#1
I need a bit of help with a question from my Engineering coursework...

Is the calculated current (Icalc) always less than the measured current (Icalc)?

For example...
My measured current was .22A, while my calculated current was .1697A .

What would cause this?
I know it's a bit sad, but I've never had any experience with current/circuits and really have no idea on the principles behind the math and how they correlate into real-life situations.

EDIT:

The chart shows...
Vs (volts) = 7
Power (watts) = .9
Measured Current I (amps) = .22
Calculated Current I (amps) = .1697
Last edited by dudey5691 at Sep 16, 2009,
#2
I'm lost...
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#3
I would have expected the measured current to be less than the calculated current due to physical imperfections in the components, which leads to higher resistance and less current.

However, I have only had basic level electronics so I don't really know.
#4
yeah im with tremolo. but yeah im just in some beginer classes in highschool.
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#5
http://wiki.answers.com/Q/Why_is_calculated_current_different_from_measured_current_in_a_circuit

Looks like measured should be lower than calculated.

Are you sure you're calculating it properly?

EDIT :

What's your resistance? Current (I) is Voltage over Resistance (in Ohms). You didn't give the ohms, do you have that number?
Last edited by dark&broken at Sep 16, 2009,
#7
Quote by dark&broken
http://wiki.answers.com/Q/Why_is_calculated_current_different_from_measured_current_in_a_circuit

Looks like measured should be lower than calculated.

Are you sure you're calculating it properly?

EDIT :

What's your resistance? Current (I) is Voltage over Resistance (in Ohms). You didn't give the ohms, do you have that number?


Unrelated. P=IV, where P = power in watts, I = current in amps, and V = voltage. According to that, the calculated seems to be wrong.
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#8
Quote by asfastasdark
Unrelated. P=IV, where P = power in watts, I = current in amps, and V = voltage. According to that, the calculated seems to be wrong.

Ah ok, I knew there were other equations, but I haven't done anything with this since high-school, and I'm now in 2nd year uni, so it's been a while. I was just going by what google had to say.
#9
Alright, after discussing things with an online colleague...

I should have mentioned that we used some quadratic equation that is meant to find theoretical current... RI^2 - VI + P = 0
Our TA said pick the solution closest to the measured, so I chose that... There is, however, a solution that is higher.
BUT, in the case of the .22A-current... the calculated Current would be .5303, which is apparently an irrealistic drop in current.

Any ideas?

It's a parallel circuit with one 10- and 20-ohm resistors, with a multimeter and watt-meter in the circuit.
I'd post a picture, but cannot right now.