Math Question. How many amps on load B ?

[gary350]

I know, idle current and load current A but my meter will not read load current B. How do I find current for load B ???

Electronic device is turned ON but has no load, idle current is 6.2 amps.

Electronic device is turned ON pulling a medium size load and current is 18.6 amps.

Electronic device is on pulling a bigger load, my meter will not read more than 20 amps?

I put a .4 ohm resistor in the circuit to read voltage across the resistor. 3 voltage readings for all 3 loads, 1.22v, 4.4v and 5.18 volts.

How do I find amps for load B ?

**broken link removed**

 

[KeepItSimpleStupid]

Couter reply:

Which is load A and Which is Load B.
No load usually means open, 6.xx Amps? Ammeter broken?

There is something your not telling us?

 

[gary350]

The .4 ohm resistor has lowered the current enough to let me keep my amp meter in the circuit.

The difference between 18.6 amps and 14.65 amps for load a = 3.95 amps.

If I add 3.95 amps to 18.6 amp load B = 22.95 amps estimated current with out the .4 ohm resistor. Wonder how accurate this is?

I had to correct the picture I forgot to draw the load.

**broken link removed**

 

[KeepItSimpleStupid]

Still not clear But I'll bet the ammeter is not an ideal ammeter. Measure the voltage across the ammeter and the resistor. They will approximately add up to the source voltage. Approximately, because the resistance of the wires are not accounted for.

 

[Paul_L]

Electronic device is turned ON but has no load, idle current is 6.2 amps. If there is NO load current should read 0.0 amps.

 

[gary350]

Electronic device is turned ON but has no load, idle current is 6.2 amps. If there is NO load current should read 0.0 amps.

Lots of electric items have idle current with no load. An electric motor that sets on the floor running has an idle current with no load. A power supply transformer that is not connected to anything is plugged into 120 vac it has an idle current. Turn on a 100 watt tube amplifier and volume is turned OFF the idle current is 5 amps . Turn volume up 50% with sound ON current is 10 amps. Turn volume up 100% with sound ON current is 15 amps. An arc welder has an idle current when it is ON but no one is welding.

 

[Ratchit]

I know, idle current and load current A but my meter will not read load current B. How do I find current for load B ???

Electronic device is turned ON but has no load, idle current is 6.2 amps.

Electronic device is turned ON pulling a medium size load and current is 18.6 amps.

Electronic device is on pulling a bigger load, my meter will not read more than 20 amps?

I put a .4 ohm resistor in the circuit to read voltage across the resistor. 3 voltage readings for all 3 loads, 1.22v, 4.4v and 5.18 volts.

How do I find amps for load B ?

**broken link removed**

There is something I am not understanding. The ammeter is in series with the resistor, load, and voltage source. So you are taking a direct measure of current through the load and other components. Isn't that obvious? What is the real question?

Ratch

 

[KeepItSimpleStupid]

Soething your not telling us. Suppose this load was a Class A amplifier at no load. **broken link removed**

Note the article states:

The efficiency of this type of circuit is very low (less than 30%) and delivers small power outputs for a large drain on the DC power supply. A Class A amplifier stage passes the same load current even when no input signal is applied so large heatsinks are needed for the output transistors.

 

[ChrisP58]

...
Electronic device is on pulling a bigger load, my meter will not read more than 20 amps?
...

You need a meter with a higher range.

The .4 ohm resistor has lowered the current enough to let me keep my amp meter in the circuit.

The difference between 18.6 amps and 14.65 amps for load a = 3.95 amps.

If I add 3.95 amps to 18.6 amp load B = 22.95 amps estimated current with out the .4 ohm resistor. Wonder how accurate this is?

I had to correct the picture I forgot to draw the load.

But your 0.4 ohm resistor has reduced the voltage to the load significantly. While using a shunt resistor is a valid way to measure current, it's value should be low enough that it's insertion loss is negligable. My standard shunt resistor is a four terminal, 50 amp, 1 milliohm, instrument grade shunt similar these.

Even with such a low value, they are accurate due to the kelvin connection the extra two terminals give. As long as your meter can read millivolts, you can get a good measurement. 1 millivolt = 1 amp.

