This article is intended for those interested in basic electrical theory. The first part reviews the laws of series and parallel circuits. The second part explains the parallel circuit of the series.
A Review: Laws Series Circuits
In a series circuit with two resistors and powered by a battery, only one electrical path exists from the plus terminal. charge to the lower end of the ball and that path passes through both resistors. If you disconnect one resistor, you will have an open circuit and no current will go through the other resistor.
Here are the instructions for problem solving in the circle series:
1) The total voltage drop across the resistor is equal to the voltage of the battery. If you have two resistors in series, the total battery voltage is the total voltage across each resistor.
2) The total resistance is the sum of the resistances.
3) The current is the same in the series circle. There is only one electric way. Therefore only one current and that current flows through both resistors.
4) The total power dissipated is the total power dissipated in each resistor.
5) The voltage across any resistor equals the current through the resistor multiplied by the resistance of that resistor. This is an application of ohms law.
6) The voltage of the battery equals the current multiplied by the total resistance.
A Review: The Laws of Parallel Circuits
In a parallel circuit containing two resistors and powered by a battery, there are two electrical paths from the plus terminal of the battery to the minus terminal of the battery. One electrical path goes through the first resistor and the other electrical path continues through the second resistor.
The rules here are to be solved on a level playing field;
1) The voltage drop across each resistor equals the battery voltage.
2) The total resistance in a parallel circuit is contained by two resistors;
1/Rt = 1/R1 + 1/R2
where
Rt is the total resistance
R1 is the first resistor
The second resistor is R2
3) The total current is the sum of the runs through each resistor. In a parallel circuit containing two resistors and powered by a battery, two electrical paths run from the positive terminal of the battery to the negative terminal of the battery. Current splits: Part of the current takes one electrical path and part of the current takes the other electrical path.
4) The total power dissipated is the total power dissipated in each resistor.
5) The voltage across any resistor equals the current through the resistor multiplied by the resistance of that resistor. Since both resistors are directly across the battery, the voltage across each resistor is the same as the battery voltage.
6) The source voltage equals the total current multiplied by the total resistance.
The Series-Parallel Circuit
Now let’s examine one area of the figure. In the figure there are one battle and three resistance.
Look at resistor R1. There is only an electrical path from point A to point B, and the electrical path is through R1. With R1 removed, there is no electrical path from point A to point C. Therefore, R1 must be in series with the other resistors in the circuit.
Two electrical paths between point B and point C. One electrical path passes through R2 and the other electrical path through R3. R2 must therefore be parallel to R3 itself.
R1 is in series with parallel aggregates R2 and R3.
A current flows through R1 and then splits. Part of the current through R2, the other part of the current through R3. Therefore, the current through R1 equals the sum of the currents through R2 and R3.
The voltage across R2 equals the voltage across R3 because R2 and R3 are in parallel. The voltage of the battery equals the voltage across R1 plus the voltage across the parallel combination of R2 and R3.
To solve for the total resistance, first calculate the resistance of the parallel combination of R2 and R3.
the parallel resistance of R2 & R3 is Rp.
Rp = (R2*R3)/(R2 + R3)
Figure Two shows the equivalent circuit with R1 and Rp.
R1 and Rp are in series. Therefore the total resistance (Rt) is the sum of R1 and Rp.
Now that we have the total resistance, we can calculate the total current. The total current (sic) is equal to the battery voltage (Vt) divided by the total resistance Rt.
It = Vt/Rt
Now we can calculate the voltage across R1. The voltage across R1 equals the current produced by R1 and the resistance of R1.
V1 across R1.
V1 = It * R1
The voltage across Rp equals the product of the current through Rp and the resistance Rp
voltage Vp across Rp
Vp = It * Rp.
Now for current through R2. The voltage across R2 is Vp.
The current through R2 is I2
I2 = Vp/R2
Also, the current through R3 is I3.
I3 = Vp/R3.
The current through R1 is equal to the sum of the currents through R2 and R3.
Is = I2 + I3
This concludes the discussion of series-parallel circuits. The next article will show solutions to problems in parallel series circuits.
Notes:
I have a Bachelor of Science in Electrical Engineering
Introductory Circuit Analysis Third Edition
ISBN 0-675-8559-4
