# Is there more information needed or am I just missing how to start this solution???

#### Rendog

Joined Feb 8, 2019
4
I am stumped where to begin. I am not looking for the answer just some help getting started!!! Do I start through resistor 4 or through the parallel resistors 1,2,3?

#### crutschow

Joined Mar 14, 2008
33,373
First I would combine the parallel R1 and R2 into their equivalent single resistor value.
Then you should be able to write the two loop equations to solve for the unknown.

#### Rendog

Joined Feb 8, 2019
4
OK I am getting 57.7 ohms for r1 and r2. can i know just do the series and bypass r3 to find a total. I think the current on R3 is throwing me off.

#### crutschow

Joined Mar 14, 2008
33,373
can i know just do the series and bypass r3 to find a total.
Nope.
If you don't now how to write loop equations so solve this problem, now's the time to learn.

#### Rendog

Joined Feb 8, 2019
4
Thank you for the link. I am brand new to all of this.

#### WBahn

Joined Mar 31, 2012
29,534
It appears that the problem is using electron flow for current. That often indicates that this is being taught in a course that is very light on theory, often taken by students whose math background is not very extensive. Of course, this isn't always the case, so it would help us help you if you could describe the type of program this is in and how strong your math skills, particularly algebra, are.

In the meantime, assume that you know the voltage on the top-right node (let's call it Vo). Can you write the equation for the currents entering and leaving that node (i.e., Kirchhoff's Current Law)?

#### The Electrician

Joined Oct 9, 2007
2,957
If the top of the 66 volt source is the positive end, as would be usual, the direction of the 1.3 amps is odd.

#### crutschow

Joined Mar 14, 2008
33,373
If the top of the 66 volt source is the positive end, as would be usual, the direction of the 1.3 amps is odd.
Not if it's electron flow.

#### WBahn

Joined Mar 31, 2012
29,534
If the top of the 66 volt source is the positive end, as would be usual, the direction of the 1.3 amps is odd.
According to the battery symbol, the top is, indeed, the positive end. The direction is only odd if it is using conventional current, which is why I suspect the course is using electron flow (which should be banished unless it's done correctly, and I've never seen it done correctly yet except by a couple of physicists I've worked with).

#### The Electrician

Joined Oct 9, 2007
2,957
But we don't know what the polarity of the 66 volt source is, since it isn't labeled. That also might be non-standard.

#### The Electrician

Joined Oct 9, 2007
2,957
According to the battery symbol, the top is, indeed, the positive end..
Only if the person who drew the diagram followed convention for battery symbols.

#### WBahn

Joined Mar 31, 2012
29,534
Only if the person who drew the diagram followed convention for battery symbols.
Most of the drawings I've seen tend to get that right. I think Occam's Razor argues in favor of electron current.

#### Rendog

Joined Feb 8, 2019
4
If the top of the 66 volt source is the positive end, as would be usual, the direction of the 1.3 amps is odd.
That is what is throwing me off i think. All of our classes he has taught the current going clockwise and labeled the resistors as so. The arrow pointing in the opposite direction is what throws me off.

#### JoeJester

Joined Apr 26, 2005
4,390
According to the symbol for the current flow in R3, the diagram is for conventional flow. It would be the other direction for electron flow.

#### The Electrician

Joined Oct 9, 2007
2,957
Wbahn , you say "most" of the drawings you've seen tend to get that right, which probably means that you have seen some that don't get it right. Also, I get the impression that you haven't seen many uses of electron current in problems for beginners. So, what is the relative occurrence of bad battery symbols versus the rate of occurrence of usage of electron current? Do we need to take those rates into account as we invoke Occam's razor?

I don't know where the "odd" aspect occurs in this problem, but something's atypical about it.

#### The Electrician

Joined Oct 9, 2007
2,957
According to the symbol for the current flow in R3, the diagram is for conventional flow. It would be the other direction for electron flow.
If the top of the battery is positive, the current arrow on R3 is wrong for conventional current (if the resistor is an ordinary passive resistor).

#### JoeJester

Joined Apr 26, 2005
4,390
If the top of the battery is positive, the current arrow on R3 is wrong for conventional current (if the resistor is an ordinary passive resistor).

#### The Electrician

Joined Oct 9, 2007
2,957
That is what is throwing me off i think. All of our classes he has taught the current going clockwise and labeled the resistors as so. The arrow pointing in the opposite direction is what throws me off.
OK. Ignore the arrow for now. What network solving methods have you learned? Can you follow WBahn's suggestion to describe what happens at the upper right hand node?

#### crutschow

Joined Mar 14, 2008
33,373
All of our classes he has taught the current going clockwise and labeled the resistors as so. The arrow pointing in the opposite direction is what throws me off.
You can draw the loop arrow in either direction as the direction is completely arbitrary.
It still gives the same answers as long as you draw all loops in the same direction and use the appropriate polarity for all the component voltages around the loop.

#### The Electrician

Joined Oct 9, 2007
2,957
Huh?

Would you agree that the diagram shows the voltage at the top-left node being 66 V higher than the voltage at the bottom-left node?

If so, then would you agree that conventional current flows clockwise around the circuit?

If so, then would you agree that the voltage (relative to the bottom-left node) is lower than 66 V (but greater than 0 V)?

If so, then how is current flowing right-to-left in the bottom resistor consistent with conventional current flow?
Apparently you missed post #17