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?
Best Answer
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.