Ohms calculations does not match meter reading

[oawad79]

Hello, I am a beginner and need help understanding an experiment I tried from the "Make: Electronics" book, the experiment I am looking at is "Experiment 2: Go with the Flow (2/2)" where I connected a 15 Ohms resistor with a LED 20mA max and a 9v battery, the goal is to measure the current passing thru the multimeter, I got a reading of 1.9mA on the meter, but according to Ohms law

V = I*R
I = 9v / 15 Ohms
= 0.6 A
= 600 mA

so, why does the meter reads 1.9 mA ?

 

[Bluejets]

Assuming connection resistor and LED was in series, you forgot the voltage drop on the LED of your choice plus........
9-(say 1.7) = 7.3v on resistor
I=E/R = 7.3v/15R = 0.486A or 486mA so LED is toast .......

Then if LED shorted 9V/15R = 600mA
AND, VxA = W (or IsquaredR) so 9 x .6 = 5.4W and resistor is not far behind the LED in the smoke department.

[mod edit: corrected 486A to 0.486A ]

 

[Harald Kapp]

While @Bluejets ' post is entirely correct, it doesn't explain the 1.9 mA. Have a look at the datasheet of your multimeter, specifically at the current measurement section and there look for burden voltage or similar.
In this example

we see a max. burden voltage of 1 V in the 200 mA range, which translates to a series resistor of 5 Ω. This can have an influence on your measurement.
Still Using 20 Ω instead of the 15 Ω you claim to have used still doesn't explain the measurement:
7.3 V / 20 Ω = 365 mA, way off from your measurement.

Next look at the pass voltage of the LED. @Bluejets assumed a typical red LED with ~1.7 V pass voltage. Assuming you used a white LED with a pass voltage of 3 V (r higher) we arrive at:
(9 V - 3 V) / 20 Ω = 300 mA.

None of these calculations explains your observation. I suggest:

1. Show us a complete diagram of your circuit. It always helps to understand what's going on.
2. Check your resistor. Is it really 15 Ω? The measurement suggest something in the kΩ range.
3. Check the pass voltage of the LED. What do you actually measure?
4. Check the battery voltage under load. What do you measure (with the LED + resistor connected)? the typical 9 V block tends to lose voltage quite rapidly.

 

[Bluejets]

it doesn't explain the 1.9 mA.

Just possible it was after the resistor died as well.

 

[Audioguru]

A little 9 battery is not made to produce more than 10mA for a few hours.
A Super-Heavy-Duty poor quality 9V battery from "over there" cannot produce anything.

I have one Name Brand 9V battery that was used in something and is dated by me "Feb/2015". It measures 7.57V with no load but its voltage drops to 1.7V with a 1k load (1.7mA).

 

[oawad79]

While @Bluejets ' post is entirely correct, it doesn't explain the 1.9 mA. Have a look at the datasheet of your multimeter, specifically at the current measurement section and there look for burden voltage or similar.
In this example
View attachment 53558
we see a max. burden voltage of 1 V in the 200 mA range, which translates to a series resistor of 5 Ω. This can have an influence on your measurement.
Still Using 20 Ω instead of the 15 Ω you claim to have used still doesn't explain the measurement:
7.3 V / 20 Ω = 365 mA, way off from your measurement.

Next look at the pass voltage of the LED. @Bluejets assumed a typical red LED with ~1.7 V pass voltage. Assuming you used a white LED with a pass voltage of 3 V (r higher) we arrive at:
(9 V - 3 V) / 20 Ω = 300 mA.

None of these calculations explains your observation. I suggest:

1. Show us a complete diagram of your circuit. It always helps to understand what's going on.
2. Check your resistor. Is it really 15 Ω? The measurement suggest something in the kΩ range.
3. Check the pass voltage of the LED. What do you actually measure?
4. Check the battery voltage under load. What do you measure (with the LED + resistor connected)? the typical 9 V block tends to lose voltage quite rapidly.

Thanks for the reply.

1 - Image attached
2 - Yes, 15 Ohms
3 - 1.8V
4 - 2.67 V

I tried again and got a reading of 2.7mA

 

[Audioguru]

We told you that the 9V battery cannot produce much current. Its 9V fell on its face to only 2.67V when you loaded it.
Then the current became reduced to only 2.7mA because the 9V battery that measured 2.67V was almost dead.

 

[oawad79]

Thank you all for the help and prompt responses...appreciate you all.

 

[davenn]

We told you that the 9V battery cannot produce much current. Its 9V fell on its face to only 2.67V when you loaded it.
Then the current became reduced to only 2.7mA because the 9V battery that measured 2.67V was almost dead.

But it is going to produce enough to light a single LED via a resistor

 

[Harald Kapp]

2 - Yes, 15 Ohms
3 - 1.8V
4 - 2.67 V

Still makes no sense:
voltage across the resistor: 2.67 V - 1.8 V = 0.87 V
current through resistor acc. to Ohm's law: I = 0.87 V / 15 Ω = 58 mA

Or computed backwards: 2.7 mA × 15 Ω = 40.4 mV across the resistor which is a contradiction to your voltage measurements.
My guess is that the instrument introduces a noticeable error due the internal burden of the milliamps measuring range.
If you do have a second instrument available, set up the circuit to measure the current as you did before. Then measure the voltage drop across the multimeter (the one which measures the current).

Another option is that your multimeter is off by a lot. Can you check it against a known good reference?

 

[Audioguru]

Specs for the meter show a burden resistance of 5 ohms on the 200mA range. The meter leads also add to the resistance. The battery is weak and is almost dead so its power is quickly draining.

 

[Harald Kapp]

The specs I showed are for demonstration only. We do not know what make and type of meter the op uses.