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I'm studying cert 3 in Electrotechnology as an electrical apprentice. My tafe lecturer is insisting that in an ac circuit, and increase in voltage will cause a decrease in current. I'm aware that it depends which voltage you talk about. Ie, supply or volt drop, but his words were that in dc circuits, voltage and current are directly proportional, but in ac circuits they are not. I don't agree with this, could anyone help me understand, maybe with some formulae or something, I'm a bit stuck.. Thanks

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  • \$\begingroup\$ There isn't enough context to formulate a concrete answer, just lots of "if"s and "but"s. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 15 '14 at 5:49
  • \$\begingroup\$ I'm just stuck on him telling me that I'm ac circuits, voltage and current have an inverse relationship. If voltage goes up, current goes down... But obviously if supply voltage goes up, current also has to \$\endgroup\$ – Daniel Smith Dec 15 '14 at 5:53
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    \$\begingroup\$ He has transformers on the brain. (Real) impedances still follow the normal transresistance scheme. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 15 '14 at 5:54
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    \$\begingroup\$ Ignoring losses, a transformer is a constant-power transducer. Any increase in voltage on the secondary results in a proportional decrease in current, and vice versa. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 15 '14 at 5:57
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    \$\begingroup\$ If you write out the transformer equation, and don't realize that changing one variable may cause another to change, then you can convince yourself that increased voltage causes smaller current. But it is sophistry. Someone posted a question about it a few weeks ago. As far as I know, the only time larger voltages cause smaller currents is when the load has negative resistance. For example, switch mode power supplies. But this applies to both AC and DC. \$\endgroup\$ – mkeith Dec 15 '14 at 8:47
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Your instructor may be talking in terms of constant power. P = I*E, so for constant P an increase in the voltage E would force a decrease in the current I. This is probably the case if the lesson involves transformers.

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Think about an inductor or capacitor. For a capacitor, the current is: -

I = \$C\dfrac{dV}{dt}\$

A simple example of this is that if voltage is rising linearly, then current is constant.

This isn't quite the case that an increase in voltage results in a decrease in current but when you combine inductors and capacitors this will certainly happen.

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In a DC circuit, Ohm's law states that:

$$\ I=\frac{E}{R} $$

so it's easy to see that - with R constant - as E grows, so does I.

In an AC circuit we have:

$$\ I=\frac{E}{Z} $$

Where

$$\ Z=\sqrt{R^2+(Xl-Xc)^2}, $$

$$\ Xl =2\pi f L, $$

and

$$\ Xc=\frac{1}{2\pi f C} $$

"Z", the putative AC resistance of a circuit, is - as shown - dependent on the resistance in the circuit as well as on the inductive (Xl) and capacitive (Xc) reactances in the circuit.

The reactances of ideal inductors and capacitors will change with frequency, but not with voltage and,assuming a non-parametric resistance in the circuit, the result will be that at any given frequency, Z will be fixed and, as in the DC case, as E increases so will I.

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  • \$\begingroup\$ Thank you all, I'm aware of these principles, and it's great to hear that what I've learned backs itself up, and now I can prove it mysekf :) thanks \$\endgroup\$ – Daniel Smith Dec 15 '14 at 12:28

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