Providing with only 9V battery, a load such as LED is used... will the current goes as high as 50A++ with power up to 1300W++? Possible? Or i have sumting wrong with it? I am testing my watt meter...

 

Unfortunately something is wrong with your wattmeter or the way in which you are taking the measurement. Is the meter something you built yourself? Post a circuit or schematic of your measurement setup.

 

Unfortunately something is wrong with your wattmeter or the way in which you are taking the measurement. Is the meter something you built yourself? Post a circuit or schematic of your measurement setup.

Oh no.... I am using the schematic provided by bhabott:
[the link is here...]

I am using a 9V battery and a 1 ohm shunt. However if i use a smaller shunt like 0.1Ohm..it short circuited. I wonder what had happened...

 

9V batteries aren't very robust, figure around 200ma max, and then their lifespan is severly limited. My personal experience is they max out at 1A with a dead short (no voltage), and get hot.

Is this circuit you show a wattmeter?

The load you describe, an LED, what limited the current in it? No current limit and the LED is toast.

 

9V batteries aren't very robust, figure around 200ma max, and then their lifespan is severly limited. My personal experience is they max out at 1A with a dead short (no voltage), and get hot.

Is this circuit you show a wattmeter?

The load you describe, an LED, what limited the current in it? No current limit and the LED is toast.

so you mean 9V may have up tp a max of 200ma? yup..that is a watt meter...i juz connected the LED across the load + and -.

 

200 ma may be pushing it for a 9V battery, less is better.

You're LED is a dead short without any current limiting. As I said, it is toast. Did you share the 9V between the wattmeter and the load?

 

200 ma may be pushing it for a 9V battery, less is better.

You're LED is a dead short without any current limiting. As I said, it is toast. Did you share the 9V between the wattmeter and the load?

currently yes....
even if i changed to use LED having external 9V battery also same result..

 

Duracell says that the internal resistance of a new 9V alkaline battery is 1.7 ohms. Then with a current of 2.65A the output voltage is 4.5V. Its current is 5.3A into a dead short.

Energizer shows a graph of the capacity of a 9V alkaline battery (when its voltage has dropped to only 4.8V). It is 625mAh when the current is 25mA and it is 280mAh when the current is 500mA.

The shunt on your wattmeter should be much less than 1 ohm.
Your LED will immediately burn out without a series current-limiting resistor.

 

You need to use a current limiting resistor in series when powering LEDs. Otherwise, the LEDs will burn up in a hurry.

The general formula for determining the resistance needed is:
Rlimit >= (Vsupply - VfLED) / DesiredCurrent
where:
Vsupply is the voltage being supplied to the LED
VfLED is the manufacturer's typical forward voltage at the desired current.
DesiredCurrent is usually 20mA for a standard LED.

For example: you're using a 9v battery, and a typical red LED may be rated for a Vf of 2.1v @ 20mA.
Rlimit >= (9v - 2.1v) / 20mA
Rlimit >= 6.9V / 0.02 A
Rlimit >= 345 Ohms. The closest E24 standard value is 360. Chart: http://www.logwell.com/tech/components/resistor_values.html
Then calculate how much current you'll actually get through the resistor (and hence the LED):
Current = Voltage / Resistance = 6.9v/360 Ohms = 19.17mA

 

Try testing your wattmeter with a 1kΩ Resistor with a 5VDC supply. With NO LED.

It should give a value around 0.025mW or 25uW

Then try only a 220Ω Resistor With NO LED., it should give a display of 113 mW

If you tested the LED without a current limiting resistor, it was trying to estimate the value of a short circuit, and the LED is probably dead.

 

thanks for all the guidance provided....
which means i had done something wrong during my measurement?
I am afraid if my wattmter goes wrong....