Article : Andy Collinson
Email me

Shortcut to Series & Parallel Circuits, Kirchhoff's Laws, Current Division, Voltage Division


Ohm's Law
In DC circuits, the relationship between the current, voltage, power and resistance may be resolved with the aid of a pie chart :-



There are four quadrants representing Voltage, V, Power, W, Resistance, R and Current, I. Knowing any two quantities allows the other two to be found.

For example, if you have a 1k resistor and apply a voltage of 10 Volts DC across its terminals, then the current flowing through the resistor will be:

V/R = 10 / 1000 = 0.01A

The power dissipated in the same 1k resistor would be:

V2/R = 102 / 1000 = 0.1 Watt.

Depending on the flow of current, a voltage will have a polarity. The direction of current is indicated with an arrow, as shown below :-

idc


The side of the resistor where the current flows into the resistor will be the positive side of the voltage, the negative side is where the current flows out. If the resistor was 5 ohms and the current ( I, not one) was 2 amps, then the voltage or potential difference would be 10 volts. In electronics, it is normal to talk about potential differences (p.d.) with reference to one point, which is usually zero. If the point was not zero, then its value would be clearly indicated, but for convenience, most systems have a common ground or earth which is usually zero volts.



Scientific Notation and Unit Shortcuts
In electronics, often units can be very large e.g. 1M = 1,000,000 ohms or very small e.g 1uF = 0.000001F

To make things easier, scientific notation is often used where the exponent is raised to the power 10. The following table contains common units and exponents

.
Unit Symbol Exponent
Mega M x106
kilo k x103
milli m x10-3
micro u x10-6
nano n x10-9
pico p x10-12

So for example if you apply 10V DC across a 2k resistor then the current flowing is 10 / 2 x103=5 x 10-3
As the exponent x10-3 is the unit "milli" then the current is 5mA. Similarly if a resistor value is in Megaohms, then the current will be in units microamps.



Series and Parallel Circuits
The current in a series circuit is the same at all points. Below, 2mA is measured at any break between two points:


The voltage in a parallel circuit will be the same when measured across any two points:


Kirchhoff's Laws
Kirchhoff's voltage and current laws are also a basic prerequisite for circuit analysis. Kirchhoff's current law simply states that the sum of currents flowing into a junction equals the sum of currents flowing away from the junction. For example:-



The arrows represent the direction of current flow, the junction is where the wires meet. I1 is flowing into the junction whereas I2 and I3 are flowing out. If I1 was 20 amp and I3 was 5amp then I2 would be 15amp, as I1=I2+I3.Kirchhoff's voltage law states that the sum of voltage drops around a closed circuit is equal to zero. This can also be expressed as the sum of voltage drops around a closed circuit is equal to the sum of voltage sources :-



If the diagram above, the voltage drops across R1, R2 and R3 must equal 10v or 10=V1+V2+V3. Here is an example :-


The currents i2 and i3 and the unknown resistance, R can all be calculated, using basic dc theory. The direction of current flow is as indicated by the arrows. The voltage on the left hand 10 ohm resistor is flowing out of the top terminal of the resistor. The p.d. across this resistor is (i1* R ) or 5 volts. This is in opposition to the 15 volt battery. By Kirchhoff's voltage law the p.d. across the centre 10 ohm resistor is thus 15-5 or 10volts. Using ohms law, the current through the 10 ohm resistor ( V/R ) is then 1 amp. Using Kirchhoff's current law and now knowing i1 and i3, i2 is found , i3=i1+i2 therefore i2=0.5 amp. Again using Kirchhoff's voltage law the p.d. across R can be calculated. 20=i2R+10. The voltage across R ( i2R) is then 10 volts. The value of R is (V/I) or 10/0.5 or 20 ohms. A few more examples are presented under the Circuit Analysis section.

Current Division
The current flow at a junction will divide into two parts, the current through the respective branch can be worked out as shown below:
The current through R1 can be found using the equation:
I1 = IT R2
R1 + R2
The current through R2 can be found using the equation:
I2 = IT R1
R1 + R2


Voltage Division
The voltage across R1 can be found using the equation:
VR1 = V R1
R1 + R2
The voltage across R2 can be found using the equation:
VR2 = V R2
R1 + R2