Hello all,
I have a question about current. I should know this. I really should, but for some reason my mind is blanking.
A circuit will only draw current that is needed, correct? So, a 5V supply at 1.6A won't actually be pushing out 1.6A of current if the circuit is simply a 150ohm resistor in series with the supply. It would be I=V/R. Add up all the resistances in a more complex circuit and I have my R. So for a circuit with 5kohms resistance and a 5V supply, 3mA.
Is that right?
I didn't do the math for your examples but this is accurate
"A circuit will only draw current that is needed, correct? So, a 5V supply at 1.6A won't actually be pushing out 1.6A of current"
One of the more popular comparisons of current and voltage is pressure (voltage) and flow (current) of a fluid.
5 volt supply, 5kohm load current = 1mA (5/5000)
Note that non-linear components such as semiconductors (transistors, diodes, LEDs etc) do not comply with ohms law (ie measuring their "resistance" using a test meter may not be representative of their "resistance" when powered via a different voltage source) so you need to exercise some degree of caution when basing your assumptions of load resistance. Resistance is in quotes to indicate that they are non-linear
card9inal:
Hello all,I have a question about current. I should know this. I really should, but for some reason my mind is blanking.
A circuit will only draw current that is needed, correct? So, a 5V supply at 1.6A won't actually be pushing out 1.6A of current if the circuit is simply a 150ohm resistor in series with the supply.
Correct, but if could be said that the single resistor is wired in parallel with the supplies output, or 'across' the power supplies output. Saying something is in series or parallel usually means in context of additional components in the network as well as the voltage source. Nit picking perhaps, but anyway only .03333.....amps would flow through the 150 ohm resistor, even though the power supply is capable of suppling up to 1.6 amps of current. Ohm's law tells us that a voltage source (your power supply) cannot 'push out' current, but rather the load resistance determines how much current will be drawn from the voltage source.
It would be I=V/R. Add up all the resistances in a more complex circuit and I have my R. So for a circuit with 5kohms resistance and a 5V supply, 3mA.
Check your math, I get 1mA of current using those values?
Is that right?
You understanding is correct, your math skills needs work.
Lefty
A "voltage" source attempts to keep the voltage across a load constant, and
the current into the load depends upon the load, while a "current" source attempts
to push its set current into the load. In the real-world, we rarely run across current
sources. 5V, 3.3V, 120VAC, etc, those are all voltage sources.
Basic theory... but how is it applicable to the OP's question???
Doc
That's what I thought... And HOW I came up with 5/5000 = 3mA, I have no idea. I chalk it up to posting this too early in the morning!
