Power in Series Combination Of Appliances

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Everything you write is correct.

You need to distinguish between the rated power of a light bulb and the actual power consumed in a specific circuit. Because of this, your last sentence (If the appliance is a bulb then in the series combination the bulb(having max power) will glow with maximum brightness?) does not contradict your calculation.

If it says 60 W on the light bulb, this is the rated power, and means it consumes 60 W only if attached to your socket's voltage (i.e. to 110 V or to 220 V depending on where you live).

In a series circuit however, on the n-th light bulb, you only have a voltage of:

$$ U_n = U_{tot} \frac{R_n}{R_{tot}} $$

Since bulbs with higher rated power have lower resistance than bulbs with lower rated power, you get that the voltage over the lower rated bulb is higher. And because of that the lower rated bulb consumes more power and will glow brighter.

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Perspicacious
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Perspicacious

Updated on May 02, 2020

Comments

  • Perspicacious
    Perspicacious over 3 years

    The general formula for power, P for an appliance is

    P= (I^2)R where I is Current flowing across a resistance R

    Thus power, P is directly proportional to resistance, R of an appliance for a given current, I.

    Now when number of appliances are connected in Series, the current flowing through each appliance will be same, I.

    So the apppliance having maximum resistance will have maximum power/wattage for which the current, I is constant.
    If the appliance is a bulb then in the series combination the bulb(having max power) will glow with maximum brightness?

    But in series bulb with lowest wattage glow with maximum brightness.

    So what's wrong in my content?

  • Perspicacious
    Perspicacious almost 7 years
    Well, is it the high resistance or high power which makes the light bulb glow with maximum brightness without considering any intrinsic property of the bulb?
  • user1583209
    user1583209 almost 7 years
    @Mritun Jay: I am not sure I understand your question. In a series circuit, the power of one bulb is $P=I^2 R$, so higher resistance means higher power. Note that this is the actual power (not rated power) and it is basically the amount of energy (per time) which is converted from electrical energy to light (and heat).
  • user1583209
    user1583209 almost 7 years
    Higher rated light bulbs are more efficient (convert a higher percentage of energy to light, vs heat). However this can not compensate for the other effect I described above (higher resistance, higher voltage at the lower rated bulb).