Kirchhoff’s Laws

Original source retained for reference; this draft does not establish AP coverage.

Original author draft. Original wording and recorded answer states are preserved without claiming reviewed scientific correctness or AP alignment.

Physics 2 original textbook · Chapter 4

Reviewed corrections and authored clarifications are applied to this displayed edition. The original manuscript remains preserved.

Archived original reading · contains source errors and unfinished material

Section Overview

Section 4

Kirchhoff’s Laws

Now that we understand how certain circuit elements function in circuits we are ready to introduce what are called the Kirchhoff Circuit Laws. These laws when used together with what we know about certain circuit elements provide us with a powerful set of tools to analyze circuits with interesting configurations.

The first of these laws is as follow. At any particular node the sum of the currents entering and leaving must sum to zero. We can write this mathematically as

This law is justified by the conservation of charge. The amount of charge entering a some must be the same as the amount of charge leaving. Since current is nothing more than the change of charge with time this translate into current also being conserved in those same areas.

The second of these rules is as follows. The sum of the voltages across any element in a closed loop must be zero. As we did with the current law we will write this law down mathematically.

When both of the above laws are used simultaneously they form a set of linear equations which can be used to solve for unknown variables in complex circuits.

Although Kirchhoff’s Laws may be relatively straightforward applying them is another game entirely. For this reason we will discuss setting up the equations in some more detail before preceding to the example problems.

The first thing we will discuss is the current at nodes. You must always set up a convention for what the sign of a current going out and what the sign of a current going in will be. For this book we will say that currents going into a node are positive while currents coming out of a node are negative.

The direction of these currents are chosen on a feeling basis. When the math is all done and we interpret the results we will see that the math corrects our guess for us by making our current positive if we are right and negative if we were wrong with our guess while leaving the magnitude unchanged.

Now we will discuss the voltage equation. In figuring out equations for loops you must take into account several different variables. The first of these is in what way are you crossing an emf. If you cross an emf starting at the lower potential going to the positive you will have a positive voltage contribution from the emf. If instead you start at the positive and go to the negative you will have a negative emf voltage contribution.

Determining the voltage across a resistor can be a little trickier because it depends on what we have chosen the direction of the current to be. When moving against the current you will be adding voltage since you are moving from a lower to a higher potential. When moving with the current you will subtract voltage since you are moving from a higher to a lower potential.

After using the above to form the desired equations you can then solve for the missing information using your knowns.

Example 4.4.1

Using the circuit given to you in BlahBlah Gallery with the given values determine what the current across each of the resistors will be.

Whenever asked to do a problem of this type your goal will always be to set up a system of liner equations which you can then use to solve for the desired values.

In this case since we have three unknowns (the currents) then we must create three equations to completely solve for their values.

In order to do this we will use Kirchhoff’s rules on one node and two different loops (using three loops would also suffice).

We will begin with the node a. Remember that in order to count the current coming in and out of a node you must first choose the direction of the currents through the different wire segments. We have chosen for the current to be clockwise in the left-most loop and counter-clockwise in the rightmost loop.

This means that the current coming through resistor one and the current coming through resistor three both enter into the node. The current going through resistor two then comes out of the node.

Since current one and three go into the node they receive positive signs and since current two is coming out it receives a negative sign. Summing them all together and equating them to zero gives the following equation.

Next we will consider the leftmost loop. Here we will use the second of Kirchhoff’s rules, that the sum of the voltage around a closed loop is zero.

We will start at the point marked d and move clockwise through the loop. The first circuit element we encounter is an emf. Since we move from a low potential to a high potential we have an increase of potential or a positive voltage.

The next element encountered is resistor one. Since we are moving through the loop with the current when we move through the resistor we will start at a higher potential and move to a lower potential so that or change in potential is negative. We therefore add the negative of resistors one’s voltage to our equation.

Resistor two’s voltage follows suit, since we are moving with the voltage through the resistor we must subtract the voltage from our equation.

Finally we encounter the second emf. Since we are moving through this emf backwards, starting with the high potential side and moving to the low we are lower the potential by a voltage equal to the emf’s voltage.

Using the about we can then form the equation.

Using Ohm’s Law to substitute for the voltages we can get the above into a more desirable form.

Starting at b and moving counter-clockwise through the rightmost loop the same process can be applied yield the equation

Which after using Ohm’s Law to replace the V’s gives.

Finally we arrive at a point where we can use our equations to solve for the currents. Below the equations have been rearranged and placed together in a more sensible manner.

Which after plugging in numerical values can be solved for each of the currents.

Since each current is negative all our guesses were incorrect.

Example 4.4.2

Consider the circuit given in gallery blahblah using kirchhoff’s rules to find an expression in terms of resistances and the current across one of the resistors in parallel which describes the voltage across the circuits emf.

