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I have the simple circuit as attached.

schematic

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?

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  • \$\begingroup\$ Are the LEDs on or off? Probably off. I have the feeling that forward voltage sum of the LEDs is more than 9V, which means there is no current going through the LEDs. You can easily check, by removing 2 or 3 LEDs (adjust the resistor to not have too much current as test). Note that if one LED has a forard voltage of 3 V, 4 LEDS reduce 12V. \$\endgroup\$ – Michel Keijzers May 3 '18 at 11:11
  • \$\begingroup\$ The LEDs are all on, so I assume some current is going through them. What would the calculation be? \$\endgroup\$ – ED9909 May 3 '18 at 11:15
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    \$\begingroup\$ The calculation for the resistor? R = (VCC - sum(forward voltages) / I). But if the LEDs are on, there should go current through the circuit, although it may be 1 mA (if they don't bright much). I assume you checked your Ammeter? \$\endgroup\$ – Michel Keijzers May 3 '18 at 11:22
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    \$\begingroup\$ The current should pulsate a lot since you're rectifying without any smoothing. How are you actually measuring this? \$\endgroup\$ – pipe May 3 '18 at 11:33
  • \$\begingroup\$ I = V/R = 9V / 16.8 = 0.536A Nope. 0.35/16.8 = 20.8 mA \$\endgroup\$ – winny May 3 '18 at 14:44
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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.

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  • \$\begingroup\$ In order to calculate the power do you take the total voltage, so in this example the 9V multiplied by the current or 20.8mA which is 0.252W? \$\endgroup\$ – ED9909 May 5 '18 at 17:24
  • \$\begingroup\$ @ED9: Yes, but note that is the total power delivered by the full wave bridge to the rest of the circuit, not just the resistor. \$\endgroup\$ – Olin Lathrop May 6 '18 at 12:09
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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

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  • \$\begingroup\$ Why is that? Why do you not take the voltage drop across the LEDs? \$\endgroup\$ – ED9909 May 3 '18 at 12:00
  • \$\begingroup\$ Inaccurate. 1/ A diode it has small Delta-V for large Delta-I. 2/ The voltage drop varies from type to type and batch to batch. \$\endgroup\$ – Oldfart May 3 '18 at 12:09
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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:

current through LEDs

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.

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