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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?

Qmechanic
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2 Answers2

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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.

user1583209
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  • 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? – Perspicacious Dec 04 '16 at 06:23
  • @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 Dec 04 '16 at 20:45
  • 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). – user1583209 Dec 04 '16 at 20:48
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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, $I$.

Bottom Line

$P=I^2R$ is not generally applicable to appliances.

It works for filament style light bulbs which have reached a steady-state temperature. The largest bulb in the series will absorb the most power and look the brightest, but increasing the resistance will make all the bulbs in series dimmer.

If you have a filament bulb with a 100 $\Omega$ resistance in series with a 200 $\Omega$ bulb connected to a 120 V RMS source, then

  • 100 $\Omega$ will absorb 16 W and
  • 200 $\Omega$ will absorb 32 W:
  • the 200 $\Omega$ will be brighter.

If you replace the 200 $\Omega$ with a 300 $\Omega$ bulb, - the 100 $\Omega$ will absorb 9 W and - the 300 $\Omega$ will absorb 27 W, brighter than the 100 but dimmer than the original 200 $\Omega$.

These bulbs would be rated in the US as

  • 144 W for the 100 $\Omega$,
  • 72 W for the 200 $\Omega$ and
  • 48 W for the 300 $\Omega$,
  • so in parallel, the 100 $\Omega$ would be brightest.

Real Appliances

If the "appliance" is a capacitor or an inductor or some primarily capacitive or inductive device like a motor, or if it is an LED or fluorescent bulb, that power formula doesn't work because you can't reduce every device to a simple resistance.

Every real appliance is a mixture of resistances, capacitances, inductances, diodes, transistors, etc. Each of these devices contributes to a frequency dependent quantity known as electrical impedance which is the AC extension of the concept of resistance. In most countries, electrical supplies are not DC, so the AC behavior is important.

Using the impendance idea, there is an Ohm-like relationship telling the relationship between the voltage across and current through an appliance: $$V_x=I_x Z_x(f)$$ where $Z_x(f)$ is the frequency dependent impedence of the appliance. The impedance also shifts the phase of the voltage relative to the current so that maximum current value does not occur at the maximum voltage value. inductance):

Power Factor

So ultimately, the average power consumed over a complete AC cycle is $$<P>=\frac{|V_{rms}|^2}{|Z|}\cos\phi$$ where $\phi$ is the total phase shift between the voltage and current. So the phase shift induced by the appliance can affect the power it absorbs regardless the actual magnitude of $Z$.

Bill N
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