I searched what I can find on resistors and beg your pardon on this question.
Given: A 12VDC battery is connected to one side of a 12VDC incandescent light. The other side of the light is connected to ground.
Question: Is there any size resistor that I could put in series between the battery and the light to COMPLETELY stop the voltage?
Completely.
Would this huge resistor stop the current flow?
What say all of you?

 

Can you work out an equation for the voltage across the light bulb that is dependent of the resistor value 'R' and voltage source 'Vdc'?

 

Voltage is not something that is stopped by a resistor, voltage doesn't flow. Voltage is like the pressure that pushes electrons down the witre. Current (electrons) flow, they are measured in amps and current flow drops off as resistance is increased. See ohms law for the relationship of resistance, current and voltage.

A resistor is not a switch so a resistor cannot ompletely stop current flow.. And, examining the bulb, there is no threshold of voltage required to initiate current flow through an incandescent bulb. Threrefore, if your question is really, "can a big enough resistor stop the flow of current?" The answer is, no.

Now we could get into all sorts of rediculous questions like, " what if I had a resistor that is such a high value that a meter cannot measure the current flow". In that case, I am done discussing. For all practical sized resistors, some current will flow. The low will be so low that your 12 volt lamp will not generate any light but current will flow. It won't even take a high resistance value to dim or prevent the filament rom glowing, but current will flow.

 

To completly stop current flow the resistor would have the value of infinity, which is essentially the same as opening the circuit with a switch. However you can use a resistor with a large value that would lower the current to a point that "for all practicle purposes" the bulb would be off.

 

alright.....if voltage = current x resistance.
say the resistor in the circuit is 0.000001 ohm (not much resistance)
then....12vdc = ?current x 0.000001
then ....?current = 12/ 0.000001
12,000,000 amps
out of a 12vdc battery
Is that right....?

 

alright.....if voltage = current x resistance.
say the resistor in the circuit is 0.000001 ohm (not much resistance)
then....12vdc = ?current x 0.000001
then ....?current = 12/ 0.000001
12,000,000 amps
out of a 12vdc battery
Is that right....?

your math is correct, but you won't get 12MA out of an ordinary battery.

 

Batteries have "internal resistance". The chemical reaction that happens inside the battery must be infinitely fast (which it is not), and the electrolyte paste in the battery must have infinitely high conductivity (which it does not).

Therefore, a real battery looks very much like an ideal battery with a resistor in series with it. The value of the internal resistance (also known as internal impedance), can be calculated by measuring the voltage of a battery with a volt meter. Then measure again with a resistor across the terminals of the battery. Example, add a 100 ohm resistor to a 9 volt battery. The voltage may drop from 9.2 to 8.2. This means the current flow through the 100 resistor is based on 8.2 volts divided by 100 ohms = 0.082 amps. That means one volt drops inside the battery (9.2 - 8.2 volts). 1 volt / 0.082 amps = 12.2 ohms.

The internal resistance changes as the battery ages (discharges).

You will have additional resistance along the wires (not perfect conductors) and any load.

I hope that helps.

 

alright.....if voltage = current x resistance.
say the resistor in the circuit is 0.000001 ohm (not much resistance)
then....12vdc = ?current x 0.000001
then ....?current = 12/ 0.000001
12,000,000 amps
out of a 12vdc battery
Is that right....?

Just one last note, the internal resistance of the battery is only part of the story.

The resistance you refer to, 0.000001 ohm, is so low that the copper wire needed to connect your circuit will have more resistance than that. For example, a 1 cm diameter wire that is 1 cm lonng will have 1 micro ohm of resistance.

If the battery had no internal impedance, that length of wire will dissipate 12 volts x 12M amps = 144 megawatts! The wire will vaporize in a fraction of a second.