Kirchoff's Laws

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Claimed by Abieshek Subramaniam - Fall 2019

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The node (or junction) rule states that the current flowing in is equal to the current that flows out. I1 + I4 = I2 + I3 + I5

Kirchoff's Laws are are two fundamental principles of electric circuits and are used to determine the behaviors of electric circuits and their components. They serve as a guide for how circuits will behave, and are accurate for all DC and low frequency AC circuits. These principles are used to measure voltage and current in voltmeters and ammeters.

Kirchoff's Node Rule, also known as Kirchoff's First Law or Kirchoff's Junction Rule, further exercises the law of Conservation of Charge and states that if current is constant, all the current that flows through one junction must be equal to all the current that flows out of the junction. This rule can be applied both to conventional and electron currents. This rule is not a fundamental principle, but rather a consequence of the fundamental principle of conservation of charge and the definition of steady state.

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The Loop rule states that the sum of voltage around a loop is equal to 0. LOOP 1: [math]\displaystyle{ \Delta {V}_{AB} + \Delta {V}_{BC} + \Delta {V}_{CF} + \Delta {V}_{FA} = 0 }[/math] LOOP 2: [math]\displaystyle{ \Delta {V}_{FC} + \Delta {V}_{CD} + \Delta {V}_{DE} + \Delta {V}_{EF} = 0 }[/math] LOOP 3: [math]\displaystyle{ \Delta {V}_{AB} + \Delta {V}_{BC} + \Delta {V}_{CD} + \Delta {V}_{DE} + \Delta {V}_{EF} + \Delta {V}_{FA} = 0 }[/math]

Kirchoff's Second Law, also known as Kirchhoff's Loop Rule or Kirchhoff's Voltage Law states that the sum of potential differences around a closed circuit is equal to zero. More simply, in a completed circuit, the voltages around a loop will sum to 0. This is because voltage is just energy per unit charge, and both energy and charge are conserved by fundamental laws.

Note that this is only true when the magnetic field is neither fluctuating nor time varying. If a changing magnetic field links the closed loop, then the principle of energy conservation does not apply to the electric field, causing the Loop Rule to be inaccurate in this scenario.

The Main Idea

Kirchoff's Node Rule

The node rule states that at any junction in an electrical circuit, the amount of current flowing into the junction is equal to the amount of current flowing out of the junction in steady state.

In the steady state, for many electrons flowing into and out of a node,

  • electron current: [math]\displaystyle{ net\ i_{in} = net\ i_{out}, }[/math] where [math]\displaystyle{ i = nA\mu }[/math][math]\displaystyle{ E }[/math]
  • conventional current: [math]\displaystyle{ net\ I_{in} = net\ I_{out}, }[/math] where [math]\displaystyle{ I = |q|nA\mu }[/math][math]\displaystyle{ E }[/math]

Conservation of Charge

This rule is an application of the conservation of electric charge, basically that charge within a circuit cannot be created or lost. During the flow process around the circuit, there is no loss of any charge, thus the total current in any cross-section of the circuit is the same. If there are no nodes in the loop, the conventional current is the same throughout the loop.

Kirchoff's Loop Rule

The loop rule simply states that in any round trip path in a circuit, Electric Potential equals zero. This applies through any round trip path; in more complex circuits, there can be multiple round trip paths. This principle is an application of the conservation of energy, specifically within a circuit. This principle is often used to solve for resistance or current passing through of light bulbs and other resistors, as well as the capacitance or charge of capacitors in a circuit.

LOOP 1: [math]\displaystyle{ \Delta {V}_1 = emf - I_1R_1 = 0 }[/math]

LOOP 2: [math]\displaystyle{ \Delta {V}_2 = -I_1R_1 + I_2R_2 = 0 }[/math]

LOOP 3: [math]\displaystyle{ \Delta {V}_3 = emf - I_2R_2 = 0 }[/math]

To figure out the sign of the voltages, act as an observer walking along the path. Start at the negative end of the emf and continue walking along the path. The emf will be positive in the loop rule because you are moving from low to high voltage. Once you reach a resistor or capacitor, this will be negative in the loop rule equation because it is high to low voltage. Continue along the path until you return to the starting position.

Bringing Both Laws Together

Find all possible node and loop equations for the following circuit.

