Complex Circuits

Complex circuits have components that are in series and some that are in parallel.  Lets start by reviewing series and parallel circuits and then see how a complex circuit works as a combination.

Series Circuits and Rules (Click Here to Go to the Lesson)

A series circuit you have no branches, current has only one path to follow.  The same current remains together through all components from the positive terminal and back to the negative of the battery.

Ohm's Law (V=IR):

You can use this at a single location (T, 1, 2, 3, and any more) as seen below

  • (VT) = (IT)(RT)
  • (V1) = (I1)(R1)
  • (V2) = (I2)(R2)
  • (V3) = (I3)(R3)

Overall Series Circuit Rules

  • VT = V1 + V2 + V3 + …
  • IT = I1 = I2 = I3 = …
  • RT = R1 + R2 + R3 + …

Parallel Circuit Rules (Click Here to Go to That Lesson)

In a parallel circuit you have branches, multiple paths to follow.  So current splits up but later comes back together as it returns to the battery.

You can use Ohm's Law (V=IR) at a single location the same as above in series circuits.

Overall Parallel Circuit Rules

  • VT = V1 = V2 = V3 = …
  • IT = I1 + I2 + I3 + …
  • 1/RT = 1/R1 + 1/R2 + 1/R3 + ...

Complex Circuit

Follow the current from the positive terminal of the battery in the animation.  Some parts of the circuit are in series and some in parallel.

  1. The battery and resistor 1 are in series: 2A of current starts at the batter and flows through both.
  2. The branch with resistor 2, the branch with Resistor 3 and 4, and the branch with Resistor 5 are in parallel:  Current splits between these branches.
  3. The resistor 3 and resistor 4 are in series: Once current goes down the branch that same current must flow from both resistors.

Solving A Complex Circuit For Resistance

When solving for the resistance in a series circuit, the goal is to break down all the different parts making act like one single series circuit.  Follow the steps or our example as we do this.

Example Problems

1. Solve for the total resistance of the complex circuit below.

First add resistor 3 and resistor 4 which are in series down a branch

RT = R1 + R2

RT = 10Ω + 5Ω = 15Ω

1

Now take resistor 2, the equivalent from resistor 3 and 4, and resistor 5 which are all on branches and put them together following parallel rules.  In this case RT will be the equivalent of these three branches.

RT = ((1/R1) + (1/R2) + (1/R3))-1

RT = ((1/10Ω) + (1/15Ω) + (1/20Ω))-1 = 4.615Ω

2

Current must flow through the 5Ω resistor and 4.615Ω resistor equivalent so they act like they are in series together.  Use series rules to put these together.

RT = R1 + R2

RT = 5Ω + 4.615Ω = 9.615Ω

This is the resistance placed at the battery equivalent to the resistance of the total circuit.

3

2. Solve for the unknown components of the following complex circuit.

Combine Resistor 2 and resistor 3 in parallel

RT = ((1/R1) + (1/R2))-1

RT = ((1/30Ω) + (1/30Ω))-1 = 15Ω

1

Combine the 15Ω equivalent from resistor 2 and 3 with resistor 1 which is in series

RT = R1 + R2

RT = 15Ω + 5Ω = 20Ω

This is the resistance of the circuit placed at the battery.  We used and no longer need the equivalent or resistor 2 and resistor 3 so you will see that equivalent dropped off the next step.

2

Now solve for current at the battery using Ohm's Law

IT=VT/RT

IT= 40V/20Ω = 2A

3

2A of current flows from the battery and does not branch before resistor 1.  This make resistor 1 have the same current

Series rule

IT = I1

I1 = 2A

Now solve for voltage at resistor 1 using Ohm's Law

V1= (IT)(RT)

V1= (2A)(5Ω) = 10V

5

The 10V is dropped crossing the first resistor leaving 30V for the rest of the circuit.  In parallel the voltage left will be dropped all and equally on both the branches which have resistor 2 and resistor 3

The parallel rule we followed was this:

VT = V1 = V2

6

Now we can use Ohm's Law to finish this problem off

At both resistor 2 and resistor 3 you would do the follwing.

I = V/R

I= 30V/30Ω = 1A

7

For our final check we can see that the 2A of current from resistor split mathematically correct and know we are right.

IT = I1 + I2

2A = 1A + 1A

3. Solve for the unknown components of the following complex circuit.

Use Ohm's Law to solve for the current at resistor 4

V1= (IT)(RT)

V1= (2A)(5Ω) = 10V

V4=( I4)(R4)

I4=(V4)/(R4)

I4=(16V)/(16Ω)= 1A

1

Resistor 2 and resistor 3 are in series.  In series current is equal

Series Circuit Rule

IT = I1 = I2 = I3 = …

So the 4A of current at resistor also passed trough resistor 2 to get there.  Since I3 = 4A, I2 = 4A.

2

Now use Ohm's Law to solve for the voltage drop of resistor 3.

V3=I3R3

V3=(4A)(1Ω) = 4V

3

  • The branch that has Resistor 2 and resistor 3 are in parallel with the branch that has resistor 4.
  • So the entire branch that has resistor 2 and resistor 3 must be equal to the 16V that drops on the other branch.
  • Now observe that resistor 2 and resistor 3 are in series and current must flow through both resistors.
  • they mus add up to 16V following the series rule.

two resistor on a branch in series must drop the same voltage as the other branch

Therefore the voltage drop at resistor 2 must be 12 V since V2 and V3 dropping an additional 4V must add to the 16V of the other branch.

4

Now you can use Ohm's Law to solve for the resistance of resistor 2.

V2=I2R2

R2=V2/I2

R2= 12V/4A = 3Ω

5

Now look at the branch that has resistor 2 and resistor 3.  4 amps of current flows down that branch.  It is the same 4 amps of current since these two are in series.  (Look at the current parallel circuit animation earlier in this lesson if you are confused)

An additional 1 amp of current flows down the separate branch with resistor 4

So the total current that split off coming from resistor 1 was 5 amps.

The battery and on the other side resistor 5 are also in series so they will be 5 amps as well.
6

Now do Ohm's law to determine voltage drop at resistor 1.

V1=I1R1

V1= (5A)(8Ω) = 40V 

and Ohm's Law for the voltage drop at resistor 5

V5=I5R5

V5= (5A)(4Ω) = 20V 

8

Lastly do Ohm's Law to determine the resistance of the entire circuit at the battery.

VT=ITRT

RT=VT/IT

RT=76V/5A = 15.2Ω

9

You can spend a little time doing your final check calculating the resistance of the entire circuit as a last check and you will get a RT = 15.2Ω when solving for total resistance of the circuit correctly.

Click on the picture below to see the animation of current flowing.  This is for understanding of the final answers for current.

Current Visual Through Complex Circuit

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