Combination Circuit Question: Determining Current

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Homework Help Overview

The discussion revolves around a combination circuit problem involving resistors, where the original poster seeks to determine the current flowing through each resistor given a voltage source of 14V and a specific resistance value of 6.3Ω. The problem involves both series and parallel resistor configurations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the overall resistance and current using Ohm's Law and series/parallel resistor rules but struggles with determining the individual currents for each resistor. Some participants question the understanding of voltage across different components and suggest using known values to find unknown currents.

Discussion Status

The discussion is ongoing, with participants providing guidance on using previously calculated values to find voltages and subsequently the currents through individual resistors. There is recognition of the need for clarification on the mathematical relationships involved.

Contextual Notes

The original poster has been given correct answers for the currents but is unable to reverse-engineer the calculations. There is an acknowledgment of potential confusion regarding the application of Ohm's Law and the relationships between voltage, current, and resistance in the circuit.

Zaphia
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Homework Statement



Find the current flowing through each of the resistors in the circuit shown, where [tex]\xi[/tex] = 14V and R=6.3[tex]\Omega[/tex].

Circuit-Fig2131.jpg


Homework Equations



I=V/R
where I=current (Amps), V=potential different (Volts), R=resistance ([tex]\Omega[/tex]).

Resistors in Series:

Req=R1+R2+...
Ieq=I1=I2=...
Veq=V1+V2+...

Resistors in Parallel:

1/Req=1/R1+1/R2+...
Ieq=I1+I2+...
Veq=V1=V2=...

The Attempt at a Solution



Step 1: Determine the overall resistance of the circuit.

6.3[tex]\Omega[/tex] and 5.8[tex]\Omega[/tex] are in series, so Req=12.1[tex]\Omega[/tex] so far.

Next, 12.1[tex]\Omega[/tex] and 3.2[tex]\Omega[/tex] are in parallel, so the resulting Req is: 1/Req=1/12.1[tex]\Omega[/tex]+1/3.2[tex]\Omega[/tex]=2.531[tex]\Omega[/tex].

Finally, 2.531[tex]\Omega[/tex], 1.0[tex]\Omega[/tex] and 4.5[tex]\Omega[/tex] are in series. So Req=2.531[tex]\Omega[/tex]+1.0[tex]\Omega[/tex]+4.5[tex]\Omega[/tex]=8.031[tex]\Omega[/tex]

The total resistance of the circuit is 8.031[tex]\Omega[/tex].

Step 2: Determine the overall current, I.

Using Ohm's equation, I=V/R, I found that I=1.743A

Step 3: Determine the individual currents at each of the resistors.

This is the part that I can't figure out. I know that resistors 4.5[tex]\Omega[/tex] and 1.0[tex]\Omega[/tex] have current, I=1.743A because the current travels from positive to negative, but I don't understand how to solve for the currents for the rest of the resistors.

I also understand that the resistors 6.3[tex]\Omega[/tex] and 5.8[tex]\Omega[/tex] will also have the same current.

(I have been given the correct answers for all of the resistors, but I can't reverse-solve the problems either:)
6.3[tex]\Omega[/tex] and 5.8[tex]\Omega[/tex], I=0.365A
3.2[tex]\Omega[/tex], I=1.38A

Thanks for any insight!
 
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Your REQ calculation and source current calculation looks correct.

Notice that the voltage across (R = 3.2 ohms) is the same as the voltage across the combination (R = 6.3 + R5.8 ohms). You combined those in series, then parallel in your earlier work and got (R = 2.53) ohms as a result. The voltage across that combination resistor will tell you the voltages across the re-expanded elements (R = 3.2 ohms) and
(R = 6.3 + 5.8 ohms). Let me know if I need to clarify, I know that probably sounds confusing.
 
I understand (I think!) what your saying, but I don't know the math in order to solve the problem. Is it a set of ratios to get the Current values for each resistor?
 
You just need to use Ohm's Law V = IR. You can determine the voltage across the resistor combination I was talking about in the previous post because you will know the current through it and the equivalent resistance. You solved for those values earlier, ISource = 1.74 A and REQ = 2.53 ohms, from your calculation below.
Zaphia said:
Next, 12.1[tex]\Omega[/tex] and 3.2[tex]\Omega[/tex] are in parallel, so the resulting Req is: 1/Req=1/12.1[tex]\Omega[/tex] +1/3.2[tex]\Omega[/tex] =2.531[tex]\Omega[/tex].


Use those values to determine the voltage across that equivalent resistive element, then expand the circuit back to its original form. Then you will have the voltages across the resistor (R = 3.2 ohms) and the resistor combination (R = 6.3 + 5.8 = 12.1 ohms). You can use the calculated voltage value and the resistor values to determine the two unknown branch currents.
 

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