You should have 3 Loop Equations and 2 Node equations.

Some examples:

[math]\displaystyle{ I_1= I_2+I_3 }[/math]

[math]\displaystyle{ I_1*R_1 + emf_1 + emf_3 + I_3*R_4 = 0 }[/math]

Matrices in Conjunction with Kirchhoff's Laws

After creating loop equations for all possible loops in a circuit using the node rule and loop rule, you can input the coefficients of the equations into a matrix. After inputting it into the matrix, you can use linear algebra and row reduction in order to solve for the current or resistances of the different parts of the circuit. This format and method of understanding complex circuits are especially useful for circuits with a large number of loops as it makes the algebra involved in getting the currents much simpler and straightforward. Simply creating the equations for every loop possible and every node and then inputting the coefficients into a row reduction calculator can give you all the values for I and/or R that you might need.

After finding the loop equations you can use a matrix calculator as well to further simply your task!: Online Matrix Row Reduciton Calculator

An example of how you would format the matrix of a Kirchhoff's law problem is shown below:

[math]\displaystyle{ \mathbf{A} = \begin{bmatrix} R1 & R2 & R3 & | EMF \\ & & & | EMF \\ & & & | EMF \end{bmatrix}. }[/math]

Simply place to coefficients or resistances of each of the different resistors as seperate elements in the columns of the matrix while keeping track of which resistance pertains to which part of the loop and place the EMF of that loop as the solution column in the augmented matrix. Then after row reducing, each of the resistors should be simplified so that there is a 1 in at least 1 row for every resistor and the current at the end of that row is the current for the resistor that is in the same column as the 1 in the row of the current.

For example in the matrix below the current going through R2 is 1/15 amps:

[math]\displaystyle{ \mathbf{A} = \begin{bmatrix} R1 & R2 & R3 & | EMF \\ & 1 & & | 1/15 \\ & & & | EMF \end{bmatrix}. }[/math]

Limitations

The Node rule assumes that current flows only in conductors, and that whenever current flows into one end of a conductor it immediately flows out the other end. This is not a safe assumption for high-frequency AC circuits. In other words, the node is valid only if the total electric charge, [math]\displaystyle{ Q }[/math], remains constant in the region being considered. In practical cases this is always so when the node rule is applied at a point.

The Loop rule is based on the assumption that there is no fluctuating magnetic field linking the closed loop. This is not a safe assumption for high-frequency AC circuits. In the presence of a changing magnetic field, the electric field is not a conservative vector field. Therefore, the electric field cannot be the gradient of any potential. That is to say, the line integral of the electric field around the loop is not zero, directly contradicting KVL.

Understanding these limitations is essential to understanding how motional emf, motors, and generators work.

A Mathematical Model

Kirchoff's Node Rule

The node rule can be stated as:

[math]\displaystyle{ \sum \Delta{I} = 0 }[/math]

where [math]\displaystyle{ I }[/math] stands for the current of the individual parts or wires in a circuit and the sign of the current that flows into the junction is opposite of the current that flows out of the junction. This simply supports the idea that charge cannot be created or destroyed because the total current must remain equal regardless of the path it takes.

[math]\displaystyle{ \sum_{k=1}^n I_k = 0 }[/math]

where [math]\displaystyle{ n }[/math] is the total number of branches with current flowing through the node, as well as along any node in a circuit:

[math]\displaystyle{ \Delta {I}_{1} + \Delta {I}_{2} + \space.... = 0 }[/math]

or more generally:

[math]\displaystyle{ \sum_{k=1}^n \tilde{I_k} }[/math] = 0

Kirchoff's Loop Rule

A mathematical representation is:

[math]\displaystyle{ \sum \Delta{V} = 0 }[/math]

where [math]\displaystyle{ V }[/math] stands for the voltage of the individual parts or wires in a circuit and the sign of the voltage that flows clock-wise is opposite of the current that flows counterclock-wise. This simply supports the idea that energy cannot be created or destroyed because the total current must remain equal regardless of the path it takes.

