Direction of current flow in a circuit

In summary, the student is confused about the direction of current flow in a circuit with a current source. They are unsure if the current flows in the direction of the arrow on the current source or if it is dictated by something else. The professor has stated that the direction of current flow does not matter, but the student is having trouble with this concept when using the loop method to solve the problem. The current is shown to be 2A and the voltage drop across the current source is determined to be 12V. The student is still unsure about the concept of voltage gain or drop and how to determine it in this circuit.
  • #1
theBEAST
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0

Homework Statement


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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  • #2
theBEAST said:

Homework Statement


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).
 
  • #3
cepheid said:
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?
 
  • #4
theBEAST said:
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?
 
  • #5
cepheid said:
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?
 
  • #6
theBEAST said:
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.
 
  • #7
cepheid said:
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.
 

1. What is the direction of current flow in a circuit?

The direction of current flow in a circuit is from the positive terminal of the voltage source to the negative terminal. This is known as conventional current flow and is based on the historical direction of electron flow.

2. Does current always flow in the same direction in a circuit?

No, in an alternating current (AC) circuit, the direction of current flow constantly changes. This is because the voltage source used in AC circuits, such as a wall outlet, changes polarity at regular intervals.

3. Can current flow in both directions in a circuit?

Yes, in a direct current (DC) circuit, current can flow in both directions. This is because the voltage source used in DC circuits, such as a battery, maintains a constant polarity.

4. How does the direction of current flow affect circuit components?

The direction of current flow determines the functioning of circuit components. Some components, such as diodes, only allow current to flow in one direction, while others, such as resistors, are not affected by the direction of current flow.

5. How can I determine the direction of current flow in a circuit?

The direction of current flow can be determined by using a multimeter to measure the voltage across a component. The positive probe should be placed on the component's positive terminal, and the negative probe on the negative terminal. The direction of current flow will be from the positive to the negative probe.

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