How to calculate Voltage? Help

[jamme61]

I was messing with the dropping resistors in the power supply of a Marshall 2204 clone amp. I wanted to know if there's a way using Math, to know what the voltage would change too? My example is, I have 443 volts going thru a 4.7k dropping resistor- which then drops the voltage to 410 volts- is there a way to figure this out using Math? I now have to guess, at what value resistor I need and then measure, with the DVM. I'd like to work smarter. Thanks for any help on this

[John_P_WI]

Ohms law is your friend.

V=IR

Given you are dropping 33 v (443-410) and know you have a 4700 ohm resistor.

Therefore, 33/4700=I, I = 0.007 amps

So, knowing current from above and the voltage drop you want, let's say 75v

V=IR or V/I=R or 75/0.007= R, R=10714 or 10k7.

Of course, the numbers will change under signal load conditions...

[jamme61]

Thank you very much Im going to talk to my daughter now get some of that tuition back LOL She's my math person.

Thanks I get it- your words and OHMS LAW are hanging on the wall in my Man Cave Thanks

[martin manning]

In the above example, increasing the dropping resistor value will reduce the current draw of the circuits downstream, and so the current through and the voltage drop across the new resistor. Assuming constant current, the calculated value will be too small to achieve the desired result.

Treating the resistor as the upper leg of a voltage divider, with all of the downstream circuitry being the lower leg, can make a better estimate. With that assumption the new resistor’s value can be found as

Rnew = R*((V1/V2new-1)/(V1/V2-1)),

where V1 and V2 are the existing upstream and downstream node voltages and R is the existing resistance.

Using this formula and the values from the example, the new resistance would be estimated at 11k9.

[tubeswell]

@jamme61 - what John and Martin said.

The equation V = I x R (where: V = Volts, I = Amps, R = Ohms)

reworks to:

I = V/R (when solving for current)

or:

R = V/I (when solving for resistance)

Since each factor (I or R) can be solved with the product (V) and the other factor (R or I respectively), then you only ever need 2 known quantities to solve any riddle.

Hence the power equation, W = V x I (Watts = Volts x Amps) can also be solved if you only know R and V, thus: W = V x (V/R); which easily rearranges to: W = (V x V)/R or (Vsquared/R).

[jamme61]

@jamme61 - what John and Martin said.

The equation V = I x R (where: V = Volts, I = Amps, R = Ohms)

reworks to:

I = V/R (when solving for current)

or:

R = V/I (when solving for resistance)

Since each factor (I or R) can be solved with the product (V) and the other factor (R or I respectively), then you only ever need 2 known quantities to solve any riddle.

Hence the power equation, W = V x I (Watts = Volts x Amps) can also be solved if you only know R and V, thus: W = V x (V/R); which easily rearranges to: W = (V x V)/R or (Vsquared/R).

wow thanks, I needed that break down, helped a lot - I love the help just a little tough to understand but I got it. Wasn't kidding about my daughter- even e-mailed her physics teacher. If I only paid attention in school. Thanks again to all you guys for the help.

[Stevem]

Many moons ago I printed out a nice big 4" ohms law wheel at stuck it up on the wall in front of me at the work bench!

[jamme61]

Many moons ago I printed out a nice big 4" ohms law wheel at stuck it up on the wall in front of me at the work bench!

I'm starting to run out of wall space - this stuff is my man cave's works of art