Series and Parallel Connected Resistors consist of both serial and parallel connected resistance networks in the same circuit, in many circuits resistances come out in this way rather than being connected directly to the series or parallel.
In previous content, we learned how to connect individual resistors together to create a Serial Resistance Network or Parallel Resistance Network, and used the Ohm Act to find the various currents and voltages flowing in each resistance combination.
However, if we want to connect various resistors in both parallel and serial combinations within the same circuit to produce more complex resistant networks, how do we calculate combined or total circuit resistance, currents and voltages for these resistant combinations?
Resistance circuits that combine serial and parallel resistance networks are often known as Resistance Combinations or mixed resistance circuits.The method of calculating the equivalent resistance of circuits is the same as for any individual series or parallel circuit, and I hope that we now know that the series-bound resistors carry exactly the same current and that the parallel connected resistors have exactly the same voltage between them.
For example, calculate the total current ( I T ) from the 12v source in the following circuit.
At first glance this may seem like a difficult task, but if we take a closer look, we can see that the two resistances, R2 and R3,are actually serially connected, so that by combining them we can pretend to be a single resistance.Therefore, the resulting resistance to this combination will be:
R 2 + R 3 = 8Ω + 4Ω = 12Ω
In this way, we actually get 2 parallel resistances and a serial resistance to them. In the two parallel resistances we achieve, we can reduce it to a single resistance:
After all the procedures we do, our circuit takes this form:
We can see that the remaining two resistances, R 1 and R (comb), are interconnected in a "SERIES" combination and can be added together again, so that the total circuit resistance between points A and B is:
R (ab) = R comb + R 1 = 6Ω + 6Ω = 12Ω
Thus, in the original circuit above we can use only a single resistance of 12Ω to replace the original four interconnected resistances.
Using the Ohm Act, we can calculate the value of the transferred current (I) as follows:
Next, we can see that any complex resistance circuit consisting of several resistances can be reduced to a simple single circuit with only one equivalent resistance, replacing all interconnected resistors in series or in parallel using the above steps.
By taking this a step further, we can use the Ohm Act to find two branch currents, I 1 and I 2, as shown.
V (R1) = I*R 1 = 1*6 = 6 volts
V (RA) = V R4 = (12 – V R1 ) = 6 volts
I 1 = 6V ÷ R A = 6 ÷ 12 = 0.5A or 500mA
I 2 = 6V ÷ R 4 = 6 ÷ 12 = 0.5A or 500mA
Since the resistance values of the two arms are the same at 12Ω, both branch currents of I 1 and I 2 are equal at 0.5A (or 500mA).Therefore, this gives the total feed current, I T : 0.5 + 0.5 = 1.0 amps, as calculated above .
With complex resistance combinations and resistant networks, it is sometimes easier to draw the new circuit after these changes are made, since this helps mathematics as a visual aid.Then continue to change any series or parallel combination until an equivalent resistance, R EQ, is found. Now let's try another complex resistance combination circuit.
Serial and Parallel Connected Resistors Question Example 1
Find the equivalent resistance, R EQfor the resistance combination circuit below.
Again, at first glance this network of resistance ladders may seem like a complex circuit, but as before it is only a combination of interconnected serial and parallel resistors.Starting from the right side and using the simplified equation for two parallel resistors, we can find the equivalent resistance of the combination R 8 – R 10 and call it R A .
The resistance of RA and R7 will be serial, so the total resistance is R A + R 7 = 4 + 8 = 12Ω.
This resistance value of 12Ω is now parallel to R6 and can be calculated as RB.
Since R B is serial with R 5, the total resistance will be R B + R 5 = 4 + 4 = 8Ω as shown.
This resistance value of 8Ω is now parallel to R 4 and can be calculated as R C as shown.
Since R C is serial with R 3, the total resistance will be R C + R 3 = 8Ω as shown.
Since the calculated 8Ω value is parallel to R 2, we can call them RD with a new calculation:
Since R D is serial with R 1, the total resistance will be R D + R 1 = 4 + 6 = 10Ω.
Then, the above complex combination-resistant network of ten separate resistors connected in serial and parallel combinations can be replaced with a single equivalent resistance (R EQ) of 10Ω.
The first step we must take when solving any combinational resistance circuit consisting of resistances in series and parallel branches is to identify simple series and parallel resistance branches and replace them with equivalent resistors.
This step will help us reduce the complexity of the circuit and convert a complex combinational resistant circuit into a single equivalent resistors, remembering that serial circuits are voltage dividers and parallel circuits are current dividers.
However, calculations of more complexT-pad debilitating and resistant bridge networks that cannot be reduced to a simple parallel or serial circuit using equivalent resistors require a different approach.These more complex circuits need to be resolved using Kirchhoff's Current Act and Kirchhoff's Voltage Act, which will be addressed in other contexts.
In the next tutorial on resistors, we will look at the electrical potential difference (voltage) between two points, including a resistance.