Current reading in a simple LED circuit

Question

I have the simple circuit as attached.

^{simulate this circuit – Schematic created using CircuitLab}

The voltage V0 measured across the bridge rectifier BR1 is 9V, the voltage V1 measured across the resistor is 0.35V (the resistor is 16.8 ohms). The voltage V2 measured across the multiple LEDs is 8.65V.

By Ohms law isn't the total current through the circuit

I = V/R = 9V / 16.8 = 0.536A

If we use

I = (V0 - V1) / R = (9V - 0.35V) / 16.8 = 0.515A

or

I = (V0 - V2) / R = (9V - 8.65V) / 16.8 = 0.021A

But why is the ammeter reading a current of (A) 0A? (The multimeter has a resolution of 1mA.)

If the calculated current going through the circuit does not include the Voltage drop across the LEDs, why not, why is this voltage not used in the I=V/R calculation?

Answer 1

By Ohms law isn't the total current through the circuit

I = V/R = 9V / 16.8 = 0.536A

No. You have the right idea but the wrong execution.

The current thru a resistor is the voltage across it divided by its resistance. Your error is in using 9 V for the voltage across the resistor. That's a voltage somewhere between two points in your circuit, but not what is across the resistor.

You have measured 350 mV across the resistor. The current thru the resistor is therefore (350 mV)/(16.8 Ω) = 20.8 mA.

Because they are all in series, this is also the current thru the string of LEDs, and thru the ammeter A1.

Answer 2

To calculate the current through the resistor you can only take the actual voltage drop of that resistor \$I=V/R = 0.35V/16.8R = 20.8mA\$ This should be what your Amp-Meter reads

Answer 3

I think it is likely that the current waveform you are getting, cannot be correctly measured by the meter you are using.

The current will look something similar (different peak value, maybe different frequency (this assumed 50 Hz)) to this:

Depending on the setting (DC/AC) of your multimeter and the capabilities it might read: the peak value, the average value, the true rms value, some random value in between or just zero.

Note that your voltage measurement over the resistor suffers from the same problem - so the value you are getting with your multimeter might be something but not an accurate representation of what is going on. The voltage across it will have the same waveform as the current.