This problem asks us to find an expression for the voltage of the emf using Kirchhoff’s rules. At the first node after applying our node rule we get that

Next we need to apply our rules to a loop. Either loops will work since both loops have exactly the same components. This gives us the equation

Luckily this equation has exactly the term we’re looking for, the emf’s voltage. We will then solve for it in terms of the other variables so that

Now we can use the first equation we found relating the currents and make the appropriate substitutions to get the wanted equation. Since

We can make a substitution for current one in our reworked loop equation to give us our answer

Section Overview

Section 4

Kirchhoff’s Laws

The first of these laws is as follow. At any particular node the sum of the currents entering and leaving must sum to zero. We can write this mathematically as

The second of these rules is as follows. The sum of the voltages across any element in a closed loop must be zero. As we did with the current law we will write this law down mathematically.

When both of the above laws are used simultaneously they form a set of linear equations which can be used to solve for unknown variables in complex circuits

Gallery 1.11 The Imaginary Circuit

After using the above to form the desired equations you can then solve for the missing information using your knowns.

Example 4.4.1

Use preserved Gallery4.4.2.1-1: three branches join common upper/lower nodes. Their upper-positive batteries are 1 V, 2 V, 1 V and resistors 2 Ω, 3 Ω, 2 Ω. Find branch currents with outer arrows upward and middle arrow downward.

Authored clarification: The unresolved placeholder is clarified by naming the actual preserved gallery and its values.

Reviewed worked solution:

Set lower potential zero and upper V.
${V-1\over2}+{V-2\over3}+{V-1\over2}=0$
Solve the independent junction equation.
$V=1.25\,\mathrm V$
Convert signed branch currents to source arrows.
$I_1=(1-V)/2=-0.125\,\mathrm A,\quad I_2=(V-2)/3=-0.25\,\mathrm A,\quad I_3=(1-V)/2=-0.125\,\mathrm A$

Answer: $I_1=-0.125\,\mathrm A,\quad I_2=-0.25\,\mathrm A,\quad I_3=-0.125\,\mathrm A$

Check: I₁+I₃=I₂ and all branches have the same node voltage.

Gallery 1.12 Example 4.4.1

Example 4.4.2

Authored explicit replacement for the unresolved diagram: an ideal emf drives R₁ in series with R₂ parallel R₃. Given I₂ through R₂, express emf.

Authored clarification: This supplied topology is an authored replacement rather than a recovered source diagram.

Reviewed worked solution:

Parallel branches have equal voltage.
$V_p=I_2R_2=I_3R_3$
Use the junction rule.
$I_1=I_2+I_3=I_2(1+R_2/R_3)$
Add series drops.
$\mathcal E=I_1R_1+I_2R_2=I_2[R_2+R_1(1+R_2/R_3)]$

Answer: $\mathcal E=I_2[R_2+R_1(1+R_2/R_3)]$

Check: All terms have units volts.

Gallery 1.13 Example 4.4.2

Now we can use the first equation we found relating the currents and make the appropriate substitutions to get the wanted equation. Since

We can make a substitution for current one in our reworked loop equation to give us our answer

Review 1.4 Kirchhoff’s Laws

This review preserves the author’s recorded prompts, choices, and answer states. Corrected items have separate reviewed explanations and keys; unchanged draft keys are marked unreviewed. Unfinished items are not scored.

Original question 1

After solving a linear system obtained from Kirchhoff’s Laws what does having a negative value for a current mean?

Nothing. It is just an artifact of your original choices.
The actual current flows opposite to the reference direction chosen for the calculation.
That the initially chosen current direction is correct.
Something went wrong applying Kirchhoff’s Laws.
None of the above

Reviewed answer and reasoning

A reference direction is a valid bookkeeping choice. A negative calculated value means the actual current flows in the opposite direction; its magnitude is the absolute value of that calculated current.

The sign gives the physical direction relative to the chosen reference arrow.

Archived original question and recorded key · contains the source error

After solving a linear system obtained from Kirchhoff’s Laws what does having a negative value for a current mean?

Nothing. It is just an artifact of your original choices.
The direction you originally chose for that current is incorrect.
That the initially chosen current direction is correct.
Something went wrong applying Kirchhoff’s Laws.
None of the above

Original recorded key: B

Original question 2

A current enters a node and is split in three unequal ways. What equation would properly describe this?

$2I_1-2I_2=0$
$I_1+I_2-2I_3=0$
$I_1+I_2-I_3-I_4=0$
$I_1-I_2-I_3-I_4=0$
None of the above

Reviewed answer and reasoning

Given one incoming I1 and outgoing I2,I3,I4, KCL I1-I2-I3-I4=0.

The independent semantic audit confirms this source prompt and recorded key under its stated conditions.

