simulate this circuit – Schematic created using CircuitLab

So the problem I've been given is to determine whether the power consumed by a string of lights increases or decreases as more lights are added in parallel (and a similar problem concerning increasing number of bulbs in series). The problem states that the source is a standard wall electrical outlet (I'll assume it's like the ones in the US since that's where I'm at).

Now, I don't know a whole lot about electrical standards anywhere in the world, so I'm not sure what exactly comes out of a wall, in terms of electricity and current. My understanding is that the current alternates (which doesn't really make sense to me because my light bulbs aren't flickering on and off like a sine wave). If resistance remains constant, that means the voltage is alternating too, since \$V = I \times R\$, so the value of \$V\$ is fluctuating proportionally with \$I\$

I was under the impression that, at least in the US, wall outlets had something like 120V and the current actually drawn is variable depending on your load. I thought that since the current is alternating, then as discussed above, the voltage must be alternating too, and maybe the 120V refers to the amplitude of the alternating voltage. Please correct me if I'm wrong, as it seems crucial to understanding the problem.

So if you add light bulbs (resistance) in parallel, the overall resistance of the load decreases. If your max voltage remains constant, then more current would be drawn, no? Similarly for bulbs in series, the total resistance increases as bulbs are added, so the drawn current would be reduced. So for parallel bulbs, your power would increases, since \$P = I^2 \times R \$ and the \$I\$ term is increasing faster than the \$R\$ term is decreasing. Likewise for series, the power seems like it would decreases, since the \$I\$ term decreases faster than the \$R\$ is increasing.

How far off am I?

  • \$\begingroup\$ Your eyes cannot dicern flicker below about 18 Hz the lights will flicker at 120 Hz so you cannot see this unless it is a Neon light and you move it in front of you without moving your eyes (in a dark room) and can see the dashed line. With a filament lamp the light does not (usually) go off due to thermal mass of the filament keeping it hot. Remember that in series the bulbs will not get as hot so the resistance will be less than a bright bulb, the equation will not be linear but more bulbs in series will result in less current. \$\endgroup\$
    – KalleMP
    Feb 14, 2016 at 23:21

4 Answers 4


You're right in thinking that current and voltage varies according to time in an alternating current circuit. However, when analysing circuits with alternating current and simple resistive loads, it is much easier to work with average power instead of instantaneous power. You can use RMS voltages and RMS currents to simplify your analysis.

The 120V voltage rating you get out of a wall outlet is given in RMS. You can multiply this number by \$\sqrt2\$ to obtain the amplitude of the sine wave.

Your circuit analysis does seem right on the most part. Assuming your input voltage remains constant, adding resistances in parallel decreases overall resistance and implies increased power. Adding resistances in series increases overall resistances and implies decreased overall power.

If you rearrange for the voltage, you can see how the resistance \$R\$ of the circuit affects the power consumption of the circuit.

\$P = \frac{V_{rms}^2}{R} = \frac{120^2}{R} = \frac{1.44\times10^4}{R}W\$

Lastly, there's a few reasons why you might not observe the lightbulb to flicker when current changes direction. One of which is that a light bulb doesn't dim instantly when power goes to zero (try switching off the lamp and you'll see the filament fades away slowly) so it still produces light as the instantaneous power momentarily crosses over zero. Furthermore, 50/60Hz is generally fast enough that the flickering is indiscernible by humans.


Everything pretty much spot on except, maybe, the last paragraph.

"The I term is increasing faster than the R term is decreasing"

I will increase with the inverse of R. You've got the idea though. A couple of calculations should clarify the rest.

Regarding alternating current: when you spin the magnetic rotor in the coil of a generator the sign of the voltage (and hence the direction of the current) reverses every time the N and S poles of the magnet swap over. This is how grid power is generated and also the alternator on a car engine.


Your conclusions are correct, but you might be missing a fundamental aspect of a domestic power supply: it is constant voltage, not constant current.

If resistance remains constant, that means the voltage is alternating too, since V=I×R, so the value of V is fluctuating proportionally with I.

Technically it's the other way round - I fluctuates [...] with V.

This misunderstanding stems from the terms "AC" and "DC", which suggest the current is the primary yardstick by which we quantify electricity. It isn't; voltage is. The voltage alternates, therefore the current alternates through a connected load (resistance). With no load the current is zero but the voltage at the terminals of the wall socket is always there, oscillating away - unless you turn it off at the fusebox or distribution board.

With domestic electricity it makes life much simpler to ignore the concept of an alternating voltage and thus we refer to the Root Mean Square value of the voltage when discussing power. RMS is, in short, the average value of the voltage. This means that the Live/Hot conductor swings between +172V and -172V, give or take some tolerance.


I'll also add that you dont notice your light bulbs flickering because the AC voltage is changing polarity 100 times a second (i.e. 50 Hz, for my country anyways); your eyes can't track this because of the effect called persistense of vision.

Also, there are 3 wires in a standard outlet: Live, Neutral and GND. The voltage on the LIVE cable is what creates the sine waveform we call AC, while voltage on the NEUTRAL cable may be seen as the horizontal axis about which the sine wave oscillates. The GND cable provides an low-impedance path to the earth for stray current, in the event that perhaps your LIVE cable gets loose within your equipment and makes contact with the equipment's metal chassis, thereby making the scary AC voltage appear on the chassis as well. Thus if you touch the chassis, your GND cable provides an alternate and, more importantly, really-low-resistance path to earth for current, compared to the relatively high resistance of your body; kind of like a 2-branch parallel circuit with way more current passing through one branch (GND cable) than the other (your body).

  • \$\begingroup\$ Your first paragraph should probably be a comment since it doesn't really answer the question. The second paragraph doesn't seem relevant to the question. \$\endgroup\$
    – tangrs
    Feb 14, 2016 at 23:38
  • \$\begingroup\$ @tangrs I dont know...it seemed like he wasnt entirely sure of the whole AC concept; it doesnt hurt to clarify things. He can always ignore it... \$\endgroup\$ Feb 14, 2016 at 23:40
  • \$\begingroup\$ All I really understand about AC is that it tends to be either voltage or current delivered with its instantaneous measurement being a sinuisoidal. The actual details about wall outlet design are a mystery to me! \$\endgroup\$
    – L P
    Feb 15, 2016 at 0:52
  • \$\begingroup\$ @LP I feel you...any particular questions about my answer? \$\endgroup\$ Feb 15, 2016 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.