[math]\displaystyle{ \sum_{i=1}^n {V}_{i} = 0 }[/math]

where [math]\displaystyle{ n }[/math] is the number of voltages being measured in the loop, as well as

[math]\displaystyle{ \Delta {V}_{1} + \Delta {V}_{2} + \space.... = 0 }[/math]

along any closed path in a circuit. The voltages may also be complex:

[math]\displaystyle{ \sum_{k=1}^n \tilde{V}_k = 0 }[/math]

Matrix Intrepretation

Using matrices can simplify the process of using node and loop rules in order to find out different parts of complex circuit. Simply write out all the loop and node rules for a circuit. Then place coefficients of the equations you created in a matrix in the format below. The matrix should be an augmented matrix with the elements on the left being the resistances and the right being the EMF of that loop.

[math]\displaystyle{ \mathbf{A} = \begin{bmatrix} R1 & R2 & R3 & | EMF \\ & & & | EMF \\ & & & | EMF \end{bmatrix}. }[/math]

To row reduce you should add, subtract, multiply, interchange or add a multiple of rows to each other until the matrix is as far simplified as possible. Otherwise, you can simply use a row reduction calculator online (Online Matrix Row Reduction Calculator) in order to do it for you and the right hand column will show the currents associated with the different resistors. To figure out which current goes with which resistor look to see which resistor has a 1 in its column and the current at the end of that row is the current of that resistor.

More information on how to row reduce by hand can be found here: Row Reduction Methodology

A Computational Model

Click here for an online circuit simulator. It opens up with an LRC circuit that has current running through it. The graphs on the bottom show the voltage as the current runs through the circuit. You can see that after a full loop the voltage is 0, verifying the loop rule. You can make all sorts of different circuits and loops and see for yourself.

Examples

Simple

Question

Figure 1

Figure 1 displays a node in a circuit. I1 is equal to 10 amps. I2 is equal to 4 amps. What is I3?

Solution

The current flowing into the node: [math]\displaystyle{ I_1 = 10A }[/math]
The current flowing out of the node: [math]\displaystyle{ I_2 + I_3 }[/math]
We know that the current flowing in must equal the current flowing out, so [math]\displaystyle{ 10A = 4A + I_3 }[/math]
Therefore [math]\displaystyle{ I_3 }[/math] must equal 6A.

Question

The circuit shown above consists of a single battery and a single resistor. The resistance of the wires is negligible for this problem.

If the [math]\displaystyle{ emf }[/math] is 5 V and the resistance of the resistor is 10 ohms, what is the current passing through the resistor?

Solution

Although we can solve this using the [math]\displaystyle{ V = IR }[/math] equation for the whole loop, let's examine this problem using the loop rule equation.

The loop rule equation would be [math]\displaystyle{ {V}_{battery} - {V}_{resistor} = 0 }[/math]

Since we know the [math]\displaystyle{ emf }[/math] of the battery we just need to find the potential difference through the resistor. For this we can use the equation of [math]\displaystyle{ V = IR }[/math].

Thus we now have an loop rule equation of [math]\displaystyle{ emf - IR = 0 }[/math] From here it is a relatively simple process to find the current. We can rewrite the loop rule equation as [math]\displaystyle{ emf = IR }[/math] and then plug in 5 for the emf and 10 for the resistance, leaving us with I = .5 amperes.

Middling

Question

The circuit shown above consists of a single battery, whose [math]\displaystyle{ emf }[/math] is 1.3 V, and three wires made of the same material, but having different cross-sectional areas. Let the length of the thin wires be [math]\displaystyle{ {L}_{thick} }[/math] and the length of the thin wire be [math]\displaystyle{ {L}_{thin} }[/math] Find a loop rule equation that starts at the negative end of the battery and goes counterclockwise through the circuit.

Solution

When beginning this problem, you must notice that the difference in cross-sectional areas affects the electric field in each wire. Because of this we will denote the electric field at D. as [math]\displaystyle{ {E}_{D} }[/math] and the electric field everywhere else as [math]\displaystyle{ {E}_{A} }[/math]. To begin we will go around the circuit clockwise and add up each component. First, we know that the [math]\displaystyle{ emf }[/math] of the battery is 1.3 V. Then, we will add up the potential voltage of each of the wires.

