe an amplifier. You can make a simple oscillator from an inverter with feeback through a crystal. Two or three components to match the antenna would be nice. What were the parts?

[/ QUOTE ]
Or you just write a note and send it home with your pigeon. The circuit and the pigeon answer were both right.