Solving Current Divider Rule with 30V Voltage Source

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In summary, Amit was trying to solve a problem from last semester regarding current over a resistor, but he did not understand how to find the current. He tried to guess it each time and did not factor in other resistors. After understanding the concept, Amit was able to solve the problem.
  • #1
amitjakob
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Hello everyone.
im amit, newbie in this forums, hopefully I am posting my question in the correct one.
anyways, i was trying to solve this qustion from last semester considering the attached image as a circut
with voltage source of 30V

the task was to find the current over the resistor position diagonally R=20 Ohm

My assumption was with a total equivalent current of 2A is flowing over the 5 Ohm resistor , then to be divided between 10/20 Ohm resistors and then again between 10/20 ... what gives to result of 2/3A which is ... wrong.

result is 0.5A , but why ? does it matter in which direction i consider the current flow ? how can I know which is the right direction if so ?

thanks alot, exam coming by the end of the week
cya around. amit.
 

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  • #2
You are correct with that 2A flows from the source through the 5 Ohm resistor. Now look at the equivalent resistance on both branches. Both paths have a 20 Ohm resistance (20 Ohm on left and 20||20 + 10 = 20 Ohm). So the current is divided equally to 1A each. So now that 1A flows through the 10 Ohm resistor and gets to the point where the diagonal 20 Ohm connects with the vertical 20 Ohm. Both paths have equal resistance so the current evenly splits again to 0.5A each.

The direction does not matter. In general when you solve for current and get a minus sign that just means your initial guess at the direction was wrong. The current flows the opposite way instead.
 
  • #3
Allllright .. i think i got it now

from what i got from your answer one should always consider the total resistance when one calculate those sorts of problem, i tried to guess it each time with a specific resistor with a specific resistance and didnt consider other resistors might be connected and influencive.

now i know

Thanks:biggrin:

Amit, Germany.
 
  • #4
Hope it helps. This stuff becomes easier as you get more experience doing these types of problems.
 
  • #5
Corneo said:
Hope it helps. This stuff becomes easier as you get more experience doing these types of problems.

Thanks man, hopefully i can just forget about it after this friday, if ill pass the exam i failed last semester togather with this semester's course exam (Electical Engineering 1+2)

:-p:-p :devil::biggrin::blushing::mad::eek::bugeye::cool::zzz::cry::approve:
 

FAQ: Solving Current Divider Rule with 30V Voltage Source

What is the current divider rule?

The current divider rule is a principle in circuit analysis that helps determine the distribution of current in a parallel circuit. It states that the total current entering a parallel circuit is divided among the branches in proportion to their individual resistances.

How does the current divider rule work?

The current divider rule works by using the ratio of each branch's resistance to the total resistance of the circuit. The current in each branch is then determined by multiplying the total current entering the circuit by the ratio.

How do you apply the current divider rule with a 30V voltage source?

To apply the current divider rule with a 30V voltage source, you first need to determine the total resistance of the circuit. Then, use the ratio of each branch's resistance to the total resistance to calculate the current in each branch. Finally, use Ohm's law (I = V/R) to determine the voltage across each branch.

Can the current divider rule be used with any voltage source?

Yes, the current divider rule can be used with any voltage source. It is a general principle that applies to parallel circuits, regardless of the voltage source.

Are there any limitations to the current divider rule?

The current divider rule assumes that the voltage source is constant and that the branches have equal resistance. If these conditions are not met, the results may not be accurate. Additionally, the rule only applies to parallel circuits and cannot be used for series circuits.

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