Calculating the Current in a Two battery Combination Circuit

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SUMMARY

The discussion focuses on calculating the current I4 through a 1.94Ω resistor in a circuit with two batteries. The circuit parameters include resistors R1 = 4.05Ω, R2 = 8.11Ω, R3 = 6.17Ω, and a current I1 = 2.97A with a voltage source E1 = 12V. The participant successfully derived the equation I4 = I1 - ((-(I1*R1) - I1R2 + E1)/R3) but initially mixed up the values for R2 and R3. The final solution confirmed that only two equations were necessary for the calculation.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Knowledge of Kirchhoff's Current Law (ƩI in a junction = ƩI out of a junction)
  • Familiarity with Kirchhoff's Voltage Law (ƩV in a loop = 0)
  • Basic circuit analysis techniques
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  • Practice solving complex circuits with multiple resistors and voltage sources
  • Learn about series and parallel resistor combinations
  • Explore the impact of battery configurations on circuit behavior
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Students studying electrical engineering, circuit designers, and anyone involved in analyzing battery-powered circuits.

BornSurvivor
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Homework Statement



For the circuit shown in the figure below, calculate the current I4 through the 1.94Ω resistor.
cutj20104.gif

As shown, R1 = 4.05Ω, R2 = 8.11Ω, R3 = 6.17Ω, R4 = 1.94Ω, I1 = 2.97A and E1 = 12V

Homework Equations


V=IR
ƩI in a junction = ƩI out of a junction
ƩV in a loop = 0

The Attempt at a Solution


I1R1+I1R2 -E1 + I2R3 =0
I1 = I2 + I4
So
I4 =I1 - ((-(I1*R1) - I1R2 + E1)/R3)

I don't have many tries left, so I want to know if my analytical solution is correct, and if I've just been making my error plugging the numbers back in.
 
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Hi BornSurvivor! :smile:
BornSurvivor said:
I1R1+I1R2 -E1 + I2R3 =0
I1 = I2 + I4

looks ok so far

but won't you need an equation involving E2 ? :confused:
 
Nope! Those two equations were all I needed. It turns out I had mixed up the numbers for R2 and R3. :)
 

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