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Direction of current flow in a circuit

by theBEAST
Tags: kirchoff circuit law
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theBEAST
#1
Mar15-12, 08:00 PM
P: 368
1. The problem statement, all variables and given/known data
I attached a picture of the problem. I am confused about how the current source will affect the direction of current flow. Does the current flow in the direction of the current source arrow? If not what does the arrow dictate? Our professor said the direction of the current flow does not matter but when I tried solving this using the loop method it does.

For example if I assume the current flows ccw my equation becomes IR + V - E = 0 which means E=10V. But if the current flows cw my equation becomes IR - V + E = 0 which means E=8V.
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cepheid
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Mar15-12, 08:49 PM
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Quote Quote by theBEAST View Post
1. The problem statement, all variables and given/known data
I attached a picture of the problem. I am confused about how the current source will affect the direction of current flow. Does the current flow in the direction of the current source arrow? If not what does the arrow dictate? Our professor said the direction of the current flow does not matter but when I tried solving this using the loop method it does.

For example if I assume the current flows ccw my equation becomes IR + V - E = 0 which means E=10V. But if the current flows cw my equation becomes IR - V + E = 0 which means E=8V.
I'm not sure what you mean by "E" here. Generally speaking, yes, current flows in the direction indicated by the arrow on the current source. And since the circuit is a single loop, the current must be the same everywhere. This means that the current flows counterclockwise through the circuit, and that it is equal to 2 A.

Let's call the voltage across the current source VI. Let's measure voltages relative to the the minus terminal of the battery (at the bottom of the loop). This is our ground reference point. Since current flows counterclockwise, the right hand side of the resistor is at a higher potential than the left hand side. (The right hand side is "upstream" and the left hand side is "downstream", after energy has been dissipated as heat in the resistor). Therefore, as you move clockwise through circuit, starting at the battery minus terminal, your voltage increases by V as you go across the battery. Then it increase by IR as you go across the resistor from left to right (opposite to the current). Then it decreases as you go down across the current source. Hence:

V + IR - VI = 0

VI= V + IR

= 10 V + (2 A)(1 Ω) = 10 V + 2 V = 12 V

The voltage drop across the current source is 12 V.

It makes sense that the potential at the top of the current source is higher than the potential at the + terminal of the battery. This is necessary if it is to "win" the fight and drive the current counterclockwise. (The battery wants to drive it clockwise).
theBEAST
#3
Mar15-12, 08:55 PM
P: 368
Quote Quote by cepheid View Post
I'm not sure what you mean by "E" here. Generally speaking, yes, current flows in the direction indicated by the arrow on the current source. And since the circuit is a single loop, the current must be the same everywhere. This means that the current flows counterclockwise through the circuit, and that it is equal to 2 A.

Let's call the voltage across the current source VI. Let's measure voltages relative to the the minus terminal of the battery (at the bottom of the loop). This is our ground reference point. Since current flows counterclockwise, the right hand side of the resistor is at a higher potential than the left hand side. (The right hand side is "upstream" and the left hand side is "downstream", after energy has been dissipated as heat in the resistor). Therefore, as you move clockwise through circuit, starting at the battery minus terminal, your voltage increases by V as you go across the battery. Then it increase by IR as you go across the resistor from left to right (opposite to the current). Then it decreases as you go down across the current source. Hence:

V + IR - VI = 0

VI= V + IR

= 10 V + (2 A)(1 Ω) = 10 V + 2 V = 12 V

The voltage drop across the current source is 12 V.

