Measuring Current wastage in Circuit

Using Arduino General Electronics
#1

Hi guys,

I started my learning with a small circuit and wanted to know whether I am measuring the current wastage in the circuit right.

I understood from some of answers in this forum that, If resistor is used, a nominal amount of current will be wasted in order to reduce the voltage to acceptable level of LED.

In this circuit I used

  1. 9v battery
  2. 1k resistor
  3. breadboard
  4. L.E.D

1st diagram attachment explains how I connected the circuit.

Battery volt = 10.2 (As per multimeter)
LED fwd volt = 2.06 (As per multimeter)
1k resistor exact ohms = 950 (As per multimeter)

As per ohms law, I (Current) = 8 mA or 0.008A

In the 2nd diagram I measured the current flowing in the circuit through multimeter and got a reading of 50 mA or 0.05A

(Also attaching the picture of mode in which I used the multimeter for measuring current of 50mA)

If current flowing in the circuit (minus) LED current = Current wastage?
i.e 50mA - 8mA = 42mA wastage?

I am new to drawing diagrams and stuff so please pardon my mistakes in it. I am just a man trying to learn some electronics.

Thanks in advance.

Diagram 1, Ohms law measurement.jpg
Diagram 1, Ohms law measurement.jpg2952×3201 831 KB

Diagram 2, multimeter Current measuring.jpg
Diagram 2, multimeter Current measuring.jpg3120×4160 872 KB

Example multimeter mode.jpg

#2

(!) You need to measure the forward voltage of the LED (or any diode) at (or at least near) the current that you are going to be using. The forward voltage is a nonlinear function of the current.

(2) In the second diagram, you do not specify the voltage of the battery. Presumably, it is much larger than the 9 volts of the first diagram.

(3) I do not know what you mean by wastage. If the LED lights appropriately at 8 mA, then I suppose that 50 mA is unnecessary. The 42 mA is not exactly wasted. The 42 mA + 8 mA is flowing through the LED, through the resistor, and through the battery.

#3

The multimeter internally uses different resistances to make current measurements. The resistance will be different depending upon which range you use. This may change the readings that you get.

#4

In the 2nd diagram I measured the current flowing in the circuit through multimeter and got a reading of 50 mA or 0.05A

If you are using the same battery, either the meter is wrong, or you are misreading it.

The meter disturbs the circuit by adding resistance, thus lowering the current, so it is possible that the reading is 5 mA and you misinterpreted it.

#5

I would imagine that when the multimeter is on the 10A range, then it won;t measure currents of 8mA accurately.
What current is indicated if you select the 20mA range?

Your calculations look correct so the current is going to be around 8mA, Putting the multimeter (on the 10A range) in place isn't going to increase this to 50mA, so I can only assume that it is an inaccuracy of the meter.
I would have expected it to read 0.01A if it was simply a rounding error.

#6

Your multimeter may not be accurate when measuring milliamps on a 10A scale.

1k resistor exact ohms = 950 (As per multimeter)

The resistance reading may not be “exact” either. No measurement is exact/perfect. Better meters are more accurate, and for example the meter I use at work gets calibrated once a year to keep it within spec. (At some point, if you can get your hands on a 1K/1% resistor you can check compare your meter’s reading to the known resistor value.)

If current flowing in the circuit (minus) LED current = Current wastage?
i.e 50mA - 8mA = 42mA wastage?

The current is a series circuit is the same through all series components the battery, the resistor, and the LED). This is one of [u]Kirchhoff’s Laws[/u].

Think of water flow through closed-pipes… The water-current flow has to be the same through all series pipes. That’s true even if there is a restriction (resistance) in one of the pipes. However, the pressure can be different on each side of the restriction. In the water analogy*, water pressure is voltage and water-flow is electrical current flow.

Another of Kirchhoff’s Laws describes how voltage is divided among series components.

Power (energy) is lost/wasted in the resistor. That power is converted to heat. Power is calculated as Voltage x Current.** So, with 8mA and 8.14V across the resistor, that’s about 64mW.

You have the same current through the wires, but since the resistance of the wires is (essentially) zero, you have (essentially) zero voltage across the wires, and (essentially) zero power is lost in the wires.

There is also some power wasted/lost in the LED, but that can’t be easily calculated. If the LED were 100% efficient all of the LED energy would be converted to light, but in reality it’s not 100% efficient and some energy is lost as heat.

  • Don’t take the water analogy too far… If you cut a pipe you get zero resistance and water spills-out all over the place. If you cut a wire you get infinite resistance and no current flows.

Also, zero resistance in water pipes doesn’t cause any problems. With zero resistance in an electrical circuit too much current can flow and things can burn-up.

** Power can also be calculated as P = I2 x R or P = V2/R.

#7

shenbak:
I understood from some of answers in this forum that, If resistor is used, a nominal amount of current will be wasted in order to reduce the voltage to acceptable level of LED.

No, the voltage (and thus power) is wasted. The current is all needed and used. The resistor
sets the amount of current for the LED, but that current is everywhere the same as charge is not created
nor destroyed, nor builds up in one place.

#8

DVDdoug:
Your multimeter may not be accurate when measuring milliamps on a 10A scale.

Yes you are right. I switched the multimeter to 20mA mode and checked it. It showed 7.92mA which is so close to the calculation.

DVDdoug:
Power (energy) is lost/wasted in the resistor. That power is converted to heat. Power is calculated as Voltage x Current.** So, with 8mA and 8.14V across the resistor, that’s about 64mW.

** Power can also be calculated as P = I2 x R or P = V2/R.

Thanks for the power calculation.

DVDdoug:
Also, zero resistance in water pipes doesn’t cause any problems. With zero resistance in an electrical circuit too much current can flow and things can burn-up.

Yup. Learn’t it the hard way by burning up a LED. :smiley:

#9

Thank you all for explaining the concept. I am delighted. :slight_smile:

#10

JohnLincoln:
I would imagine that when the multimeter is on the 10A range, then it won;t measure currents of 8mA accurately.

It is beyond "inaccurate", that cheap meter looks like it only goes to 2 decimal places. 8 mA shouldn't even be enough to make it to the first tick, but since it's a cheap piece of garbage it just reads 6 times higher than it should.

DVDdoug:
Also, zero resistance in water pipes doesn't cause any problems.

Yeah...no. Trying running a pump dry, see how well that "zero resistance" is tolerated.

#11

Jiggy-Ninja:
Yeah...no. Trying running a pump dry, see how well that "zero resistance" is tolerated.

If you consider the pump=battery/power source, it more fits in. Open water line/0 resistance (in the pipe at least) = max flow/0 pressure, = inf I, min V.

The "open water line" really doesn't make much sense though. It more resembles a short circuit, while an open circuit is more like a capped water line. I accidentally pulled a power jumper off a breadboard, yet it didn't spill any power on the desk. It just stopped working

#12

I’m not sure what your point is.

#13

Jiggy-Ninja:
I'm not sure what your point is.

Just a different view for a hammer-fit analogy. I guess a valve could be similar to a variable resistor, and an expansion tank a crude capacitor, but what is a sprinkler???

#14