Download Electrical-and-Electronic-Principles-P1

Electrical and Electronic Principles P1
Calculation and use of Ohm’s law, for this circuit the applied voltage will be 12 volts.
Find:
The total resistance RT
The total current IT
The current through R3
The voltage across R5
The total power dissipated
Related formulae
R = V/I
I = V/R
V =I x R
Power in Watts = I2 x R = V x I = V2/R
To find RT
Use Product/Sum to find resistances in parallel. Warning, this technique only works with 2 resistors at a
time.
The product simply means to multiply the values of the resistors. Therefore the product of R1 x R2:
R1 x R2 = 220 X 100 = 22,000
We now take the sum of R1 and R2 and divide it into the product to find the parallel resistance of R1 and
R2 together.
Hence 22,000/R1 + R2
= 22,000/320
= 68.75 Ω
To help you understand the process we now redraw the circuit:
You should be able to see that R1 and R2 have been replaced by a single resistance of 68.75Ω which
multisim has labeled R2
We now carry out exactly the same procedure with R2 and R3. When this is complete all of the 3 original
resistances will be replaced by a single resistance.
(R2 x R3)/(R2 + R3)
(68.75 x 47)/(68.75 + 47)
= 27.91 Ω
And we redraw the circuit again
Now do the same for R4 and R5 and redraw the circuit:
So RT = R1 + R2 = 36.7Ω
You can check this with the results of the simulation. You will see that it is completely correct.
To find IT
Using the triangle or the related formulae, finding the total current flowing through the circuit is easy
enough.
I = V/R
= 12/36.7
IT = 0.326 Amps or 326 mA
Although we could find the current through R3 at this stage, it is much easier to find the voltage
dropped across each of the parallel branches first. From the formulae shown above we note that:
V=IxR
We have just worked out the total current flowing into the circuit, so the voltage drop across both banks
is given by:
Vbank 1 = 0.326 x 27.9 = 9.122 volts
Vbank 2 = 0.326 x 8.8 = 2.86 volts
If we have worked it out correctly then Kirchoff’s law tells us that the sum of voltage drops in the
external circuit must = the applied voltage ( 12 volts)
Sum of external volt drops = 9.122 + 2.86 = 11.99 volts
That’s close enough for me
Now we have worked out two different things here.
First you now know the voltage drop across R5.
VR5 = 2.86 volts = Vbank 2
Second you now know the volt drop across R3 = 9.122 volts = Vbank 1
Now applying Ohms law we can find the current flowing through R3
IR3 = 9.122/47
IR3 = 0.194 Amps
Finally you can work out the total power consumed by the circuit in 3 different ways.
P = I2 x R
= 0.3262 x 36.7
= 3.9 watts