AtomSoft
Well-Known Member
Quoted from : http://www.bcae1.com/resistrs.htmWe know:
# The total voltage
# The individual resistor values
We can calculate the current flow and then the voltage drop across the individual resistors. From the Ohm's law page, we will use the formula:
I = V/R
I = 12/3000 ohms
I = 0.004 amps or 4 milliamps
The current flow through the resistors is 4 milliamps. Since the resistors are in series, we know that the current flow through each resistor is the same.
Then, to find the voltage drop across the 1000 ohm resistor, we can use the formula:
V = I*R
V = .004*1000
V = 4 volts across the 1000 ohm resistor
And to find the voltage drop across the 2000 ohm resistor, we can use the formula:
V = I*R
V = .004*2000
V = 8 volts across the 2000 ohm resistor
Around the middle of page
so i got to :
I = V/R
then
V = I*R
This would sum out to be 5volts / 4 resistors regardless since all resistor values are equal.
I = 5 / 4000 = 0.00125
then
V1 = 0.00125*1000 = 1.25V
V2 = 0.00125*1000 = 1.25V
V3 = 0.00125*1000 = 1.25V
V4 = 0.00125*1000 = 1.25V
Now if i had
R1 = 1k
R2 = 1.5k
R3 = 2k
R4 = 2.5k
1 + 1.5 + 2 + 2.5 = 7k
I = 5 / 7000 = 5.5555555555555555555555555555556e-4
then
V1 = 5.5555555555555555555555555555556e-4*1000 = 0.55555555555555555555555555555556v
V2 = 5.5555555555555555555555555555556e-4*1500 = 0.83333333333333333333333333333325v
V3 = 5.5555555555555555555555555555556e-4*2000 = 1.111111111111111111111111111111v
V4 = 5.5555555555555555555555555555556e-4*2500 = 1.3888888888888888888888888888888
would this mean i need more power?
how the hell would this work ?
Tried in a ohms law calc and got
0.00071 A over 7K
so
V1 = 0.00071 * 1000 = 0.71
V2 = 0.00071 * 1500 = 1.065
V3 = 0.00071 * 2000 = 1.42
V4 = 0.00071 * 2500 = 1.775
=4.97V
why not 5V ? Will it still work due to the +/-5% thing on resistors?
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