# equivalent resistance

### Asked by TABISH | 1st Jun, 2012, 04:17: PM

Calculate the equivalent resistance, series current, voltage drop and power for each resistor of the following resistors in series circuit.

All the data can be found by using *Ohm's Law*, and to make life a little easier we can present this data in tabular form.

Resistance | Current | Voltage | Power |
---|---|---|---|

R_{1} = 10? |
I_{1} = 200mA |
V_{1} = 2V |
P_{1} = 0.4W |

R_{2} = 20? |
I_{2} = 200mA |
V_{2} = 4V |
P_{2} = 0.8W |

R_{3} = 30? |
I_{3} = 200mA |
V_{3} = 6V |
P_{3} = 1.2W |

R_{T} = 60? |
I_{T} = 200mA |
V_{S} = 12V |
P_{T} = 2.4W |

Find the total resistance, R_{T} of the following resistors in parallel network.

The total resistance R_{T} across the two terminals A and B is calculated as:

## Resistor Combinations

Resistor circuits that combine series and parallel resistors circuits together are generally known as**Resistor Combination** or mixed circuits and the method of calculating their 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 ? + 4 ? = 12 ?

So now we can replace both the resistors R_{2} and R_{3} with a single resistor of resistance value 12 ?

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 ? + 6 ? = 12 ?.

and a single resistance of just 12 ? 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 circuit consisting of several resistors can be reduced to a simple circuit with only one equivalent resistor by replacing the resistors in series or in parallel using the steps above. It is sometimes easier with complex resistor combinations to sketch or redraw the new circuit after these changes have been made as a visual aid to the maths. Then continue to replace any series or parallel combinations until one equivalent resistance 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 circuit 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 = 12? as shown.

This resistive value of 12? 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 = 8? as shown.

This resistive value of 8? 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} = 8? as shown.

This resistive value of 8? 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 = 10? as shown.

Then the complex combinational resistive network above can be replaced with one single equivalent resistance ( R_{EQ} ) of value 10?.

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.

### Answered by | 2nd Jun, 2012, 09:52: AM

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