It makes sense that the potential at the top of the current source is higher than the potential at the + terminal of the battery. This is necessary if it is to "win" the fight and drive the current counterclockwise. (The battery wants to drive it clockwise).
Thanks, so if the current source arrow was pointing the other direction would the voltage drop be 8V?

cepheid
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Mar15-12, 09:13 PM
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Direction of current flow in a circuit

Quote Quote by theBEAST View Post
Thanks, so if the current source arrow was pointing the other direction would the voltage drop be 8V?
Follow the rules and work it out. If the current flows from left to right across the resistor, what is the direction of the voltage drop across the resistor?
theBEAST
#5
Mar15-12, 09:26 PM
P: 368
Quote Quote by cepheid View Post
Follow the rules and work it out. If the current flows from left to right across the resistor, what is the direction of the voltage drop across the resistor?
Alright I am very confused now, my professor has not gone over these concepts as this is a MATH course. He says that it's not about the physics but for this question the answer changes depending on how you look at the physics. So I am confused about how you tell if it is a voltage gain or drop. I noticed that you said it is a drop of 12V how do you know this? Why isn't it a gain of 12V?
cepheid
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Mar15-12, 10:20 PM
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Quote Quote by theBEAST View Post
Alright I am very confused now, my professor has not gone over these concepts as this is a MATH course. He says that it's not about the physics but for this question the answer changes depending on how you look at the physics. So I am confused about how you tell if it is a voltage gain or drop. I noticed that you said it is a drop of 12V how do you know this? Why isn't it a gain of 12V?
From Kirchoff's loop rule: the sum of voltages around a closed loop in a circuit is zero.

This is also known as "Kirchoff's voltage law" or KVL, but fundamentally it is just an expression of the conservation of energy. When you go around a closed loop, the amount of energy gained by the charges has to be equal to the amount of energy lost, otherwise you are creating energy from nothing.

Conservation of energy is very much a physics principle. You do need to know physics in order to do circuit analysis.

How do I know it's a voltage drop when you go from the top side of the current source to the bottom side? Conservation of energy. Let's say I start at the battery minus terminal and move clockwise around the circuit (this is the opposite direction from the current flow, but let's say we go in this direction, just for the sake of argument). I gain potential energy in moving up across the battery, since the electric potential is higher at the top (+ terminal) than at the bottom. I gain even more energy in moving across the resistor from left to right, because the potential is lower on the left side of the resistor than on the right side. (The potential drop across a resistor is always in the direction of current flow). So, now, all the energy I've gained up to this point has to be lost in going across the current source from top to bottom. So the electric potential at the bottom end of the current source must be lower than it is at the top end.

If you don't know what electric potential is, then you don't know what voltage is, so I'd highly recommend you brush up on this concept in that case.

EDIT: It makes no sense that this is for a math course, unless it's a very low-level math course (middle school or early high school). I can't think of what math this problem would be testing other than basic algebra.
theBEAST
#7
Mar15-12, 10:27 PM
P: 368
Quote Quote by cepheid View Post
From Kirchoff's loop rule: the sum of voltages around a closed loop in a circuit is zero.

This is also known as "Kirchoff's voltage law" or KVL, but fundamentally it is just an expression of the conservation of energy. When you go around a closed loop, the amount of energy gained by the charges has to be equal to the amount of energy lost, otherwise you are creating energy from nothing.

Conservation of energy is very much a physics principle. You do need to know physics in order to do circuit analysis.

How do I know it's a voltage drop when you go from the top side of the current source to the bottom side? Conservation of energy. Let's say I start at the battery minus terminal and move clockwise around the circuit (this is the opposite direction from the current flow, but let's say we go in this direction, just for the sake of argument). I gain potential energy in moving up across the battery, since the electric potential is higher at the top (+ terminal) than at the bottom. I gain even more energy in moving across the resistor from left to right, because the potential is lower on the left side of the resistor than on the right side. (The potential drop across a resistor is always in the direction of current flow). So, now, all the energy I've gained up to this point has to be lost in going across the current source from top to bottom. So the electric potential at the bottom end of the current source must be lower than it is at the top end.

If you don't know what electric potential is, then you don't know what voltage is, so I'd highly recommend you brush up on this concept in that case.

EDIT: It makes no sense that this is for a math course, unless it's a very low-level math course (middle school or early high school). I can't think of what math this problem would be testing other than basic algebra.
Oh that is just a simple circuit that confused me. The other circuits have many resistors/other components so we have write a system of linear equations and use Gaussian elimination to solve it

EDIT: I think I understand it now. So if we moved counter clockwise it would be a voltage gain? Since we are now moving from the negative (low potential) to the positive (high potential) terminal of the current source.


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