2007-02-11 12:35:38 · 5 answers · asked by maximum30ft 1 in Science & Mathematics ➔ Engineering
E=IR
12=.0012*10000
You would drop the 12 volts across the 10k resistor by passing .0012 amps. No matter what the resistor, the voltage across it would be 12V. What changes with the resistor is the current, so that I*R always equals 12.
2007-02-11 13:22:32 · answer #1 · answered by Anonymous · 0⤊ 1⤋
The total voltage drops in a series circuit always equals the source voltage. Since there is only one component in your circuit, it would be equal to the source, or 12 VDC (irregardless of the resister value).
The current would equal 1.2 mA
2007-02-11 21:27:36 · answer #2 · answered by LeAnne 7 · 0⤊ 0⤋
If it is a "good" voltage source, i.e. with a very low source resistance, then the voltage will still be 12 when you connect the resistor.
2007-02-11 21:20:27 · answer #3 · answered by dmb06851 7 · 1⤊ 0⤋
It would be 12 VDC
2007-02-11 21:17:53 · answer #4 · answered by Cupcake 2 · 0⤊ 0⤋
12?
are you looking for amperage maybe?
2007-02-11 21:11:22 · answer #5 · answered by hidingbehindthisemailaddy 3 · 1⤊ 0⤋