As for just extrapolating the current draw when you device has the B load, that will only be accurate if the voltage to current transfer function of your device is known. If you don't know that, then you need to make a current measurement of all three states with your the correct operating voltage at your devices input terminals.

I expect that, when feed the full 15 volts, your idle, Load A, and Load B currents will all change.

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**broken link removed**
As for the confusion others, including myself had until I re-read your post a few times, it's because your drawings and your text don't match. Your schematic just shows a simple load, but your text describes some unknown device in between the power and one of two loads. It would be more clear if your drawing showed [device?] where you now show the load, and two more wires from it to your A or B end load.

 

[gary350]

There is something I am not understanding. The ammeter is in series with the resistor, load, and voltage source. So you are taking a direct measure of current through the load and other components. Isn't that obvious? What is the real question?

Ratch

Someone on the forum told me to put a resistor between power supply and the circuit then take a voltage reading across the resistor to calculate current in the circuit. So far I have not figured out how to do the math to calculate current in the circuit? What you said about the resistor makes sence. If I had a 1/4 HP motor setting on the floor running on 120 vac with a .4 ohm resistor in the circuit I dont see any way to calculate current flow through the motor from a voltage reading across the resistor?

 

[Ratchit]

Someone on the forum told me to put a resistor between power supply and the circuit then take a voltage reading across the resistor to calculate current in the circuit. So far I have not figured out how to do the math to calculate current in the circuit? What you said about the resistor makes sence. If I had a 1/4 HP motor setting on the floor running on 120 vac with a .4 ohm resistor in the circuit I dont see any way to calculate current flow through the motor from a voltage reading across the resistor?

Well, you use the resistance formula I=V/R . Say the resistance is 0.4 ohms and the voltage across the resistance is 8 volts. Then the current through both the motor and resistance is 8/0.4 = 20 amps. The current existing through all components in series is equal. Otherwise there would be a pile up of charge at a point within the circuit.

Ratch

 

[Wade Hassler]

A real quick look at your circuit with a .4Ω resistor says that 'Load B' acts like a .52Ω resistor.
If it were alone across a 15V source, the current would be 15/.52= 29A.
You need a smaller series resistor, as already noted, to make sure that you are not qualitatively changing the way that whatever-it-is acts.

The series resistor should, as a rule I just made up dictates, be less than 10% of the circuit under test. That would be .05 ohms. That's why ammeter shunts are so small in resistance.

 

[gary350]

Ok, I have 1.22v, 4.4v and 5.18v

1.22/.4=3.05a

4.4/.4=11a

5.18/.4=12.95a

But that is NOT what my meter says. With the resistor removed and my Meter in its place I get 6.2a. 18.6a and the 20a meter fuse blows on the next reading.

 

[Ratchit]

gary350,

If you are using a motor for a load, remember that motors sometimes have high initial surge currents at startup. Their current also changes depending on the rpm's and torque. In other words, a motor is a variable electrical load.

Ratch

 

[ChrisP58]

Ok, I have 1.22v, 4.4v and 5.18v

1.22/.4=3.05a

4.4/.4=11a

5.18/.4=12.95a

But that is NOT what my meter says. With the resistor removed and my Meter in its place I get 6.2a. 18.6a and the 20a meter fuse blows on the next reading.

That's because your sense resistor is to large with respect to the load. If there is 5.18 volts across the sense resistor, there is only 9.82 volts at the input of your device. So it's getting less than 2/3rds of the voltage it should be running at.

 

[Reloadron]

Hi Gary

Won't work for the reasons mentioned. Make another 2 drawings. Just drawings using a load of 1.0 Ohm in both. Make one series shunt of 0.4 Ohm as you have and the second drawing make the shunt 0.001 Ohm. Make both Vsupply 15 Volts. Now in one circuit Rtotal is 1.4 Ohms and in the other Rtotal is 1.001 Ohm. This is especially what ChrisP58 is getting at. When you toss in a .4 Ohm shunt you create a large change and subsequent error.

Ron

 

[KeepItSimpleStupid]

Your meter has in internal resistor as well. When you insert it, the circuit is no longer the same.

 

[KeepItSimpleStupid]

R17-R20 looks the resistances of the shunt added to your circuit. https://www.eserviceinfo.com/preview.php?fileid=66019&mode=direct&ext=gif