ohms law question

Out of a 9v battery - Is a Megohm resistor less volts than a 10 ohm resistor, or is it just the current that changes?

If you apply the same voltage to either of those resistors, the applied voltage doesn't change.

But there will be less current flowing in the 1Meg resistor.

Volts / ohms = current.

Yep. thanks for that.
That means you can spark out of a capacitor, even if you resist it off, cause the voltage still stays high.

richngreen said:

That means you can spark out of a capacitor, even if you resist it off, cause the voltage still stays high.

Click to expand...

Sorry I don't know what that means.
What is "spark out of a capacitor"?
What is "resist if off"?

When you resist things off, the voltage still stays high, and then you beat the dielectric (which means *NON*-conductive.) breakdown voltage of things and the electricity conducts thru it, even air.

or... is that wrong? and if you add a resistor, theres no way a spark is gettin' thru that mate... and sparks are caused by power, not voltage...

richngreen said:

When you resist things off, the voltage still stays high, and then you beat the dielectric (which means *NON*-conductive.) breakdown voltage of things and the electricity conducts thru it, even air.

Click to expand...

You should really get your terms straight.... I know what you are trying to say but it is still difficult to decipher.

With a high value resistor, the capacitor will still charge vs a low resistor. What changes is the rate at which you can charge the capacitor due to the current limited through the resistor. Higher current yields a higher charge rate. As far as "sparking out" the capacitor, you are basically just shorting the capacitor terminals to create a spark creating huge potential currents from the capacitor.... OHMs law is still at play here. The wire used to short the capacitor has a very small resistance relative to the 1Meg or the 10 Ohm resistor you mention above in your example, but the voltage remains the same (assuming the capacitor is allowed to charge to 9V).

V = I x R
I = V / R
R = V / I
P = I x V = V^2 / R

In simple terms, you can compare voltage, current and resistance in electronics to pressure, flow and restrictions in fluid flow.

Voltage = pressure; how strongly flow is forced through things.

Current = flow.

Resistance is like restrictions, or thin tubes rather than thick pipes. Less flow for a given pressure, or needing more pressure to maintain the same flow (current).

A high resistance will show the same voltage (pressure) as a low one, when there is no flow.
As soon as there is a low resistance elsewhere to complete the circuit, the pressure / voltage is dropped or lost across the high resistance part.

A capacitor is a bit like a hydraulic accumulator, or an elastic disc across a pipe; it will allow some flow as pressure changes and it stretches or releases, but no flow with a steady pressure / voltage.


If you have two or more resistors in series in a circuit, the voltage will be divided in proportion to the resistance; eg. 10K and 1K would have 10/11 of the voltage across the 10K and 1/11 across the 1K.

However, as soon as anything else is connected, you also have to allow for the resistance of the other parts; eg. two 1M resistors in series would have half the supply at the junction with nothing else connected. Add a 1K resistor from the junction to 0V and the voltage is then less than 1/1000th the supply...

You have to allow for all current paths in the end total circuit, you cannot work one bit out then connect something else and expect things to not be changed.

Get your head around this, 2 capacitors in parallel have the same output as 1 with the same battery... confusing as hell mate...
What if current was voltage and voltage was current, how are we supposed to tell?

Yeh 10/11 11/10 thats how you design circuitrs. i like the old american amplifiers, they were different kind of dude back then.

Thats how my oscillator works, it 2/.5 .5/2's on each side then the battery stays mutual exclusive on the two pathways.