A few circuits with resistors and Ohm meters to represent Ohm's law.
Some resistors are connected in series and some are connected parallel.
Also a few are combined.
Ohm's law has the following formula:
U=I×R
This formula can be used to calculate the voltage, when current and resistance are known.
When calculating current, the formula looks like this:
I=U÷R
And when calculating resistance, the formula looks like this:
R=U÷I
When having multiple resistors in series, or parallel, the above mentioned formulas are a little more difficult to use.
When having resistors in series, its just adding up the Ω values. For example:
R1=100Ω, R2=200Ω.
The equivalent resistance will be:
Req=R1+R2=100Ω+200Ω=300Ω.
When resistors are connected parallel, its a little more difficult to calculate the equivalent resistance.
2 formulas can be used.
The 1st one is for only 2 parallel resistors:
Req=(R1×R2)÷(R1+R2).
With the Ω values of previous used resistors this will be:
Req=(100Ω×200Ω)÷(100Ω+200Ω)=
(20.000Ω)÷(300Ω)=66,67Ω.
The 2nd formula can always be used, also with more than 2 parallel resistors:
1÷Req=(1÷R1)+(1÷R2)=answ.
1÷answ.=Req.
This can also be writtin as:
Req=1÷((1÷R1)+(1÷R2)=answ.).
The calculation would be:
1÷Req=(1÷100Ω)+(1÷200Ω)=0,015.
1÷0,015=66,67Ω.
Easier would be:
Req=1÷((1÷100Ω)+(1÷200Ω)=0,015)=66,67Ω.
The 2nd formula with the easier way of writing down the used methode for on coming calculations.
In the simulated circuits the next calculations are made regarding to equivalent resistance:
1st circuit (top left):
No equivalent resistance.
2nd circuit (top):
Req=R1+R2=10Ω+10Ω=20Ω.
3rd circuit (top):
Req=1÷((1÷R1)+(1÷R2)=answ.)=
1÷((1÷10Ω)+(1÷10Ω)=0,2)=5Ω.
4th circuit (top right):
Req=1÷((1÷R1)+(1÷R2)+(1÷R3)=answ.)=
1÷((1÷10Ω)+(1÷5Ω)+(1÷5Ω)=0,5)=2Ω.
5th circuit (bottom left):
Req=R1+R2+R3=10Ω+5Ω+5Ω=20Ω.
6th circuit (bottom):
Req1=R1+R2=5Ω+10Ω=15Ω;
Req2=R3+R4=5Ω+5Ω=10Ω;
Req3=1÷((1÷Req1)+(1÷Req2)=answ.)=
1÷((1÷15Ω)+(1÷10Ω)=0,1667)=6Ω;
Req4=Req3+R5=6Ω+10Ω=16Ω.
7th circuit (bottom right):
Req1=R1+R2=5Ω+10Ω=15Ω;
Req2=1÷((1÷Req1)+(1÷R3)=answ.)=
1÷((1÷15Ω)+(1÷10Ω)=0,1667)=6Ω;
Req3=Req2+R4=6Ω+5Ω=11Ω;
Req4=1÷((1÷Req3)+(1÷R5)=answ.)=
1÷((1÷11Ω)+(1÷5Ω)=0,2909)=3,4Ω;
Req5=Req4+R6=3,4Ω+10Ω=13,4Ω.
