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My textbook tells me that in order to calculate the power (in watts) for an electrical circuit then the equation is $P=VI$ (volts times amps equals watts). However, I have read online that this generally only applies to DC circuits. My first question is are there any exceptions to this regarding DC circuits? Are there any components that this equation would not apply to and if so, is there an alternative?

I have also read that this same equation can be used in AC circuits providing that it is a purely resistive load and not an inductive or capacitive load. I'm assuming then that things that simply produce heat or light such as a toaster, a kettle or a light bulb would count and the P=VI equation is valid here in AC providing that the RMS values for volts and amps are used. Is this correct? If so, what are some other common appliances where this is true? In regards to capacitive or inductive loads I read that you need to account for the phase difference by adding cos of the angle to the equation. How do I find that? Or are there any ways to at least get an estimate for a particular device?

Also, my textbook says that V=IR and I have read elsewhere that this is true only in some cases but I'm having a hard time finding some straight answers as to what these special cases are. To add to the confusion, my textbook goes on to combine the $P=VI$ and the $V=IR$ equations to get $P=I^2R$ and $P=V^2/R$. It then uses these equations in examples including a hairdryer plugged into a 230 volt main socket but from my reading online, I get the impression that the $V=IR$ equation does not nessesarily work for AC circuits. If true then how can it be used in combination with the power equation and then used on an AC circuit? Also, it specifically mentions the motor in the hairdryer which I thought was an inductive device and so the simple $P=VI$ equation should not hold. So what gives? Should it not need the power factor added on too? And does the power equation hold for a DC motor if not for an AC motor? What about something like a vacuum cleaner or even a blender?

I guess what my questions boil down to is where are these various equations valid and when are they not? I want to be a able to use them at home and at work but I don't want to use them when they're not valid. Any information would be greatly appreciated.

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  • $\begingroup$ For a hair dryer, $P = VI$ will be a good approximation, because most of the electrical power is used in the heating element of the dryer, which is essentially a resistor. The power used by the electric motor (which is not equivalent to a resistor) is small. To understand how to use the equations correctly for more powerful electric motors and generators, you need to study alternating current theory, i.e. what happens to alternating currents and voltages in circuits containing inductors and capacitors, and also the effects caused by the fact that motors and generators have moving parts. $\endgroup$ – alephzero Oct 8 '16 at 21:11
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The equation $P=IV$ is fine for steady currents and for a resistor $V=IR$ gives you two other variations of the equation for power $P = I^2R = \dfrac {V^2} {R}$.

For alternating currents life is a little more difficult in that the product $IV$ changes with time.

In other words the power changes with time and so it is often the case that the average power over a cycle is a better parameter to use.

For sinusoidal waveforms it is found that the average power over a cycle $<VI>= \dfrac{V_{\rm peak}I_{\rm peak}}{2}$ where $V_{\rm peak}$ and I_{\rm peak} are the peak values of the voltage and current.

To simplify things this equation is chnage to $<VI>= V_{\rm rms}I_{\rm rms}$ where "rms" stands for root mean square and these are the voltages and currents which are quoted on appliances.

$V_{\rm rms}= \dfrac{V_{\rm peak}}{\sqrt 2}$ and $I_{\rm rms}= \dfrac{I_{\rm peak}}{\sqrt 2}$

So your $3$ kW kettle which works off the $230$ volts mains supply and has a current of $13$ amps passing through it uses an average power of $3000$ W and is supplied with an rms voltage of $230$ volts and the rms current is $13$ amps.

Note that the mains voltage actually fluctuates between $+\sqrt{2} \times 230 = +325$ V and $-325$ V.

With inductors and capacitors in the circuit there are further complications because the current and voltage are not in phase and in fact there are three types of poer one has to deal with - active power, reactive power and apparent power.
You will no doubt learn about these when you cover ac theory.

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To be precise, the power delivered to a circuit element is, assuming the passive sign convention, the product of the voltage across and current through. That is, when the power is positive, power is delivered to the circuit element and when the power is negative, power is supplied by the circuit element.

So, in general, the instantaneous power delivered to a circuit element is

$$p(t) = v(t) \cdot i(t)$$

For a (ideal) resistor, the voltage across is proportional to the current through, $v_R(t) = R \cdot i_R(t)$ and thus

$$p_R(t) = i^2_R(t) \cdot R = \frac{v^2_R(t)}{R}$$

However, it is important to note that this only applies to resistors (or ohmic devices).

For AC circuits in AC steady state, we can generalize the resistive power equation by moving to the phasor domain and where we can define impedance as the ratio of the phasor voltage across and phasor current through:

$$Z = \frac{\tilde V_Z}{\tilde I_Z}$$

In the phasor domain, voltages and currents are represented by complex numbers that encode the amplitude and phase of sinusoids. So, it isn't surprising that, in this domain, power is complex (but not a phasor)

$$S = \tilde V \cdot \tilde I^*$$

where the star denotes complex conjugation. The real part of the complex power $S$ is the time average power delivered while the imaginary part is the peak reactive power (power with zero time average).

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  • $\begingroup$ The average power derived here with complex phasors is, indeed, important in practical AC circuits. $\endgroup$ – freecharly Oct 9 '16 at 0:18
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The instantaneous power entering an electrical circuit is always given by $P(t)= I(t)\cdot V(t),$ where $P,~ I$ are given at a time $t.$ This holds for dc and ac circuits. So it it not restricted only to a resistance. It equally holds for circuits with inductances, capacitors and even motors. However, $P=I\cdot V=R\cdot I^2 = V^2/R$ hold only for a resistor when Ohm's law is applicable.

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  • $\begingroup$ I do not see any factual error in this answer and it seems relevant so the down vote, in my judgment, reflects poorly on the downvoter. $\endgroup$ – Alfred Centauri Oct 8 '16 at 22:07

protected by Community Oct 8 '16 at 17:12

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