MESH Current, Voltage Polarities

In summary: This has been so frustrating.The polarities within each loop for each resistor are determined by the assumed direction of current.Therefore, if current is assumed to flow clockwise through a resistor, the voltage across the resistor will be positive. If current is assumed to flow counter-clockwise through a resistor, the voltage across the resistor will be negative.
  • #1
tylercormier2
5
0
Hello,

I'm trying to figure out MESH Analysis but for the life of me I can't seem to get the Voltage polarities on the resistors right. I usually like to assign my loop currents clockwise so I can get in some sort of a routine(I know it makes no difference). Can someone please help me explain the voltage polarities across resistors?

Here is a DC example,
http://www.art-sci.udel.edu/ghw/phys245/05S/examples/images/Mesh-ex1.gif

Can someone please explain to me a method for choosing voltage polarities in these loops and give the MESH equations? I think I am just missing something basic, so a detailed explanation would be nice.
Thanks
 
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  • #2
tylercormier2 said:
Hello,

I'm trying to figure out MESH Analysis but for the life of me I can't seem to get the Voltage polarities on the resistors right. I usually like to assign my loop currents clockwise so I can get in some sort of a routine(I know it makes no difference). Can someone please help me explain the voltage polarities across resistors?

Here is a DC example,
http://www.art-sci.udel.edu/ghw/phys245/05S/examples/images/Mesh-ex1.gif

Can someone please explain to me a method for choosing voltage polarities in these loops and give the MESH equations? I think I am just missing something basic, so a detailed explanation would be nice.
Thanks

Please show us what you think they are first, and then we can comment. Them's the PF Rules.

Welcome to the PF!
 
  • #3
Alright,

Here`s how I would approach this problem,
Left Loop : -9v + I1(5k) +(I1-I2)(6k)=0 (Clockwise Current Direction)
Right Loop: -9v - (I1-I2)(6K) + I2(12K) = 0 (Clockwise Current Direction)

I`m not sure if this is right or not, it`s what I came up with. I think my problem is I am not 100% confident in my methods which would lead me to mistakes on tougher questions. Anyways, I`d appreciate your input.
Thanks again.
 
  • #4
The thing I'm getting stuck on is always hearing "The polarities within each loop for each resistor are determined by the assumed direction of current."
Can someone explain this concept for me please?
 
  • #5
The part where current flows into to a resistor is more positive than where it leaves.

In mesh analysis, assume that current flows in a loop in some direction.
 
  • #6
Ahhhhh. Thank you. Thank you soooo much. I knew it was something basic I was missing.
 

Related to MESH Current, Voltage Polarities

1. What is the definition of "MESH current"?

MESH current refers to the current that flows in a closed loop or circuit. It is also known as loop current or Kirchhoff's current.

2. How is MESH current different from other types of currents?

MESH current is unique because it only flows within a closed loop or circuit, while other types of currents can flow through multiple paths or branches.

3. What is the significance of understanding voltage polarities in a MESH current analysis?

Understanding voltage polarities is crucial in a MESH current analysis because it helps determine the direction of the current flow and allows for the accurate calculation of voltage drops across different components in the circuit.

4. How do you determine the polarity of a voltage source in a MESH current analysis?

The polarity of a voltage source can be determined by labeling the positive terminal with a plus sign (+) and the negative terminal with a minus sign (-). The polarity of the voltage source should align with the direction of current flow in the circuit.

5. Can voltage polarities change in a MESH current analysis?

Yes, voltage polarities can change in a MESH current analysis if the direction of current flow in the circuit changes. In this case, the polarity of the voltage source will also change to align with the new direction of current flow.

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