### Resistor Combinations

**Resistor Combinations**

**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 currents and voltages flowing in each resistor combination. But what if we want to connect resistors together in "BOTH" parallel and series combinations within the same circuit to produce more complex circuits that use resistor combinations, how do we calculate the combined circuit resistance, currents and voltages for this.

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, 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 both connected together in a "SERIES" combination so we can add them together. The resultant resistance for this combination would therefore be,
R

_{2}+ R_{3}= 8 Ω + 4 Ω = 12 Ω
So now we can replace both the resistors R

_{2}and R_{3}with a single resistor of resistance value 12 Ω
Now we have single resistor R

_{A}in "PARALLEL" with the resistor R_{4}, (resistors in parallel) and again we can reduce this combination to a single resistor value of R_{(combination)}using the formula for two parallel connected resistors as follows.
The resultant circuit now looks something like this:

The two remaining resistances, R

_{1}and R_{(comb)}are connected together in a "SERIES" combination and again they can be added together so 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 4
resistor combinations circuit above.

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. 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.

## Example No2

Find the equivalent resistance, R

_{EQ}for the following resistor combination circuit.
Again, at first glance this resistor ladder network may seem complicated but as before it is 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 can be replaced with one single equivalent resistance ( R

_{EQ}) of value 10Ω.

When solving any

*combinational resistor*circuit, 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.
However, calculations of complex

*T-pad Attenuator*and resistive bridge networks which cannot be reduced to a simple parallel or series circuit using equivalent resistances. 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.

# Resistor Combinations

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 currents and voltages flowing in each resistor combination. But what if we want to connect resistors together in "BOTH" parallel and series combinations within the same circuit to produce more complex circuits that use resistor combinations, how do we calculate the combined circuit resistance, currents and voltages for this.
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, 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 both connected together in a "SERIES" combination so we can add them together. The resultant resistance for this combination would therefore be,
R

_{2}+ R_{3}= 8 Ω + 4 Ω = 12 Ω
So now we can replace both the resistors R

_{2}and R_{3}with a single resistor of resistance value 12 Ω
Now we have single resistor R

_{A}in "PARALLEL" with the resistor R_{4}, (resistors in parallel) and again we can reduce this combination to a single resistor value of R_{(combination)}using the formula for two parallel connected resistors as follows.
The resultant circuit now looks something like this:

The two remaining resistances, R

_{1}and R_{(comb)}are connected together in a "SERIES" combination and again they can be added together so 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 4
resistor combinations circuit above.

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. 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.

## Example No2

Find the equivalent resistance, R

_{EQ}for the following resistor combination circuit.
Again, at first glance this resistor ladder network may seem complicated but as before it is 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 can be replaced with one single equivalent resistance ( R

_{EQ}) of value 10Ω.

When solving any

*combinational resistor*circuit, 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.
However, calculations of complex

*T-pad Attenuator*and resistive bridge networks which cannot be reduced to a simple parallel or series circuit using equivalent resistances. 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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