Note: The resistance calculation formula of the parallel circuit 1 / Rtotal = 1 / R1 + 1 / R2
The resistors can be connected together in an unlimited number of series and parallel combinations to form a complex resistance circuit
In the previous tutorial, we learned how to connect various resistors together to form a series resistor network or a parallel resistor network. We used Ohm's law to find the various currents flowing in and the voltage of each resistor combination.
But what if we want to connect various resistors in parallel and in series with "BOTH"? The combination in the same circuit produces a more complex resistance network. How do we calculate the sum of the circuit resistance, current and voltage of these resistance combinations.
Resistor circuits that combine series and parallel resistance networks are generally called resistor combinations or hybrid resistor circuits. The method of calculating the equivalent resistance of the circuit is the same as that of any single series or parallel circuit. We hope that we now know that the series resistors carry exactly the same current, and the parallel resistors have the same voltage.
For example, calculate the total current (IT) obtained from a 12v power supply in the following circuit.
At first glance, this may seem like a difficult task, but if we look closely, we can see that the two resistors, R 2 and R 3 are actually connected together with the “SERIES” combination so we can add them in The same resistance is produced together as we did in the series resistance tutorial. Therefore, the combined resistance of the combination is:
R 2 + R 3 =8Ω+4Ω=12Ω
So we can replace resistors R 2 and R 3 with a resistance value of 12Ω
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So our circuit now has a resistance RA and resistance R 4 in “PARALLEL”. Using the resistance in the parallel equation, we can reduce this parallel combination to a single equivalent resistance value R (combination) using the following two parallel resistance formulas.
The resulting resistance circuit now looks like this:
We can see the remaining two resistances, R 1 and R (combing) are connected together with the “SERIES” combination, and they can be added together again (resistance in series), so that points A and B therefore give:
R = R comb + R 1 =6Ω+6Ω= 12Ω
A single resistor of 12Ω can be used to replace the original four resistors connected together in the original circuit.
Now using Ohm's law, the value current (I) of the circuit is simply calculated as follows:
Therefore, any complex resistance circuit can simplify the resistance composed of several resistances into a simple single circuit with only one equivalent resistance by using the above steps to replace all the resistances connected together in series or parallel.
We can find two branch currents, I 1 and I 2, by using the Ohm method one step further, as shown.
V （R1） = I * R 1 = 1 * 6 = 6伏
V （RA） = V R4 = （12 - V R1） = 6 伏
therefore:
I 1 = 6V÷R A = 6÷12 = 0.5A或500mA
I 2 = 6V ÷ R 4 = 6 ÷ 12 = 0.5A or 500mA
Since the resistance values of the two branches are the same at 12Ω, I 1 and I 2 are each equal to 0.5A (or 500mA). Therefore, the total supply current IT: 0.5 + 0.5 = 1.0 ampere, as described above.
After making these changes, it is sometimes easier to draw or redraw new circuits using complex resistor combinations and resistor networks, as this helps with the visual aid of mathematics. Then continue to replace any series or parallel combination until an equivalent resistance R EQ is found. Let's try another more complicated combination circuit of resistors.
Series resistance and parallel resistance No2
Find the equivalent resistance, R EQ is used in the following resistor combination circuit.
Again, at first glance this trapezoidal resistor network may seem a complex task, but as before, it is just a combination of series and parallel resistors connected together. Starting from the right and using the simplified formula of two parallel resistances, we can find the combination of R 8 and R 10 <equivalent resistance / sub> and call it RA.
RA and R 7 therefore the total resistance RA + R 7 = 4 + 8 = 12Ω as shown.
The resistance value of 12Ω is now parallel to R 6 and can be calculated as RB.
RB and R 5 therefore the total resistance RB + R 5 = 4 + 4 = 8Ω, as shown.
The 8Ω resistance value is now in parallel with R 4 and can be calculated as RC as shown.
RC and R 3 are connected in series, so the total resistance RC + R 3 = 8Ω is shown in the figure.
The resistance value of 8Ω is now in parallel with R 2, from which we can calculate RD as:
RD is in series with R 1, so the total resistance RD + R 1 = 4 + 6 = 10Ω as shown.
Then the complex combination of the above resistor network includes ten independent resistors connected together in series. The parallel combination can use an equivalent resistance (REQ) instead of 10Ω.
When solving any combination of resists or circuits composed of series and parallel branch resistances, the first step we need to take is to identify simple series and parallel resistance branches and replace them with equivalent resistances.
This step will enable us to reduce the complexity of the circuit and help us convert the complex combined resistance circuit into a single equivalent resistance. Remember that the series circuit is a voltage divider and the parallel circuit is a current divider.
However, the calculation of more complex T-pad attenuator and resistance bridge networks cannot be simplified into simple parallel or series circuits using equivalent resistance, and different methods are required. These more complex circuits need to be solved using Kirchhoff's current law and Kirchhoff's voltage law, which will be discussed in another tutorial.
In the next tutorial on resistors, we will see the potential difference (voltage) at two points (including the resistor).
