Does Current Direction Affect Voltage and Power Calculations in a Circuit?

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In summary, in this simplified 5V battery and 2000Ω resistor circuit, when the battery voltage drops below -5V, the current through the resistor reverses direction so I=-2.5mA. This is consistent with the direction I=V/R. If I flip the 'I' arrow around, the relevant formulas still apply.
  • #1
emcsquared
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1. I know this is really simple but I'm missing some little piece of the puzzle

No problem getting to vab=
  • +1.5V for a)
  • +1.5V for b)i.
  • -2.5V vor b)ii.
  • +0.5V for b)iii.
upload_2016-9-9_4-8-23.png


But what about current I?

2. I=V/R

3. so for IR2
  • b)i. I have 1.5V/1kΩ=1.5mA (sure I agreee with that)
  • b)ii. I have -2.5V/1kΩ=-2.5mA (hmmm)
  • b)iii. I have 0.5V/1kΩ=0.5mA
does b)ii. make sense? If the voltage is reversed to -5V does the current also reverse? What if I just chose to draw the arrow for current the other way round? How do I account for that when analysing the circuit?

Really simple I know, but it's doing my head in!
 

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  • #2
What do you know about reverse current through diodes?
 
  • #3
Only that as far as this introductory course is concerned, it doesn't exist!

I'm just trying to understand how to determine if the current through the resistor (#2) has a positive or negative value. Does it always have to be positive when entering the positive reference of the resistor?
 
  • #4
So if no current is flowing through the reverse biased diodes would anything change if you removed them?
 
  • #5
No not in the case of V(in)=-5V. They're effectively an open circuit. What I'm trying to understand is if the voltage is negative so therefore V(R2) is negative (using the polarity references I have drawn in) does it then follow that I (being V/R) =-2.5V/1k-ohm in the direction given, or does some other convention apply?

Or If I flipped the 'I' arrow around, how would that be reflected in the relevant formulas?

Or to put it another way, can a reference direction of current be arbitrarily assigned or is it determined by other variables?
 
  • #6
On paper you can assign or assume any direction for the current for your calculations. But if your calculations give a negative current, then when relating that to the hardware you would say current is in the direction opposite to your assumed direction.

In easy circuits where it is obvious which direction the current goes, it makes sense to assume that obvious direction right from the beginning (unless you just wanted to prove to yourself that it will all work out right in the end even when you assume the opposite direction to what you know is going to be right). :smile:
 
  • #7
+1

Note that the +/- marks next to R2 are probably intended to define what is meant by a positive voltage across R2. This is consistent with the direction defined for positive current I.

Normally you define these things at the outset and don't change them just because Vin become negative.
 
  • #8
OK thanks for all the replies, that's exactly what I thought. Actually I think I've got it now, but here's my reasoning (prepared earlier) and my solution at the end.

n.b, the references drawn in are mine, because we have been admonished enough times to give a magnitiude, unit and reference for every value :)

Let's simplify the circuit to a 5V battery and a 2000Ω resistor, since the diodes are not relevant here. The current enters our positive reference of our resistor R

VR is +5V and I is 2.5mA - simple.

upload_2016-9-9_17-59-1.png


Let's now say the battery voltage is -5V which can be expressed as

upload_2016-9-9_17-59-19.png


or more intuitively:

upload_2016-9-9_17-59-39.png


so now by KVL, +5V+VR=0, so VR is -5V. If I still enters the positive reference of R, do we still simply use I=V/R and get -2.5mA?

My check here is to look at the power relationships. The sign convention I prefer, because it seems simple and universal, is to
  • Take the values given for V and I, be they negative or positive.
  • Multiply them to give a result (negative or positive),
  • which indicates energy 'travelling' in the direction indicated for conventional current.
  • A negative result simply means energy traveling in the opposite direction.
  • Then, a positive value of energy entering the positive reference of a circuit element is energy consumed by that element, and:
  • energy entering the negative potential reference is energy supplied by that element.
  • A negative value of energy entering the positive potential reference is negative energy consumed, which is interchangeable with energy supplied.
  • Likewise with 'negative energy' into a negative reference, we have negative energy supplied or positive energy consumed.
So going back to the example, if VR is -5V and I=-2.5mA, P=VI=+12.5mW. So far so good, because while voltage and current are both -ve, our conventional current reference (as shown by 'I') enters the positive reference of R so energy is consumed as expected.

But what if I got bored of the clockwise current reference and just switched it? It's arbitrary after all:

upload_2016-9-9_18-0-5.png


If I still =V/R we get -2.5mA into the negative reference of R so P=-2.5mA*-5V=+12.5mW supplied by the resistor. Clearly nonsense, so where have I gone wrong?

Obviously in this case I could 'just' reverse the sign, because it is a trivial example I already know the answer to. But in a more general sense, in cases where the answer is less obvoius, can I still operate with my convention as described above and add the caveat that if the conventional current reference enters the negative reference then we use V=-IR and P=-VI?

BRAINWAVE: can I use, instead of V as given, Vd; 'voltage drop in the reference direction of conventional current'? That would mean that in the last case, since I is anticlockwise, Vd would be -(-5V)=5V, which would satisfy my sign convention. Please tell me that's right, I like my convention above, it's simple like me :)
 

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FAQ: Does Current Direction Affect Voltage and Power Calculations in a Circuit?

What is the direction/sign of electric current?

The direction of electric current is the direction in which positive charges flow. By convention, the direction of current is considered to be from positive to negative.

How is the direction of current determined?

The direction of current is determined by the flow of electrons. Electrons are negatively charged particles, so their movement is opposite to the direction of current flow.

Does current always flow in the same direction?

No, the direction of current can change depending on the type of circuit. In direct current (DC) circuits, current flows in one direction, while in alternating current (AC) circuits, the direction of current changes periodically.

What is the difference between positive and negative current?

Positive current refers to the flow of positive charges, while negative current refers to the flow of negative charges. In reality, only negative charges (electrons) flow in a circuit, but the convention is to consider the direction of current as the flow of positive charges.

How is the direction of current represented in a circuit diagram?

The direction of current is represented by arrows in a circuit diagram. The arrow points in the direction of current flow, which is from positive to negative.

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