I have the simple circuit as attached.
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
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?