It appears that either all the people over the internet have either not thought about this question or that I am mistaken in my thought somewhere:

Confusion appears when I try to apply ohms law to the power supply coming to our homes. Taking it as simply as it can be, why and how a thicker/longer resistance element of a 100W bulb draws more current than a thinner/shorter element of a say 60 W or 0W bulb when ohms law suggests something else, that current through smaller resistance would be greater. I know that the system tries to keep its voltage constant and hence has to feed more power into the load as it demands but Im not getting the result which I should on paper.

Now some or most people would like to suggest that a 100 W light bulb has a smaller resistance than a 60 W. Please think again....have you seen a heating element room heater, one which burns a hole in your pocket? or better try to visualise putting many 100 W bulbs in series...resistance increses with each light bulb introduced in circuit as well as the current!how does a load draw power from the system

  • 1
    \$\begingroup\$ If a conductor is thicker, its resistance is lower. Bulbs are put in parallel, not series. \$\endgroup\$ Nov 10, 2014 at 6:07
  • \$\begingroup\$ Incandescent light bulbs are not resistors -- their resistance is not constant, and varies substantially with temperature. And the given watts rating is only valid when connected at full voltage (i.e. parallel, not series). \$\endgroup\$
    – MarkU
    Nov 10, 2014 at 9:10

3 Answers 3


Somehow you are confused, but the principle of ohms law is very simple and still valid.

The resistance of a thinner wire is larger than that of a thick wire, so the current trough the 100W bulb will be larger than the current trough the 60W bulb.

Furthermore the (tungsten) wires of a light bulb are coiled, to have them evaporate at a slower rate. This can be seen in this picture from wikipedia:

enter image description here

Therefore it's hard to say something about the length. The lenght might appear shorter, but can still have more turns in the coil.

  • \$\begingroup\$ Ok that is right about the resistance of a thicker conductor being lesser. But thats not the point Im confused at. Consider any resistance element in the circuit. If more of such resistances are added in series then the current would increase or decrease? \$\endgroup\$ Nov 10, 2014 at 8:27
  • \$\begingroup\$ the energy meters in our houses are driven by the current flowing in our phase as per my knowledge. Greater the current greater is our consumption and the bill. Now if I connect one bulb one day, and on another day I put many bulbs(say in series) in the circuit, then which day my consumption would be greater? \$\endgroup\$ Nov 10, 2014 at 8:40
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    \$\begingroup\$ If you connect the bulbs in series, they will not use the number of watts that they specifiy. Because the voltage is not the same. Say you connect two bulbs in series, they will both use half the voltage and thus emit less light. If you use two bulbs in series you will use less power than with a single bulb (assumed the resistance stays the smae for different voltages), because P = U^2/R. For a single bulb you get P=U^2/R and for 2 in series you get P=U^2/(2R). \$\endgroup\$
    – Douwe66
    Nov 10, 2014 at 9:04

You seem to be confused about parallel and series connections. All appliances(bulbs, heaters, fans etc) in a power system like your house are connected in parallel and never in series.

When you connect in series, the total resistance increases and thus the current decreases as you correctly stated. If you were to connect say four 100w bulbs in series, they would barely light up.


Let us keep it simple.

Ohms law states that U = R x I [1]. We may freely move the parts of the equation around. This might give us I = U / R [2]. So, if we know that U is fixed, if we decrease R then I will increase.

There is as well a power equation: P = U x I [3].

In the equation [3] we may substitut I with using the formula I = U / R. So we get: P = U x U / R [4].

P is somehow related to how bright the lamp is, higher P means a brighter lamp. So in order to get a 100 W light bulb compared to a 60W, we need to make R smaller. We might rearrange equation [4] as R = U x U / P

Lets do some checks. To make numbers easier, assume that U is fixed at 100V.

Now let us find R for a 100W lamp. R = 100 V x 100 V / 100 W --> R = 100 Ohm.

What if P = 60 W? We get R = 100V x 100V / 60 W --> R around 167 Ohms.

So, it is true that a higher W lamp has a lower Ohm!

Lamps in a house are connected in parallell. This is a different way of saying that all lamps see the same U.

Now, we might complicate things by noting that a cold lamp has a different R from a hot one. And to further complicate, it generelly is alternating current AC. Regardless, the relative R-s still hold.


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