Kirchhoff's Rules and Systems of Equations

In summary, to find the current in a series/parallel circuit, you can use the equations for outer current, parallel loop current, and junction rule. However, when substituting variables and dividing, be careful to use the correct values to avoid errors.
  • #1
succubus
33
0

Homework Statement


To find the current traveling through a circuit with parallel and series connections. The circuit looks something like this.

*Right now my webserver is giving me fits and won't let me login. I will try and get a picture to those who cannot understand what I am about to describe below, and with my poor description skills, will probably be everyone.

It's a series/parallel circuit. First there is R1(97.8), and then the circuit splits into parallel with R3(553) on the outside and R2(325) on the inside, and then continues around to a battery that has a measured voltage of 2.62 volts.

We measured the current in the lab, so we have the answers. The problem is, were supposed to use systems of equations to discover the answer as if we didn't measure it. IO have tried my best to do something about, and these are the equations I have come up with (we were only told to use 3, not the 4th for the inner current) and tried solving for each individual variable to no avail. This is what I've done.

Homework Equations



Outer current
(1) V-R1I1 - R3I3 = 0
Parallel Loop Current (Inside parallel circuit from R3 to R2)
(2) -R3I3 + R2I2 = 0
Junction Rule
(3) I1 = I2 +I3

I'm pretty sure my equations are correct.

I substituted I1 in equation 3 into equation 1 to get

V-R1(I2 + I3) - R3I3 = 0
Then I solved for I2
-R1I2 - R1I3 - R3I3 = -V
-R1I2 - I3(R1 + R3) = -V
-V + I3(R1 + R3) = -R1I2
(Divide by R1)

-V + I3(R3) = - I2

I2 = V - R3I3

Then I substituted what I got for I2 into equation 2

-R3I3 + R2(V-R3I3) = 0

-R3I3 - R2V - R2R3I3 = 0

-R3I3 - R2R3I3 = R2V

-I3(R3 + R2R3) = R2V
Divide by R3 + R2R3 yields

-I3 = V/2R3

which gives me something similar to the current I measured for I3, but 1 decimal off.

Measured currents are below

I1 = 7.43 mA
I2 = 24.3 mA
I3 = 24 mAWhere did I go wrong? When I plug in the answers they don't come anywhere close to what I measured across the circuit. I'm pretty sure we measured the current correctly. :/
 
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  • #2
succubus said:

Homework Statement


To find the current traveling through a circuit with parallel and series connections. The circuit looks something like this.

*Right now my webserver is giving me fits and won't let me login. I will try and get a picture to those who cannot understand what I am about to describe below, and with my poor description skills, will probably be everyone.

It's a series/parallel circuit. First there is R1(97.8), and then the circuit splits into parallel with R3(553) on the outside and R2(325) on the inside, and then continues around to a battery that has a measured voltage of 2.62 volts.

We measured the current in the lab, so we have the answers. The problem is, were supposed to use systems of equations to discover the answer as if we didn't measure it. IO have tried my best to do something about, and these are the equations I have come up with (we were only told to use 3, not the 4th for the inner current) and tried solving for each individual variable to no avail. This is what I've done.

Homework Equations



Outer current
(1) V-R1I1 - R3I3 = 0
Parallel Loop Current (Inside parallel circuit from R3 to R2)
(2) -R3I3 + R2I2 = 0
Junction Rule
(3) I1 = I2 +I3

I'm pretty sure my equations are correct.

I substituted I1 in equation 3 into equation 1 to get

V-R1(I2 + I3) - R3I3 = 0
Then I solved for I2
-R1I2 - R1I3 - R3I3 = -V
-R1I2 - I3(R1 + R3) = -V
-V + I3(R1 + R3) = -R1I2
(Divide by R1)

-V + I3(R3) = - I2

I2 = V - R3I3

Then I substituted what I got for I2 into equation 2

-R3I3 + R2(V-R3I3) = 0

-R3I3 - R2V - R2R3I3 = 0

-R3I3 - R2R3I3 = R2V

-I3(R3 + R2R3) = R2V
Divide by R3 + R2R3 yields

-I3 = V/2R3

which gives me something similar to the current I measured for I3, but 1 decimal off.

Measured currents are below

I1 = 7.43 mA
I2 = 24.3 mA
I3 = 24 mA


Where did I go wrong? When I plug in the answers they don't come anywhere close to what I measured across the circuit. I'm pretty sure we measured the current correctly. :/

You went wrong here:
-V + I3(R1 + R3) = -R1I2
(Divide by R1)

-V + I3(R3) = - I2

R1 + R3 divided by R1 is not R3.
 
  • #3

It seems like you have the right approach in using Kirchhoff's rules and setting up a system of equations to solve for the current in the circuit. However, there could be a few potential sources of error in your calculations. For example, when you substitute I1 from equation 3 into equation 1, you should get V-R1(I2+I3)-R3I3 = 0 instead of V-R1(I2+I3)-R3I3 = -V. This could be a simple mistake that is throwing off your final answer.

Another potential source of error could be in the measured values for the resistors. Make sure you are using the correct values for R1, R2, and R3 in your calculations. Also, double check your units to make sure they are consistent throughout your equations.

Overall, it seems like you are on the right track and just need to carefully check your calculations and measurements to find the source of the discrepancy. Keep in mind that it is common to have some small differences between calculated and measured values due to experimental error. As long as your calculations are correct and your values are close, your approach should be valid.
 

1. What are Kirchhoff's Rules and how are they used?

Kirchhoff's Rules, also known as Kirchhoff's Circuit Laws, are two principles used to analyze electrical circuits: the Kirchhoff's Current Law (KCL) and the Kirchhoff's Voltage Law (KVL). KCL states that the sum of currents entering a junction in a circuit must equal the sum of the currents leaving that junction. KVL states that the sum of voltages around a closed loop in a circuit must equal zero.

2. How do Kirchhoff's Rules relate to systems of equations?

Kirchhoff's Rules can be applied to a circuit with multiple components to create a system of equations. Each component in the circuit can be represented by an equation using KCL and KVL, and these equations can be solved simultaneously to find the values of currents and voltages in the circuit. This approach is especially useful for analyzing complex circuits with multiple loops and branches.

3. What are some real-world applications of Kirchhoff's Rules and systems of equations?

Kirchhoff's Rules and systems of equations are widely used in the field of electrical engineering to analyze and design circuits. They are also used in various industries such as telecommunications, power systems, and electronics manufacturing. They are essential for understanding and troubleshooting electrical systems.

4. Are Kirchhoff's Rules always applicable in circuit analysis?

No, Kirchhoff's Rules have some limitations and assumptions that must be considered. For example, they assume that the circuit is in a steady-state and that the components are linear. They also do not apply to circuits with non-conservative fields, such as circuits with capacitors and inductors.

5. How can I solve a system of equations using Kirchhoff's Rules?

To solve a system of equations using Kirchhoff's Rules, you can follow these steps: 1) Identify the components and junctions in the circuit and assign variables to represent currents and voltages. 2) Apply KCL and KVL to each junction and loop to create a system of equations. 3) Solve the equations simultaneously using algebraic methods or numerical techniques. 4) Check your solution by ensuring that the sum of currents at each junction and the sum of voltages around each loop is equal to zero.

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