# Combination circuit analysis

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**Do not fear the combination circuit!**

When you are first faced with a series/ parallel combination circuit they seem almost down right impossible to solve to begin with. As with most things this is not true.

In order to solve them we have to remember a few rules from what we learned about series and parallel circuits.

**In series:**

Current remains constant throughout the whole circuit.

I_{t}= I_{1}=I_{2}=I_{3} …

The total circuit resistance is the sum of the individual resistors.

R_{t} = R_{1}+ R_{2} + R_{3} …

The volt drops across all resistors must add up to the source voltage

V_{Source} = V_{R1} + V_{R2} + V_{R3} …

**In parallel:**

Voltage remains constant throughout the branches.

Vsupply = V_{R1 }= V_{R2} = V_{R3} …

The sum of the branch circuit currents adds up to the total line current.

I_{t }= I_{1 }+ I_{2} + I_{3} …

The total circuit resistance is the reciprocal sum of the individual branches resistance.

^{1}/_{Rt} = ^{1}/_{R1} + ^{1}/_{R2} = ^{1}/_{R3} …

** **

**Let’s take a look at the circuit shall we?**

If you look at the far right we see that R7, R8 and R9 are in series. This means we can add them to get one resistance:

Now we have R6 and R (789) in parallel. This means we add them reciprocally. 1/R(6789) = 1/R6 = 1/R(789)

Are you starting to see it? Now we have R4, R(6789) and R5 in series. Lets add them up.

R3 and R (456789) are in parallel so we add them up reciprocally.

Now to get the total circuit resistance we add up the remaining resistors in series.

**OK, Now what???**

Once you have the RT you can determine the line current (E_{supply}/R_{T} = I_{Line}). From there we start working our way back out.

If you look back you can see that R3 and R(456789) are in parallel. Therefore they will have the same voltage.

Our next step is to determine the current in the branch containing VR(456789). This is so we can determine volt drops across R4, R5, and R(6789).

We take the current determined and multiply it against the individual resistors.

VR6 and VR(789) are in parallel and will then have the same voltage.

Now all that is left is to determine the current through the branch containing VR(789) by diving it by R(789).

Once you have that current you can determine the volt drops across R7,R8, and R9.

You’re done!

I like to call it the accordion method. You squish the circuit in to determine the total circuit resistance. From there you calculate the current. After the current is determined you can stretch the circuit back out determining volt drops as you go.

To get a feel for it make sure you watch the video and follow along by making your own drawings.

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The dreaded broken neutral - The Electric AcademyApril 6, 2018 at 8:11 pm[…] From here you will calculate the new line current and then new voltages. This is done just as you would in a combination circuit. […]