Archived original question and recorded key · contains the source error

A current enters a node and is split in three unequal ways. What equation would properly describe this?

None of the above

Original recorded key: D

Original question 3

When passing through a resistor moving in the direction of the current what sign would you expect to have on your voltage?

Negative
Positive

Reviewed answer and reasoning

Potential change through positive resistance in current direction is -IR.

The independent semantic audit confirms this source prompt and recorded key under its stated conditions.

Archived original question and recorded key · contains the source error

When passing through a resistor moving in the direction of the current what sign would you expect to have on your voltage?

Negative
Positive

Original recorded key: A

Original question 4

What sign will you get for your emf voltage if you move from the positive to the negative side.

Positive
Negative

Reviewed answer and reasoning

Traversing ideal battery from positive to negative gives -emf.

The independent semantic audit confirms this source prompt and recorded key under its stated conditions.

Archived original question and recorded key · contains the source error

What sign will you get for your emf voltage if you move from the positive to the negative side.

Positive
Negative

Original recorded key: B

Original question 5

Traverse two ideal opposing sources: cross ε₁ from − to + and ε₂ from + to −. What is the algebraic emf change?

$\epsilon_1-\epsilon_2$
$\epsilon_1+\epsilon_2$
$|\epsilon_1-\epsilon_2|$

Reviewed answer and reasoning

The traversal sets the signs. If only the strength were requested, the magnitude would be |ε₁−ε₂|.

This reviewed version states the assumptions or convention needed for one determinate answer. The archived source prompt and key are retained.

Archived original question and recorded key · contains the source error

Consider a circuit composed of a single loop, two oppositely facing batteries, and a resistor. Find an expression for the circuits total emf in terms of the batteries emfs.

None of the above

Original recorded key: A

Reviewed worked examples

These authored derivations are paired with the original example identities. Added assumptions are stated.

Example 4.4.1 · reviewed solution

Authored clarification: The unresolved placeholder is clarified by naming the actual preserved gallery and its values.

Set lower potential zero and upper V.
${V-1\over2}+{V-2\over3}+{V-1\over2}=0$
Solve the independent junction equation.
$V=1.25\,\mathrm V$
Convert signed branch currents to source arrows.
$I_1=(1-V)/2=-0.125\,\mathrm A,\quad I_2=(V-2)/3=-0.25\,\mathrm A,\quad I_3=(1-V)/2=-0.125\,\mathrm A$

$I_1=-0.125\,\mathrm A,\quad I_2=-0.25\,\mathrm A,\quad I_3=-0.125\,\mathrm A$

Check

I₁+I₃=I₂ and all branches have the same node voltage.

Example 4.4.2 · reviewed solution

Authored explicit replacement for the unresolved diagram: an ideal emf drives R₁ in series with R₂ parallel R₃. Given I₂ through R₂, express emf.

Authored clarification: This supplied topology is an authored replacement rather than a recovered source diagram.

Parallel branches have equal voltage.
$V_p=I_2R_2=I_3R_3$
Use the junction rule.
$I_1=I_2+I_3=I_2(1+R_2/R_3)$
Add series drops.
$\mathcal E=I_1R_1+I_2R_2=I_2[R_2+R_1(1+R_2/R_3)]$

$\mathcal E=I_2[R_2+R_1(1+R_2/R_3)]$

Check

All terms have units volts.

Glossary and deeper reading · 4 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Electric charge
Electric charge is a conserved property that determines electric interactions. It can be positive or negative and is measured in coulombs; ordinary net charges arise from an imbalance of charged particles. For example, transferring electrons between initially neutral objects leaves one negatively charged and the other positively charged.
Read more: Electric charge in detail →
Electric current
Electric current is the net rate of charge flow through a specified cross section. It is measured in amperes, or coulombs per second, with direction defined by positive-charge motion. For example, the same steady current passes through each element of a simple unbranched series loop.
Read more: Electric current in detail →
Potential difference
Potential difference is the change in electric potential between two locations. It represents electric potential-energy change per unit charge and is measured in volts. For example, the potential difference across a resistor describes the energy transferred per coulomb passing through it.
Read more: Potential difference in detail →
Electromotive force
Electromotive force is energy supplied per unit charge by a source or induction process. Despite its name it is measured in volts, rather than newtons, and can exist even when no current flows. For example, an ideal battery supplies an emf while an induced emf can be produced by changing magnetic flux.
Read more: Electromotive force in detail →

Gallery 1.11 The Imaginary Circuit

Gallery 1.12 Example 4.4.1

Gallery 1.13 Example 4.4.2

Review 1.4 Kirchhoff’s Laws

Original question 1

Original question 2

Original question 3

Original question 4

Original question 5

Reviewed worked examples

Electric charge

Electric current

Potential difference

Electromotive force

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