How do you recreate this schematic on a breadboard?

Hi,
I'm new to the Arduino and building circuits in general so I realize that this question may seem easy to figure out but I'm struggling to translate this schematic onto my breadboard. I have the Arduino Uno Starter kit. I have attached an image of the schematic below. Any help is greatly appreciated.

This circuit shows you need two 5 volt power supplies.
Do you have two?

Trying to find the current i?

Pretty sure I've seen that in a textbook... :sunglasses:

.

You only need one 5V supply for that circuit since they share the negative terminal.

You only need one 5V supply for that circuit

I didnt know if the OP was told to use two or one power supply in this circuit.

If you use one 5V supply, with the 4700 ohm and the 560 ohm in parallel, the parallel combination is =500.3802 ohms.
500.3802 ohms in SERIES with the 1 k ohm = 1500.380228136882129277566539924 ohms.

IR = 5V /1500.380228136882129277566539924 ohms = 0.00333248859604662949822605169792 A

= 3.333 mA.

(I hope I wasn't supposed to not give him the answer)

(I hope I wasn't supposed to not give him the answer)

He will have to give you the marks now. :sunglasses:

I think the point might be he has to work out what it should be and then measure it to see what it actually is.
Given that each resistor has a tolerance I would give top marks if he worked out the maximum and minimum it could be given the fact he is using real resistors. Also he power supply has a tolerance so that needs to be measured and factored in.
There is a lot more in this assignment still left to do.

It was MarkT who suggested one power supply .
After that the rest was a given. I think he should get the point.

I don't see what difference it makes whether there is one, two, or a dozen 5V supplies. The current is the same but sooner or later someone is going to bring up the point that if there were more than two, they would have to be shorted to each other and then they would. be fighting each other so practically speaking , the limit is two.