In the previous two tutorials we have learnt how to connect individual resistors together to form either a Series Resistor Network or a Parallel Resistor Network and we used Ohms Law to find the various currents flowing in and voltages across each resistor combination.

But what if we want to connect various resistors together in “BOTH” parallel and series combinations within the same circuit to produce more complex resistive networks, how do we calculate the combined or total circuit resistance, currents and voltages for these resistive combinations.

Resistor circuits that combine series and parallel resistors networks together are generally known as **Resistor Combination** or mixed resistor circuits. The method of calculating the circuits equivalent resistance is the same as that for any individual series or parallel circuit and hopefully we now know that resistors in series carry exactly the same current and that resistors in parallel have exactly the same voltage across them.

For example, in the following circuit calculate the total current ( I_{T} ) taken from the 12v supply.

At first glance this may seem a difficult task, but if we look a little closer we can see that the two resistors, R_{2} and R_{3} are actually both connected together in a “SERIES” combination so we can add them together to produce an equivalent resistance the same as we did in the series resistor tutorial. The resultant resistance for this combination would therefore be:

R_{2} + R_{3} = 8Ω + 4 Ω = 12 Ω

So we can replace both resistor R_{2} and R_{3} above with a single resistor of resistance value 12 Ω

So our circuit now has a single resistor R_{A} in “PARALLEL” with the resistor R_{4}. Using our resistors in parallel equation we can reduce this parallel combination to a single equivalent resistor value of R_{(combination)} using the formula for two parallel connected resistors as follows.

The resultant resistive circuit now looks something like this:

We can see that the two remaining resistances, R_{1} and R_{(comb)} are connected together in a “SERIES” combination and again they can be added together (resistors in series) so that the total circuit resistance between points A and B is therefore given as:

R_{( A B )} = R_{comb} + R_{1} = 6 Ω + 6 Ω = 12 Ω.

and a single resistance of just 12 Ω can be used to replace the original four resistors connected together in the original circuit.

Now by using Ohm´s Law, the value of the circuit current ( I ) is simply calculated as:

So any complicated resistive circuit consisting of several resistors can be reduced to a simple single circuit with only one equivalent resistor by replacing all the resistors connected together in series or in parallel using the steps above.

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It is sometimes easier with complex resistor combinations and resistive networks to sketch or redraw the new circuit after these changes have been made, as this helps as a visual aid to the maths. Then continue to replace any series or parallel combinations until one equivalent resistance, R_{EQ} is found. Lets try another more complex resistor combination circuit.

Find the equivalent resistance, R_{EQ} for the following resistor combination circuit.

Again, at first glance this resistor ladder network may seem a complicated task, but as before it is just a combination of series and parallel resistors connected together. Starting from the right hand side and using the simplified equation for two parallel resistors, we can find the equivalent resistance of the R_{8} to R_{10} combination and call it R_{A}.

R_{A} is in series with R_{7} therefore the total resistance will be R_{A} + R_{7} = 4 + 8 = 12Ω as shown.

This resistive value of 12Ω is now in parallel with R_{6} and can be calculated as R_{B}.

R_{B} is in series with R_{5} therefore the total resistance will be R_{B} + R_{5} = 4 + 4 = 8Ω as shown.

This resistive value of 8Ω is now in parallel with R_{4} and can be calculated as R_{C} as shown.

R_{C} is in series with R_{3} therefore the total resistance will be R_{C} + R_{3} = 8Ω as shown.

This resistive value of 8Ω is now in parallel with R_{2} from which we can calculated R_{D} as:

R_{D} is in series with R_{1} therefore the total resistance will be R_{D} + R_{1} = 4 + 6 = 10Ω as shown.

Then the complex combinational resistive network above comprising of ten individual resistors connected together in series and parallel combinations can be replaced with just one single equivalent resistance ( R_{EQ} ) of value 10Ω.

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When solving any combinational resistor circuit that is made up of resistors in series and parallel branches, the first step we need to take is to identify the simple series and parallel resistor branches and replace them with equivalent resistors. This step will allow us to reduce the complexity of the circuit and help us transform a complex combinational resistive circuit into a single equivalent resistance remembering that series circuits are voltage dividers and parallel circuits are current dividers.

However, calculations of more complex T-pad Attenuator and resistive bridge networks which cannot be reduced to a simple parallel or series circuit using equivalent resistances require a different approach. These more complex circuits need to be solved using Kirchoff’s Current Law, and Kirchoff’s Voltage Law which will be dealt with in another tutorial.

In the next tutorial about Resistors, we will look at the electrical potential difference (voltage) across two points including a resistor.

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Thank u ……now it seems to be easy for me..

Nice explanation

Thanks very much

Thanks for explain.It is very easy to understand.Awesome website.

These were very easy all were very simple i did them only in 2 min.

X 4=9; X=9-4; x=5

THANKU so much 4 posting such detailed explanations… Will help me a lot in cracking FTRE

But can u post some practice questions also in this topic… Will be greatly helped if such…😊😊

hope your FTRE goes well !!

Thanks!!!

I was solving the same circut from last 1 hour & believe me i was about to tear my clothes off, it was really a difficult one to solve

Thanks Guys

You are lifesavers! Our physics teacher explains hardly anything to us, but i finally understood it with you help. It gets confusing when there are both series and parallels. I guess 10th grade will be rough.

How to calculate current for It, I1 and I2 before simplifying the number of resistors ? (original drawing)

You still need to find Rcomb to give It, (It = Vab/(R1+Rcomb))

Vcomb = Vab – (It x Rcomb)

I1 = Vcomb/(R2+R3) = (It – I2)

I2 = Vcomb/R4 = (It – I1)

It = I1+I2