Remember that the electric potential of a wire is equal to the product of the electric field and the length of the wire. From this we can now find the potential difference of each section of the wires. The electric potential of location A - C is [math]\displaystyle{ {E}_{A} * {L}_{thick} }[/math]. This is the same for the electric potential of locations E - G of the wire. For the thin section of the wire, the electric potential is [math]\displaystyle{ {E}_{D} * {L}_{thin} }[/math]. From here we just go around the circuit counterclockwise and add each potential difference to the loop rule equation.

Thus we can find that a loop rule equation is: [math]\displaystyle{ emf - 2 ({E}_{A} * {L}_{thick}) - {E}_{D} * {L}_{thin} = 0 }[/math]

This can also be rewritten as: [math]\displaystyle{ emf = 2 ({E}_{A} * {L}_{thick}) + {E}_{D} * {L}_{thin} = 0 }[/math]

Difficult

Question

Figure 2

In Figure 2, I1 equal 23 amps, I2 equals 5 amps and I3 equals 42 amps. What is I4?

Solution

The current flowing into the node: [math]\displaystyle{ I_1 + I_2 = 23A + 5A = 28A }[/math]
The current flowing out of the node: [math]\displaystyle{ I_3 + I_4 = 42A + I_4 }[/math]
Applying the node rule by using substitution, [math]\displaystyle{ I_4 = -14A }[/math]
But how could we get a negative current? This negative current implies that [math]\displaystyle{ I_4 }[/math] flows in the opposite direction of what we assumed. A negative current in a loop rule problem implies that a current is in the opposite direction.

Question

For the circuit above, imagine a situation where the switch has been closed for a long time. Calculate the current at a,b,c,d,e and charge Q of the capacitor. Answer these using [math]\displaystyle{ emf, {R}_{1}, {R}_{2}, and \space C }[/math]

Solution

First, write loop rule equations for each of the possible loops in the circuit. There are 3 loop equations that are possible.

[math]\displaystyle{ emf - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} = 0 }[/math]

[math]\displaystyle{ emf - {I}_{1}{R}_{1} - Q/C = 0 }[/math]

[math]\displaystyle{ {I}_{2}{R}_{2} - Q/C = 0 }[/math]

From here, we can then solve for the current passing through a,b,d and e. We also know that the current passing through these points must be the same so [math]\displaystyle{ {I}_{1} = {I}_{2} }[/math]

[math]\displaystyle{ emf - {I}_{1}{R}_{1} - {I}_{1}{R}_{2} = 0 }[/math]

[math]\displaystyle{ emf = {I}_{1}({R}_{1} + {R}_{2}) = 0 }[/math]

[math]\displaystyle{ emf/({R}_{1} + {R}_{2}) = {I}_{1} }[/math]

So the current at [math]\displaystyle{ a,b,d,e = emf/({R}_{1} + {R}_{2}) }[/math]

You must also know that once a capacitor is charging for a long time, current no longer flows through the capacitor. We can then easily solve for c because since current is no longer flowing through the capacitor, the current at c = 0.

We will use the loop rule equation of [math]\displaystyle{ {I}_{2}{R}_{2} - Q/C = 0 }[/math] to solve for Q.

[math]\displaystyle{ {I}_{2}{R}_{2} = Q/C }[/math]

[math]\displaystyle{ C*({I}_{2}{R}_{2}) = Q }[/math] Since [math]\displaystyle{ {I}_{1} = {I}_{2} }[/math]

[math]\displaystyle{ C*({I}_{1}{R}_{2}) = Q }[/math]

We will plug in what we found {I}_{1} equals from before.

[math]\displaystyle{ C* (emf/({R}_{1} + {R}_{2})){R}_{2}) = Q }[/math]

Lastly, [math]\displaystyle{ Q = C* (emf/({R}_{1} + {R}_{2})){R}_{2} }[/math] and we have now solved the problem.

Current at [math]\displaystyle{ a,b,d,e = emf/({R}_{1} + {R}_{2}) }[/math]

Current at [math]\displaystyle{ c = 0 }[/math]

[math]\displaystyle{ Q = C* (emf/({R}_{1} + {R}_{2})){R}_{2} }[/math]

Question (USING MATRICES)

Six identical resistors and two different batteries are connected in a circuit as show in the diagram. One of thebatters has a potential difference of 20 V and the other has a potential difference of 5 V. The direction of thecurrents in the circuit are indicated and labeled. Solve your loop and node equations to determine the current through resistorR6. The resistance of eachresistor is 100 Ω.

