Kirchoff's Equation w/ 2 Batteries and 4 resistors

In summary, the current through R1 is I1, the current through R3 is I3, and the voltage across R3 is VR3.
  • #1
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Homework Statement



The circuit in the figure is composed of two batteries (ε1 = 9.0 V and ε2 = 5.0 V) and four resistors (R1 = 110.0 Ω, R2 = 40.0 Ω, R3 = 30.0 Ω, and R4 = 50.0 Ω) as shown.

1). What is the current I1 which flows through R1?

2). What is the current I3 which flows through R3?

vgW4K.gif

Homework Equations



I used

ε1-R1I1+I2R2-R4I1-ε2=0

and

ε1-R1I1-R3I3-R4I1=0

The Attempt at a Solution



I've tried plugging in variables and attempting to solve for I3 but it's not the correct answer.
 
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  • #2
wizzleman said:

Homework Statement



The circuit in the figure is composed of two batteries (ε1 = 9.0 V and ε2 = 5.0 V) and four resistors (R1 = 110.0 Ω, R2 = 40.0 Ω, R3 = 30.0 Ω, and R4 = 50.0 Ω) as shown.

1). What is the current I1 which flows through R1?

2). What is the current I3 which flows through R3?

vgW4K.gif

Homework Equations



I used

ε1-R1I1+I2R2-R4I1-ε2=0

and

ε1-R1I1-R3I3-R4I1=0

The Attempt at a Solution



I've tried plugging in variables and attempting to solve for I3 but it's not the correct answer.
Hello wizzleman. Welcome to PF !

What you have looks correct.

You need another equation --- or a method to avoid using I2.

How is I2 related to I1 and I3 ?
 
  • #3
SammyS said:
Hello wizzleman. Welcome to PF !

What you have looks correct.

You need another equation --- or a method to avoid using I2.

How is I2 related to I1 and I3 ?

For the junction equation is I1= I2+I3 ?
 
  • #4
wizzleman said:
For the junction equation is I1= I2+I3 ?
To make your equation, ε1-R1I1+I2R2-R4I1-ε2=0, correct, you need I1 + I2 = I3 .
 
  • #5
SammyS said:
To make your equation, ε1-R1I1+I2R2-R4I1-ε2=0, correct, you need I1 + I2 = I3 .

Thanks for all the help so far, but I still can't get the answer.

I decided to make three loops

Left loop was ε1-R1I1+R2I22-R4I1=0

Right loop was ε2-R2I2-R3I3=0

And the whole loop was ε1-R1I1-R3I3-R4I1=0

Junction was I1=I2+I3

Solving through and using elimination I got up to 8V-320I1+80I2=0

How am I supposed to use the junction equations in here?
 
  • #6
wizzleman said:
Thanks for all the help so far, but I still can't get the answer.

I decided to make three loops

Left loop was ε1-R1I1+R2I22-R4I1=0

Right loop was ε2-R2I2-R3I3=0

And the whole loop was ε1-R1I1-R3I3-R4I1=0

Junction was I1=I2+I3

Solving through and using elimination I got up to 8V-320I1+80I2=0

How am I supposed to use the junction equations in here?
Two loop equations plus the junction equation give three equations in three unknowns. That is enough to solve for the three unknowns. The addition loop equation is not independent of the other two.

Your junction equation is inconsistent with your Left Loop & Right Loop equations. Both loop equations are consistent with I2 flowing in the upward direction. That implies that I[1/SUB] + I2 = I3, as I said in post #4.

Solve that for I2: I2 = I3 - I1 .

Plug that into the equation for the right loop.

ε2 - R2(I3 - I1) - R3I3=0

This can be written:
ε2 + R2I1 - (R2+R3)I3 = 0

Use that with the "whole loop" equation:
ε1 - (R1 + R4)I1 - R3I3 = 0
 
Last edited:
  • #7
The circuit is equivalent to this one. You can apply KCL to the bottom node to get an equation for the voltage across R3. Call that VR3

I3 + I2 + I1 = 0

VR3/R3 + (VR32)/R2 + (VR31)/R1+4 = 0

Solve for VR3

It's then easy to calculate all three currents.

Edite to make it clearer.
 

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  • #8
Thanks for all the help guys, I figured out the answers for I1 and I3.
 

1. What is Kirchoff's Equation with 2 Batteries and 4 resistors?

Kirchoff's Equation with 2 Batteries and 4 resistors is a mathematical formula used to analyze complex electrical circuits with multiple batteries and resistors. It is based on Kirchoff's laws, which state that the sum of currents entering a node is equal to the sum of currents leaving a node, and the sum of voltage drops in a closed loop is equal to the sum of the voltage sources in that loop.

2. How do you apply Kirchoff's Equation to a circuit with 2 Batteries and 4 resistors?

To apply Kirchoff's Equation to a circuit with 2 Batteries and 4 resistors, you must first identify all the nodes and loops in the circuit. Then, you can use the equations to calculate the currents and voltage drops at each node and loop. Finally, you can use these values to solve for the unknown variables in the circuit.

3. What are the limitations of Kirchoff's Equation with 2 Batteries and 4 resistors?

Kirchoff's Equation with 2 Batteries and 4 resistors assumes ideal conditions, such as negligible resistance in wires and perfect battery behavior. In real-life circuits, these assumptions may not hold true and can affect the accuracy of the calculated values. Additionally, the equation can become complex and difficult to solve for circuits with a large number of components.

4. Can Kirchoff's Equation with 2 Batteries and 4 resistors be applied to AC circuits?

No, Kirchoff's Equation with 2 Batteries and 4 resistors is only applicable to DC circuits. In AC circuits, the variables, such as current and voltage, are constantly changing over time, making the equations for Kirchoff's laws more complex. For AC circuits, other methods, such as phasor analysis, are used to analyze the circuit.

5. Why is Kirchoff's Equation with 2 Batteries and 4 resistors important in circuit analysis?

Kirchoff's Equation with 2 Batteries and 4 resistors is essential in circuit analysis as it allows us to calculate the currents and voltage drops in a complex circuit. This information is crucial in designing and troubleshooting circuits, as well as understanding how different components affect the overall behavior of the circuit.

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