High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Physics Textbook (Master) > DC Circuits

Kirchhoff's Rules

Resistors in Series and ParallelPrevious Section
Power and BrightnessNext Section
Popular Courses
Find My Course
0:00 / 0:00

Kirchhoff’s Rules


Kirchhoff’s Rules describe how currents and voltages behave in electrical circuits. They are useful for solving a circuit where the elements are wired in more complicated ways, neither in series nor in parallel.

Junction Rule

At any junction in the circuit, the sum of the currents going in is equal to the sum of the currents going out:

 ∑Iin=∑Iout \boxed{ \ \sum I_{in}=\sum I_{out} \ }



This rule follows from conservation of charge: as current moves around the circuit, there is no accumulation of charge anywhere.

Loop Rule

For any closed loop in the circuit, the voltage drops add up to zero:

 ∑iΔVi=0 \boxed{\ \sum_i\Delta V_i=0\ }



This rule follows from conservation of energy. When a charge makes a complete circuit around a loop, it has the same potential energy that it started out with, i.e. there is no change in potential energy: ΔU=q ΣiΔVi=0\Delta U=q\ \Sigma_i\Delta V_i=0

Steps for Problem Solving

  • Choose the direction of the current for the entire circuit, and split/recombine it at the junctions. Any direction is fine: at the end, a positive current means that it's in the same direction as your label, and a negative current is in the opposite direction.

  • Choose the direction of the loops. (On the one hand, you need to have enough loops to solve for all the variables. On the other hand, some loops will be redundant and won't give you any new information.)

  • Outside the batteries, voltage drops are negative if the loop is in the direction of the current, and positive if the loop is opposite to the direction of current. Use V=IRV=IR for the voltage drop across each resistor.

  • As the loop moves from the negative to the positive terminal of a battery, the voltage drop will be positive (and vice-versa).

  • At the end, you should have a current equation for each junction (either the split or the recombination), and a voltage equation for each loop (making sure that all portions of the circuit have been accounted for).

  • Algebraically combine the equations (using substitution or elimination) to solve for the currents.


Exam Tip
All sign conventions can be switched around. The only thing that matters is to be consistent with your choice throughout the problem.

0:00 / 0:00

Example: Three Resistors and Two Batteries


Find the current in each part of the circuit below.


First choose the direction of the current and the loops:



At the junction we have:

I=I1+I2I=I_1+I_2

Now for the loops, using V=IRV=IR for each voltage drop:

Loop 1:

20−I(200)−I2(100)−10=020-I(200)-I_2(100)-10=0

Loop 2:

−I1(200)+10+I2(100)=0-I_1(200)+10+ I_2(100)=0


Let's substitute I=I1+I2I=I_1+I_2 into the Loop 1 equation:

20−200I−100I2−10=020-200I-100I_2-10=0

20−200(I1+I2)−100I2−10=020-200(I_1+I_2)-100I_2-10=0

20−200I1−200I2−100I2−10=020-200I_1-200I_2-100I_2-10=0

10−200I1−300I2=010-200I_1-300I_2=0


Also simplify the Loop 2 equation:

10−200I1+100I2=010-200I_1+ 100I_2=0

Subtract these two equations to eliminate the I2I_2 current:

10−200I1+100I2=010-200I_1+ 100I_2=0
10−200I1−300I2=010-200I_1-300I_2=0
_________________________

400I2=0400 I_2=0

I2=0I_2=0 (A)


Let's substitute this into one of the equations above to solve for the I1I_1 current:

10−200I1+100I2=010-200I_1+ 100I_2=0

10−200I1=010-200I_1=0

I1=120I_1=\dfrac{1}{20} (A)

Now we can use the junction equation to solve for the II current:

I=I1+I2=120+0=120I=I_1+I_2=\dfrac{1}{20}+0=\dfrac{1}{20} (A)

checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.

Practice: Three Resistors and Two Batteries


Find the current flowing through each battery and each resistor in the circuit below.