Confused on concept of finding voltage drop across a resistor

In summary, the problem is that the directions of the current are not always clear. They used Kirchoff's law to find the equations for the current going into and out of the nodes, and then chose the path with the fewest unknowns.
  • #1
mr_coffee
1,629
1
Hello everyone I have a slight problem. I am confused on how you
know what path current is going to travel and also what voltage sources
your suppose to add up to a specific resistor. For example here is my picture, already solved:
The directions where to find the value of ix,
the orignal problem only had the values of each voltage source and the values of the ressitors
and it already had ix labeled.

http://img326.imageshack.us/img326/681/lastscan8rj.jpg

I'm confused on how they found I1, I1 and I3. I understand current travels
from high potential to low potential and also always comes out of the postive
side of a voltage source. So i understand the drawing of the current arrows.
But for example, when they found I1 = (20V+5V-10V)/5 OHMS = 3A.
Why did they add up those voltage sources? how did they know the
current was going to go that path and not just go around the top smaller loop?
so I1 = (5V)/5OHMS ? then they found I2 = (20V-10V)/10OHMS = 1A.
Also confused. how did they know I2's ressitor was going to have 20V
pass through it and 10V's but not the 5 V's?
Then finally I3 = (10V)/2 OHMS = 5A.
Why did they only use 10V? why wouldn't u use all of them?
See this is where I'm very confused. I get the concept of kcl,
Currents at a junction = Currents out a junction.
I also understand kvl, where the sum of higher potential = lower potential.
Thanks, any help would be great! :biggrin:
 
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  • #2
But for example, when they found I1 = (20V+5V-10V)/5 OHMS = 3A.
Why did they add up those voltage sources? how did they know the
current was going to go that path and not just go around the top smaller loop?

This is an application of Kirchoff's law which simply states that the sum of the potentials around a loop is 0, i.e. energy is conserved. The loop being considered is the largest loop which contains 3 batteries (5, 10 an 20 Volts) and a single resistor (5 Ohms).
 
  • #3
Thanks for the responce, but why did they choose the longest loop? that's my confusion, how do you know which path to choose to figure it out? Perhaps someone can explain it easier on this picture:

t3-7.gif


okay they say:
Below are the 5 Ohm's Law equations needed.
V3 = 2 x 3
V4 = 4I4
V7 = 7I7
V8 = 8I7
V9 = 9I6

If the top node and the bottom node are selected, these are the resulting KCL equations:
Top Node: -2 - I5 - I7 = 0
Bottom Node: 2 + I4 + __ = 0

Okay it all makes sense but why does V8 = 8*I7 and not 8*I5? how did they know I5 didn't take a right and go through V8? also V9 = 9I6, why couldn't I4 go through V9? And what are they talking about Top Node and bottom node?

KCL says, current in a junction = current out of junction. So why are they saying -2-I5-I7 = 0? all those currents are going INTO the junction so wouldn't those be postive? also when they say bottom node, 2+I4 + ? = 0, why are those postive? the question mark should be I6, which is correct, 2+I4+I6 = 0;
 
Last edited:
  • #4
In this format, EACH branch (the part of a circuit from one node to another) gets its own label for current ... the Voltage across R8 is BY DEFINITION = I7*R8.

The statement -I7 = I5 + I2 means that I2 and I5 go into this node (solder spot) , and -I7 comes out.
I like to use "reasonable guess" for the directions of the current, but
it's okay to use the format you have started ... in words, it means that
the total current coming out of a node = 0 ... or the total into a node = 0.

ANY path you choose will give you a "correct equation" (if you do it right).
with practice, you tend to choose a path with the fewest unknowns.
Otherwise, just choose systematically different paths, and deal with the
multiple equations with multiple unknowns as simultaneous equations.
It works, but is harder math.
 

1. What is voltage drop?

Voltage drop is the difference in electric potential between two points in an electrical circuit. It occurs when current passes through a resistor or other component, causing a decrease in voltage.

2. How is voltage drop calculated?

Voltage drop can be calculated using Ohm's law, which states that voltage drop (V) is equal to current (I) multiplied by resistance (R). In equation form, this is written as V = IR.

3. Why is voltage drop important to understand?

Understanding voltage drop is important because it can affect the performance and safety of electrical circuits. Excessive voltage drop can lead to a decrease in power and overheating of components, while too little voltage drop can cause malfunctions.

4. How does the concept of voltage drop apply to resistors?

Resistors are designed to have a specific amount of resistance in an electrical circuit. When current passes through a resistor, it causes a voltage drop. The amount of voltage drop across a resistor is directly proportional to its resistance.

5. What factors can affect voltage drop across a resistor?

The amount of voltage drop across a resistor can be affected by the resistance value of the resistor, the amount of current flowing through it, and the type of material the resistor is made of. Additionally, the length and temperature of the resistor can also impact the voltage drop.

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