Appliances in Different AC Mains

1a. It depends upon the voltage ratings of each bulb.

1b. The mixer will run slower - if it starts at all. If it doesn't start, it will behave like a convection heater instead.

2. It depends upon the voltage ratings of each bulb.

consider the bulbs have the same voltage ratings too - 230 V. The point was comparing identical bulbs in different voltages. My question is why would the 230 V, 60 W bulb glow just as bright in 110 V?

Given that its resistance is constant in both voltage connections, wouldn't it draw more current in 110 V and turn out to give the same energy dissipated as light? Or is that the ENERGY dissipated will be the same, but the energy dissipated as LIGHT will be different? (Identical bulbs in terms of material, structure, wattage and voltage).

A 60W 230V bulb is about 882Ω and draws about 0.26A. If you feed this bulb with 110V, it will draw about 0.125A, produce about 13.7W, and be very dim.

Since the resistance is the same, the bulb draws less current with 110V than with 230V; as in previous paragraph.

When a questions like this come into your mind you need to find the answers.The real answers.

The wattage of bulb helps us to find current in it hence the resistance for constant voltage.

eg. 120W bulb across 240V supply means current = 120/240 = 0.5A

so the filament resistance = 240/0.5 = 480 ohms

similarly for a 60W bulb across 240V,current = 0.25A

so resistance = 960 ohms

The brightness purely depends on efficiency of energy conversion of filaments else i would have designed a bulb which glows for life long.

Motor torque is proportional to voltage applied.I remember a dialogue said by an Engineer here "There is no free lunch in this world"

P (power) = V (volts) * I (amps) in resistive circuits such as light bulbs.

Also V = I(current) * R (resistance) only in the case of a light bulb R is fixed. So if you apply twice the voltage, then the current will also attempt to double. The little filament, however, will become too hot and vaporize if you apply a voltage much higher than its rating.

So far, I have avoided the square of some factor. But if you substitute the second equation into the first you will have two relationships. First P = V*V/R and subsequently P = I*I*R. Unfortunately, the equations must be matched up to the case or description in order to apply correctly.

Considering your newfound relationships, which bulb has the higher resistance? These questions are designed to make you think. When you apply the equations correctly, you will see that Ohms law and Kirchhoff's laws are satisfied. Those rules and the above relationships are fundamental to any electrical problem. Incidentally, these relationships are repeated (in their own form) in other fields such as pneumatics and hydraulics.

You don't have to know the exact resistance to solve this problem. But you can approximate it based on knowing the rated power and applied voltage. Keep in mind that a coil changes resistance with the voltage applied to it but when comparing two similar devices, the relationship between the two remains approximately constant.

"a coil changes resistance with the voltage applied" its the frequency.

In single phase Power=Voltage x Current x P.F

here lets assume the filament is purely resistive.

In this discussion, a coil is not the same as an inductor. Filaments are wound around a mandrel which is then dissolved away in an acid bath. The resultant filament is a long piece of tungsten which has been shortened by shaping it like a spring or coil. It has insignificant inductance so the power factor is always very close to 1.0.

Consequently, you missed the point. Power is still equal to the applied voltage times the current through it. But if you measure the voltage and current using a variety of voltages and then calculate the resistance, you will find that you have a moving target. And, the resistance value change is not linear.

As it turns out, it is very difficult to obtain resistance values on some resistance coils or filaments without changing the value while reading it. You quickly find out how difficult it is when you try reading it with different instruments. It makes a great lab experiment for engineering students to study. Maybe I'll start a discussion on it!

Replying to Part 2.

Doing the math, the current in the series condition is about 0.34 amps.

(V = i * R1 + i * R2); R1 for the 60W lamp is 202 Ohms, R2 for the 100W lamp is 121 Ohms, V = 110 Volts.

Since Power goes as the square of the current times the resistance there will be more power dissipated through the 60W lamp (68.7 W) in the series condition as compared to the 100W lamp (41.1W).

Error. Power=IxIxR (.34x.34x202) =23.34W at 68.7V and 11.67W at 41.3V Hence 60Watt bulb is in its 38.9% brightness and 100Watt in its 11.67% brightness.

In the last paragraph, the 60W bulb has about 68.8V across it, and the 100w bulb has about 41.2V. The respective wattages are about 23W and 14W.

Thanks to hithuanand and Tornado for catching my math error in my 2nd post.

But from where do you get this figure that the resistance of a 60 W lamp is 202 ohms, while that of a 100 W lamp is 121 ohms? Is that a fact from manfacturing standards or is there a definite (mathematical) relation between wattage and resistance?

I could have a 60 W bulb with less resistance than a 100 W too, can't I?

60 W ÷ 110 V = 6/11 A; 110 V ÷ (6/11) A = 1210/6 Ω ≈ 202 Ω.

100 W ÷ 110 V = 10/11 A; 110 V ÷ (10/11) A = 1210/10 Ω = 121 Ω.

W = EI (By definition, watts = volts x amps.)
E = IR (Ohm's Law; volts = amps x resistance.)
For two resistances in series, R_total = R₁ + R₂.
For two resistances in parallel, 1/R_total = 1/R₁ + 1/R₂. [Or R_total = R₁R₂/(R₁ + R₂).]

These equations can solve all of the situations raised in the posted topic.

I read all Reply, but i want to know What is the design factor of a bulb , i mean bulb is designed according to wattage or voltage?

Actually everybody is wrong to some extent because incandescent bulbs or coils are designed to obtain a targeted blackbody temperature, thereby producing the color of light desired. With proper design, the coil resistance will increase with temperature from some cold resistance to a point around 5 or 6 times its cold resistance to achieve the desired color.

At that elevated temperature, there will be a tendency to evaporate the coil so it must be quite robust and stand up to rapid expansion and some vibration as well. The length of the coil may be all that is required to change the wattage at a given voltage. But other things change with it such as the rate of material loss due to evaporation. You will find that two 60 Watt bulbs from different manufacturers may be quite different. And measuring the coil resistance of anything other than a cold coil requires knowledge of applied voltage and current.

Lastly, this type of lighting is nearly extinct but the exercise is still kind of interesting because there is a lot more to consider if you want repeatable results.

Typically voltage is the fixed design parameter and wattage is the targeted result of the assembly based on how long the coil or wire is between clamps. It is difficult to get exact results because the tension in the wire or coil is not much of a consideration when it is clamped during assembly.