Why this Simple Watt Meter for a 220 Volt Circuit works?

In summary, the conversation discusses a circuit for a wattmeter that uses resistors to measure power. The circuit includes two 470m resistors as current shunts and a resistive divider with values of 33k and 7k5. There is also a version of the circuit with one resistor value changed. The conversation also mentions the calculation for the voltage drop across the 33k resistor and the possibility of using a different value for the 7k5 resistor to improve accuracy. It is also noted that the current shown in the circuit may not accurately reflect the power being measured. Overall, the conversation highlights the complexities of creating a true RMS wattmeter.
  • #1
henriquevpp
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TL;DR Summary
Help with a wattmeter for 220V circuits that have only resistors.
Hello guys!

I'm just a begginer student. I found this circuit of a wattmeter that works only with resistors. Can someone explain to me how the 33k resistor give a voltage of 1 mV to 1 Watt? I did not understand why these resistor values has been chosen and why the it measures only in the 33k resistor.

Link: https://everycircuit.com/circuit/5290189996163072/simple-watt-meter-for-220-volt-circuit

I made a version (only changing one resistor value) at Fasltad: http://tinyurl.com/y7uqu8n4 (the url was too big).

Can someone help me? Thank you!
 

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  • #2
The two 470m resistors are 0.47 ohm current shunts in parallel, so they drop a voltage proportional to the current flowing in the light globe load. Vshunt = I * 0.235

That small AC voltage is attenuated by the resistive divider, 33K and 7k5 to be measured.
The supply voltage is reasonably fixed so the small AC voltage across the 33k is proportional to the load power.
 
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  • #3
1 watt will have a load current of 1 / 220 = 4.545 milliamp.
Since then V * I = 220 V * 4.545 mA = 1 watt. (edited units on this line).
The two 470m current shunts make 0.235 ohm.
0.235 ohm * 4.545 mA = 1.068 mV per watt.
1.068 * 33k/(33k+7k5) = 0.870 mV per watt. Which is wrong.

If we assume the AC voltmeter has a very high impedance.
The divider needs a ratio of 1 / 1.068 = 0.936
What value should be used in place of 7k5 ?
33k / ( 33k + x ) = 0.936
x = ( 33k / 0.936 ) - 33k = 2.25k, not 7k5
 
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  • #4
Odd that the current is shown as 524 mA, which from a 220 V supply sounds like around 115 W not 100.

And now Baluncore has done the rest
 
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  • #5
And then again such a "wattmeter" assumes that the voltage and current are in phase. Creating a true RMS wattmeter is tricky (but possible, a colleague and I dreamed up such a circuit 15 years ago).
 

Related to Why this Simple Watt Meter for a 220 Volt Circuit works?

1. How does the watt meter measure power in a 220 volt circuit?

The watt meter works by measuring the voltage and current in the circuit and multiplying them together to calculate the power. In a 220 volt circuit, the voltage is constant at 220 volts, so the watt meter only needs to measure the current to determine the power.

2. Can this watt meter be used for circuits with different voltages?

Yes, this watt meter can be used for circuits with different voltages. As long as the watt meter is able to measure both the voltage and current in the circuit, it can calculate the power regardless of the voltage.

3. Is it safe to use this watt meter with a 220 volt circuit?

Yes, it is safe to use this watt meter with a 220 volt circuit. The watt meter is designed to handle the high voltage and is equipped with safety features to protect the user.

4. How accurate is this watt meter in measuring power?

The accuracy of this watt meter will depend on the specific model and brand. However, most watt meters have an accuracy of around 1-2%, which is sufficient for most applications.

5. Can this watt meter be used for both AC and DC circuits?

It depends on the specific watt meter. Some watt meters are designed to measure only AC circuits, while others can measure both AC and DC circuits. It is important to check the specifications of the watt meter before use.

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