A 5V, 5A supply should be capable of supplying up to 5A at 5V.
Therefore it is fine to connect it to a board requiring 5V, 500mA.
Yes, you can always use a power supply of the correct voltage with a higher current rating (and that is what you want for design margin). The circuit will only take the current it needs.
Yes you can.
The power supply CAN deliver 5A if needed , but it will only deliver the current required by your circuit.
In your case your circuit requires 500mA to operate , so that is what the power supply deliver.
If your circuit however required 6A , then your power supply may not be able to deliver that much current.
Cheers
Ned
I have heard that excess current to the circuit will burn the circuit
Remember that current is always 'drawn by' a circuit, not 'supplied to' a circuit.
A scenario where an excess current is made to flow through a circuit would be when say some part failed or when the wrong part is inserted. Lets say that to light an LED, you are supposed to put a resistor in series with it. But instead of putting a typical 4.7k you put 47Ohms. This would cause a large current to flow and will damage the LED and if the current is large enough, it can also damage the supply.
A circuit if designed and assembled well should not have such problems.
Yah, thats true....
Theory = you can connect small load to to high current source with no problem as amount of current is 'drawn by' the circuit....
Practice= You need to Limit current drawn from high current (Very Low Impedance source) or you will burn your components.
Demo= Connect your circuit with a 12V-100AH Battery and see what happens use 7805 regulator if your circuit is at 5V....
1)But in which case more heat will be generated? low resistance or high resistance?
2)I have heard about the shunt resistance concept. why do we use it to pass excess current? if it is the circuit who decide how much to consume so, we can supply 5A or 500mA there should be no use of shunt resistance?
You can find that out for yourselves.
Power Dissipated by Resistor is given by
P = I2R or P = V2/R
In this case since your voltage source is fixed, use the second equation.
Higher Resistance means more power dissipation i.e. more heat.
With a fixed voltage supplying the current, that would be true. To better see that use the equation V²/R for power.I have a confusion regarding this
as we know in case of low resistance the value of current will be high as compared to case of high resistance.
so,
I1 - High current with R1 low resistance.
I2 - low current with R2 high resistance.
using the expression,
P=I^2*R
P1=I1^2*R1
P2=I2^2*R2
So the net result will be P1>P2??
With a fixed voltage supplying the current, that would be true. To better see that use the equation V²/R for power.
The equations given nitishn5 are correct... but the conclusion he yielded baffles me..
maybe this is due to my lack of coffee today and lack of sleep last night but... R is in the denominator, meaning large R = less power, small R = more power. Hence why you get more heat being dissipated by a short then with a 1Meg ohm.
grab a large power resistor and connect it to the 5v 5a supply, set the resistance to max, tap off point being at the opposite end of the power resistor. then slide the tap point closer and closer, soon you will 'feel' the warmth of the resistor.
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Now if you had a fixed I and a variable Voltage (current source), you would see that as you increase the resistance you would dissipate more power yielding more power. just remember you control 2 of the 3 variables, V I R. use the equation that does not use the variable you do not control. if controlling the voltage and load, use P=V^2/R. If controlling Current and load, use P=I^2*R. If you know your voltage and can measure your current, but unsure of the load (active loading, ex mos devices) use P=V*I
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