Can you help me find the current through the 15 Ohm resistor branch?

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SUMMARY

The discussion centers on calculating the current through a 15 Ohm resistor in a circuit involving a 1 Ohm resistor. The participants confirm that the 1 Ohm and 15 Ohm resistors are in parallel, leading to a total resistance of 15/16 Ohm. The correct current through the 15 Ohm resistor is calculated to be 0.105 A using the formula I = V/R, where V is derived from the total current of 1.68 A. The conversation emphasizes the importance of understanding series and parallel circuits and the correct application of Ohm's Law.

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  • Understanding of Ohm's Law (I = V/R)
  • Knowledge of series and parallel resistor configurations
  • Ability to calculate total resistance in parallel circuits
  • Familiarity with circuit analysis techniques
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  • Study the calculation of total resistance in parallel circuits using the formula R = 1/(1/R1 + 1/R2).
  • Learn about Kirchhoff's laws for circuit analysis.
  • Practice solving circuit problems involving multiple resistors in series and parallel.
  • Explore advanced circuit analysis techniques such as Thevenin's and Norton's theorems.
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Students learning electrical engineering concepts, educators teaching circuit analysis, and anyone seeking to improve their understanding of resistor networks and current calculations.

Helly123
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Homework Statement


jkx0yv.jpg

Homework Equations


I = VR
Series = current the same, voltage different
Parallel = current different, voltage the same

The Attempt at a Solution


Is the 1 ohm and 15 ohm parallel? I'm still learning to differentiate it.
If yes, then R = 15/16 ohm
Is the R series to 5ohm? If yes, R total become 95/16 ohm
I = 10 / (95/16) = 160/95 A = 1,68A

For Current in resistor 15 = 15/16 x 1,68 A = 1,58 A

Is it correct?
 

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Helly123 said:
Is the 1 ohm and 15 ohm parallel? I'm still learning to differentiate it.
If yes, then R = 15/16 ohm
Yes.
Helly123 said:
For Current in resistor 15 = 15/16 x 1,68 A = 1,58 A

Is it correct?
No.
 
Asymptotic said:
No.

V for 15and1ohm = I.R = 1.68 x 15/16 = 1.57 v
I for 15 = V/R = 1.57 v / 15 ohm = 0.105 A
 
Helly123 said:
V for 15and1ohm = I.R = 1.68 x 15/16 = 1.57 v
I for 15 = V/R = 1.57 v / 15 ohm = 0.105 A
Yes.
Do you understand, and can you explain why this answer is correct, and the previous answer was wrong?
 
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Asymptotic said:
Yes.
Do you understand, and can you explain why this answer is correct, and the previous answer was wrong?
maybe because, the current cannot be calculated according to the fraction of total resistance.
 
Helly123 said:
maybe because, the current cannot be calculated according to the fraction of total resistance.
It can, but not quite as you did it. Would you expect the larger resistance to have the greater current or the lesser?
 
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haruspex said:
It can, but not quite as you did it. Would you expect the larger resistance to have the greater current or the lesser?
According to formula I = V/R
The greater the resistance the lesser the current.. to be honest, i try to think why the first one is wrong...
the total R for 15 and 1 ohm are 15/16 ohm
And the use the fraction method R1/Rtotal * I total
15/(15/16) * 1.68 A
But its wrong.. why is that..?
 
Since you've gotten the correct answers, but are not quite sure from whence they've come, allow me to suggest a methodology I call the "fully canonical with integrated smell test" approach.

Step 1: Label everything on the drawing including all the known values.
Step 2: Now begins an iterative procedure of identifying subsets of the overall problem that can be solved, solving them, updating the drawing with this new information, and leveraging each increase in circuit knowledge to solve for the other unknowns.
Step 3: Smell test phase. After all the voltages and currents have been solved for, and the drawing has been marked up with these values, inspect the drawing for obvious inconsistencies. In this example, R2 and R3 are a parallel pair, and by definition R2 and R3 must have the same voltage dropped across them - if they don't, something has gone wrong. In a similar vein, the current through R1 must be the same as total current, voltage across R1 added to voltage across the R2|R3 parallel pair must add up to total voltage, and so on.

CktCalc3.jpg


Doing it this way is a fair amount of work, and isn't as valuable a technique after one gets the hang of how to solve circuits, but is a useful 'sanity check' for beginners.
 

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Helly123 said:
According to formula I = V/R
The greater the resistance the lesser the current.. to be honest, i try to think why the first one is wrong...
the total R for 15 and 1 ohm are 15/16 ohm
And the use the fraction method R1/Rtotal * I total
15/(15/16) * 1.68 A
But its wrong.. why is that..?
I=V(1/R1+1/R2)=V/Rnet
I1=V/R1
I1/I=?/Rnet
Deduce what ? Is.
 
  • #10
Asymptotic said:
Since you've gotten the correct answers, but are not quite sure from whence they've come, allow me to suggest a methodology I call the "fully canonical with integrated smell test" approach.

Step 1: Label everything on the drawing including all the known values.
Step 2: Now begins an iterative procedure of identifying subsets of the overall problem that can be solved, solving them, updating the drawing with this new information, and leveraging each increase in circuit knowledge to solve for the other unknowns.
Step 3: Smell test phase. After all the voltages and currents have been solved for, and the drawing has been marked up with these values, inspect the drawing for obvious inconsistencies. In this example, R2 and R3 are a parallel pair, and by definition R2 and R3 must have the same voltage dropped across them - if they don't, something has gone wrong. In a similar vein, the current through R1 must be the same as total current, voltage across R1 added to voltage across the R2|R3 parallel pair must add up to total voltage, and so on.

View attachment 216833

Doing it this way is a fair amount of work, and isn't as valuable a technique after one gets the hang of how to solve circuits, but is a useful 'sanity check' for beginners.
I get it! 15 and 1 are parallel. Fractions used in series circuit , since total R = R1+R2...etc
So, in 15 ohm wire, resistance series to it is only 15 ohm itself, so the fraction is 15/15 instead of 15/(15/16)
 
  • #11
haruspex said:
I=V(1/R1+1/R2)=V/Rnet
I1=V/R1
I1/I=?/Rnet
Deduce what ? Is.
Maybe I1 = (Rnet/R1) * I net
Since I and R is inversely propotional
 
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  • #12
I'm not quite following.
Helly123 said:
15 and 1 are parallel.
Yes, they are.
Helly123 said:
Fractions used in series circuit , since total R = R1+R2...etc
I don't get this. "R = R1+R2...etc" is true for a series circuit, but the values aren't necessarily fractions.
Helly123 said:
So, in 15 ohm wire, resistance series to it is only 15 ohm itself, so the fraction is 15/15 instead of 15/(15/16)
I'm not following this at all.
 
  • #13
Asymptotic said:
I'm not quite following.

Yes, they are.

I don't get this. "R = R1+R2...etc" is true for a series circuit, but the values aren't necessarily fractions.
I'm not following this at all.
I missed something.. i meant
I = V * R
Whereas the R is 15ohm

V is the voltage of 1&15 ohm parallel pair.
 
  • #14
Helly123 said:
Maybe I1 = (Rnet/R1) * I net
Since I and R is inversely propotional
Right, since V=I1R1=I2R2=InetRnet.
This is the inverse of what you had in post #7.
 
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  • #15
haruspex said:
Right, since V=I1R1=I2R2=InetRnet.
This is the inverse of what you had in post #7.
Yes, i missed that before. Thanks
 

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