Formulas for Voltage, Current and Resistors

[Electro132]

Hi,

Its been a long while i've been on here since i was busy with my projects. However, I'm trying to understand this idea where the current input (I) is different to the voltage input (V). Because my multimeter is damaged from when i opened it up to repair it, i am pretty much using I = V / R to find out the current. I am using a 9v battery and either 10,100 or 1K ohm resistors for this.

What i'm trying to get my head around is whether i could use a resistor to limit the amount of current going into my circuit and whether i could use a different resistor to limit the amount of voltage going in. So it goes like this:

I = V / R = 9V / 100 = 0.09 A going into the circuit

and

V = I x R = 0.09 x 10 = 0.9 V going into the circuit

So i'm just wondering is this possible? The reason why i ask this is because if i put in a different current amount to the voltage input then doesn't that place the voltage amount in line with the current amount and not on 2 different values?

 

[Harald Kapp]

Sorry, I can't get my head around what you're trying to express.

use a different resistor to limit the amount of voltage going in

Resistors usually are used to limit current. This reduces the voltage available by the amount of voltage drop across the resistor. Consider this equivalent circuit:



Vbat is your battery voltage, V1 is the voltage drop across R1, V2 is the voltage drop across R2.
R1 is your current limiting resistor, R2 is the equivalent of your circuit under test.
From Kirchhoffs laws: Vbat=V1+V2
From Ohms law: V1= I*R1 and V2=I*R2 and also I= V1/R1=V2/R2=Vbat/(R1+R2)

These are the relevant equations for this circuit.

if i put in a different current amount to the voltage input then doesn't that place the voltage amount in line with the current amount and not on 2 different values?

In my view this doesn't make sense. You don't put in current into a voltage input and vice versa. Maybe it would help if you indicate where in above schematic you connect the multimeter and which settings you use.

 

[Herschel Peeler]

Hi,

Its been a long while i've been on here since i was busy with my projects. However, I'm trying to understand this idea where the current input (I) is different to the voltage input (V). Because my multimeter is damaged from when i opened it up to repair it, i am pretty much using I = V / R to find out the current. I am using a 9v battery and either 10,100 or 1K ohm resistors for this.

What i'm trying to get my head around is whether i could use a resistor to limit the amount of current going into my circuit and whether i could use a different resistor to limit the amount of voltage going in. So it goes like this:

I = V / R = 9V / 100 = 0.09 A going into the circuit

and

V = I x R = 0.09 x 10 = 0.9 V going into the circuit

So i'm just wondering is this possible? The reason why i ask this is because if i put in a different current amount to the voltage input then doesn't that place the voltage amount in line with the current amount and not on 2 different values?

One example uses a 100 ohm resistor. The other uses a 10 ohm resistor. Either approach is good.

 

[BobK]

Because my multimeter is damaged from when i opened it up to repair it, i am pretty much using I = V / R to find out the current.

You probably just blew the fuse. There is a fuse in current measuring path. Usually there are 2, one for the Amp range and another for the lower ranges.

Bob

 

[Electro132]

Thanks everyone for your responses.