Solution

From the diagram, it is evident that this particular problem will require quite a few loop and node rule equations, 11 total to be exact. To figure out the current through resistor R6 we will need to setup the node and loop rule equations in order to have a system of equations that will have enough variables and coefficients in order to be able to solve for this current.

The node equations are:

0 = I1 - I3 - I6

0 = I1 - I2 - I4

0 = I2 - I3 - I5

0 = I4 + I5 - I6

The loop rule equations are:

0 = -emf1 + (R1 + R4)I1 + (R5)I4 + (R4)I6

0 = -emf1 + (R1 + R6)I1 + (R2)I2 + (R3)I3

0 = -emf1 + (R1 + R6)I1 + (R2)I2 + emf2 + (R4)I6

0 = (R2)I2 + (R3)I3 - (R4)I6 - (R5)I4

0 = (R2)I2 - emf2 - (R5)I4

0 = (R3)I3 - (R4)I6 + emf2

0 = -emf1 + (R1 + R6)I1 + (R5)I4 + emf2 + (R3)I3

Now that we have the equations instead of taking the time to solve the system of equations for all 11 of these equations, we can simply use a matrix and row reduce in order to find the current is going through R6 as we should have enough relationships to deduce this. The matrix for this particular problem would be the following:

In the matrix above, the headers at the top represent what the coefficients in the matrix pertain too, being the emf and the resistances associated with each of the currents form I1 through I6. Now using a row reduction calculator we can make an augmented matrix with the results being the emf column and the currents and associated resistances being on the LHS and we can find the I in the right hand column of the augmented matrix that is associated with the correct part of the circuit we are looking for and that will be our result for the current through the circuit at that resistor. In this case, that ends up being 1/15 amps.

In this particular problem, using traditional system of equation methods would prove problematic and time consuming as there are 11 equations that we have to work with. Thus, by using a matrix and row reduction, we can simply place the coefficients into a matrix, row reduce, and look at the number that is in the column of the resistor we are trying to find the current for and that will be our solution.

Connectedness

The Node Rule is connected to a lot of other topics in physics. The loop rule is the most important one, as the node rule and loop rule in conjunction allow us to solve circuits. The node rule is also connected to other concepts such as voltage, current and electricity. The Node Rule also supports the law of conservation of energy because in essence, current is simply the flow of electric charge, and since you cannot (at least as of now) create energy or electrons out of nothing, everything that is put into the system must come out somehow. Therefore, all the current that is applied, must come out the other end.

The Loop Rule is simply an extension of the conservation of energy applied to circuits. Circuits are ubiquitous as they are featured in almost every technology today. Our understanding of these technologies is rooted in the empirical discoveries made in the mid 19th century, and further boosted by the the advancement of theoretical knowledge due to the likes of Faraday and Maxwell.

One interesting note about Loop Rule is that is does not apply universally to all circuits. In particular, AC (alternating current) circuits at high frequencies, have a fluctuating electric charge that changes direction. This causes the electric potential of a round trip path around the circuit to no longer be zero. However, DC (direct current) circuits, and low frequency circuits in general, still follow the loop rule.

These laws allow for voltmeters and ammeters to work, with voltmeters having a high resistance and being connected in parallel and ammeters having a low resistance and being connected in series.

History

Gustav Kirchhoff

Gustav Kirchhoff was a German physicist who lived during the 19th century. There are many equations and laws named after him that he helped to discover. His circuit laws (the node rule and loop rule) were the first laws that he conceived, and discovered this during his time as a student at Albertus University of Königsberg in 1845; he later wrote his doctoral dissertation on these laws. Kirchoff went on to explore the topics of spectroscopy and black body radiation after his graduation from Albertus. In addition to his circuit laws, he is also known for his law of thermochemistry and three laws of spectroscopy, the latter of which helped lead to quantum mechanics.

See Also

Further Reading

External Links

https://www.boundless.com/physics/textbooks/boundless-physics-textbook/circuits-and-direct-currents-20/kirchhoff-s-rules-152/the-junction-rule-539-6331/

https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s

http://www.tutorvista.com/content/physics/physics-iv/current-electricity/kirchhoffs-rules.php

http://physics.bu.edu/~duffy/py106/Kirchoff.html

References

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