Current through a resistor in a network

In summary: Later you'll learn about Kirchhoff's Current Law (KCL). It's essentially what you're doing here.yea but .. isn't the current different in other resistors besides R4?Yes, the current will be different in other resistors besides R4. However, in the given circuit, since R1, R2, R3, and R5 are all equal, the current will be evenly divided among them. This means that the current through R4 will be the same as the current through R1, R2, R3, and R5.
  • #1
arl146
343
1

Homework Statement


Consider the electric circuit shown in the figure. Assume that V = 21.7 V, R1 = R2 = R3 = R4 = R5 = 6.00 Ohm.
Part 1) What is the equivalent resistance of the circuit?
Part 2) what is the electric current flowing through R4?


Homework Equations


Series and parallel equations


The Attempt at a Solution


I don't really know where to even start on this at all ?
 

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  • #2
Finding the equivalent resistance of the entire circuit would be a good place to start.
 
  • #3
Any electrical charge which flows through the battery can follow one of two paths on its way back to the battery. What are those two paths?
 
  • #4
fss said:
Finding the equivalent resistance of the entire circuit would be a good place to start.

Yea... that's half the point of even asking this question.

SammyS said:
Any electrical charge which flows through the battery can follow one of two paths on its way back to the battery. What are those two paths?

I think I might be able to get the second question. But I need help on the first part. finding the equivalence, I keep trying but not the right answer.
 
  • #5
arl146 said:
I think I might be able to get the second question. But I need help on the first part. finding the equivalence, I keep trying but not the right answer.

Can you show an attempt?
 
  • #6
I repeat: What are those two paths?
 
  • #7
Ok I got the first part. It was 16 Ohm. I realized I was using the wrong eq's. Now for the second part...
@SammyS .. I don't really know what you mean?
 
  • #8
You have the equivalent resistance (your 16 Ohms), so what is the total current that the battery is delivering?
 
  • #9
arl146 said:
Ok I got the first part. It was 16 Ohm. I realized I was using the wrong eq's. Now for the second part...
@SammyS .. I don't really know what you mean?
That was a way get you thinking thinking about analyzing the circuit as series/parallel combinations of resistors.
 
  • #10
i have a question about this problem .. (i know it's old but hey)

if i have the total current.. which in this case is 1.35625A, how do you approach getting the current across just R4?

can you think of it the same way as you would if they were capacitors? i mean like break it down .. resistors in series have the same current on it while resistors in parallel have the same voltage. so that being said, combinations of R2, R3, and R5 are in series with combos of R1 and R4.
so I235=I5=I4 no? so wouldn't I4 just be 1/3 of the total I ?
 
  • #11
arl146 said:
i have a question about this problem .. (i know it's old but hey)

if i have the total current.. which in this case is 1.35625A, how do you approach getting the current across just R4?

can you think of it the same way as you would if they were capacitors? i mean like break it down .. resistors in series have the same current on it while resistors in parallel have the same voltage. so that being said, combinations of R2, R3, and R5 are in series with combos of R1 and R4.
so I235=I5=I4 no? so wouldn't I4 just be 1/3 of the total I ?
What do you mean by total current? Where is that total current flowing?

attachment.php?attachmentid=32641&d=1298863335.gif
 
  • #12
ummm total current flowing throughout the whole entire circuit, right ..

the sum of the current that flows thought each resistor and everything
 
  • #13
Just imaging 10 cars departed from position pitstop V, So 10 cars in R1. We call Route#1
At junction R1,R2,R3, 4 cars goings route R2/5.
How many cars going thru R3
And finally how many car passing thru R4 before reentering pitstop V.

Hope you understand the analogy.
 
  • #14
arl146 said:
i have a question about this problem .. (i know it's old but hey)

if i have the total current.. which in this case is 1.35625A, how do you approach getting the current across just R4?

can you think of it the same way as you would if they were capacitors? i mean like break it down .. resistors in series have the same current on it while resistors in parallel have the same voltage. so that being said, combinations of R2, R3, and R5 are in series with combos of R1 and R4.
so I235=I5=I4 no? so wouldn't I4 just be 1/3 of the total I ?

If you look at the diagram, you will see that the current through R4 is exactly the same as the current through the battery. You have already determined that answer.

Chet
 
  • #15
is that because you can start at the battery and travel across is towards its negative side and since nothing is before R4 that makes it the same current?
 
  • #16
Just like water in river current. Amount of water remains the same.
In electricity it is just the flows of electrons.
 
  • #17
yea but .. isn't the current different in other resistors besides R4?
 
  • #18
If it branches to 2 different tributaries,then sum of them equal to main flow.
At the end of tributaries where they are rejoined to form main river, the amount of flow as the water enter will be equal to the amount when they rejoin.

No water ever lost in the process.
Only the pressure of flow decreases.
 
Last edited:
  • #19
oh .. ok i get it but ill see in the future if i can apply it myself
 
  • #20
arl146 said:
is that because you can start at the battery and travel across is towards its negative side and since nothing is before R4 that makes it the same current?

Sure. Where do you think the current that passes from the + side to the - side R4 goes next (rhetorical question)?
 

Related to Current through a resistor in a network

1. What is current through a resistor in a network?

Current through a resistor in a network refers to the flow of electric charge through a resistor, which is a passive electrical component that resists the flow of current. It is measured in units of amperes (A).

2. How is current through a resistor in a network calculated?

The current through a resistor in a network is calculated using Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). This can be represented by the equation I = V/R.

3. What factors affect the current through a resistor in a network?

The current through a resistor in a network is affected by the voltage applied to the circuit, the resistance of the resistor, and the overall design of the network. Other factors such as temperature, material properties, and surrounding components may also have an impact.

4. How does the current through a resistor in a network impact the overall circuit?

The current through a resistor in a network can affect the overall circuit in various ways, depending on the specific design and purpose of the circuit. In general, the current through a resistor will cause a voltage drop and can also generate heat, which may need to be considered in the circuit design.

5. How can the current through a resistor in a network be controlled or adjusted?

The current through a resistor in a network can be controlled or adjusted by changing the voltage applied to the circuit, adjusting the resistance of the resistor, or by using other components such as diodes or transistors. Additionally, the overall design and configuration of the network can also impact the current flow through the resistor.

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