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I couldn't find an equation for this on Google/Wikipedia, but does the frequency of a function generator (we'll say that it's a sine wave) impact the current output. Example, if I had a 5 volt source and applied it to a 10 ohm load at 50Hz what would be the current seen by the load?

Thanks,

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  • \$\begingroup\$ What is the model of your function generator? \$\endgroup\$ – tyblu Dec 27 '10 at 21:33
  • \$\begingroup\$ The one I would be using would be the PI-8127 - store.pasco.com/pascostore/… \$\endgroup\$ – eWizardII Dec 27 '10 at 21:35
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Ideally the function generator is a perfect voltage source for any load. A 5V 50Hz sine wave has the form V(t) = 5*sin(2*pi*50Hz*t) [V]. V = IR, giving I(t) = 0.5*sin(2*pi*50Hz*t), or a 0.5A-peak, 50Hz sine wave. It looks like your linked device behaves this way.

The source and load will be in phase so long as the impedance is real (no inductors, capacitors, etc.).

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  • \$\begingroup\$ Okay thanks, that explains what I was missing, was the p-p part. \$\endgroup\$ – eWizardII Dec 27 '10 at 22:15
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if I had a 5 volt source and applied it to a 10 ohm load at 50Hz what would be the current seen by the load?

This is a simple Ohm's law problem.

5V / 10 ohms = 500 mA.

If the 5 V are RMS, then the mA are RMS. If the 5V are peak-to-peak, then the mA are peak-to-peak.

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  • \$\begingroup\$ Okay thanks, I assumed so, however I wasn't sure how the frequency affected the current. \$\endgroup\$ – eWizardII Dec 27 '10 at 22:14
  • \$\begingroup\$ If you had a capacitor or inductor instead of just a resistor, the calculation gets trickier. \$\endgroup\$ – markrages Dec 27 '10 at 22:24
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This comes down to a simple i=v/r. There are 2 things you will most likely need to worry about assuming you aren't at very high frequencies:

  1. What is the max current your function generator can output?
  2. Is your load frequency dependent.

For point 1, you just need to check to see if you are hitting a current limit.

For point 2, if your circuit is frequency dependent (has capacitance and/or inductance) then with a fixed voltage, the magnitude of your resistance, and thus your current, will change with frequency.

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  • \$\begingroup\$ Okay, thanks, the R_load is this case would just be a helmholtz coil setup so it shouldn't be frequency dependent as you said. \$\endgroup\$ – eWizardII Dec 27 '10 at 22:16
  • \$\begingroup\$ A Helmholtz coil is definitely frequency dependent, but I wouldn't worry about it. \$\endgroup\$ – tyblu Dec 27 '10 at 22:41

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