In the previous 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 ( IT ) 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, R2 and R3 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:
R2 + R3 = 8Ω + 4Ω = 12Ω
So we can replace both resistor R2 and R3 above with a single resistor of resistance value 12Ω
So our circuit now has a single resistor RA in “PARALLEL” with the resistor R4. 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, R1 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(ab) = Rcomb + R1 = 6Ω + 6Ω = 12Ω
Thus a single resistor of just 12Ω can be used to replace the original four resistors connected together in the original circuit above.
By using Ohm’s Law, the value of the current ( I ) flowing around the circuit is calculated as:
Then we can see that 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.
We can take this one step further by using Ohms Law to find the two branch currents, I1 and I2 as shown.
V(R1) = I*R1 = 1*6 = 6 volts
V(RA) = VR4 = (12 – VR1) = 6 volts
I1 = 6V ÷ RA = 6 ÷ 12 = 0.5A or 500mA
I2 = 6V ÷ R4 = 6 ÷ 12 = 0.5A or 500mA
Since the resistive values of the two branches are the same at 12Ω, the two branch currents of I1 and I2 are also equal at 0.5A (or 500mA) each. This therefore gives a total supply current, IT of: 0.5 + 0.5 = 1.0 amperes as calculated above.
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, REQ is found. Lets try another more complex resistor combination circuit.
Resistors in Series and Parallel Example No2
Find the equivalent resistance, REQ 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 R8 to R10 combination and call it RA.
RA is in series with R7 therefore the total resistance will be RA + R7 = 4 + 8 = 12Ω as shown.
This resistive value of 12Ω is now in parallel with R6 and can be calculated as RB.
RB is in series with R5 therefore the total resistance will be RB + R5 = 4 + 4 = 8Ω as shown.
This resistive value of 8Ω is now in parallel with R4 and can be calculated as RC as shown.
RC is in series with R3 therefore the total resistance will be RC + R3 = 8Ω as shown.
This resistive value of 8Ω is now in parallel with R2 from which we can calculated RD as:
RD is in series with R1 therefore the total resistance will be RD + R1 = 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 ( REQ ) of value 10Ω.
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 Kirchhoff’s Current Law, and Kirchhoff’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.