Two bulbs of 100w and 40w respectively connected in series?

Two bulbs of 100w and 40w respectively connected in series across a 230v supply which
bulb will glow bright and why?

Because 40w bulb draw more current than 100w bulb,so in series arrangement, the luminicity of bulb increase with their load current but in parallel, the bulbs has the same current but different voltage so their luminosity increase with the voltage across them. Their voltages when in parallel still depends on their resistance

You got great confusion actually in series current remains same for both the bulbs,& likely intensity is same for both if rated same.

In parallel connection voltage drop is same

40W bulb will glow efficiently. Because, Current in series connection is always same and voltage is divided. here, resistance of 40W bulb is higher than 100W bulb.

so current required to glow a 40W bulb is less than 100W bulb(it is quite appeared from ratings).so current flowing from the cicuit is according to the eq.resistance of both bulb and that would not exceed that current,which can blow off 40W bulb and less than current which fully glow the 100W bulb.remember formula I=V\R. (note that bulb is purely resistive load)

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Hey here is your answer-

There’s this formula of power versus voltage and resistance:

P = V^2/R

For the 100W bulb:

100 = 230^2/R

100 = 52900/R

R = 529 ohms

For the 40W bulb:

40 = 230^2/R

40 = 52900/R

R = 1322.5 ohms

The total current produced by the resistor in series:

I = V/R

I = 230/(529+1322.5)

I = 0.124 Amps

Now we will calculate the power in each bulb by using voltage divider.

100W bulb:

V = V(100W) / (V(100W) + V(40W)) * 230

V = 529/(1322.5+529) * 230 = 65.71 Volts

P = V * I = 65.71 *0.124 = 8.14 Watts

40W bulb:

V = V(40W) / (V(100W) + V(40W)) * 230

V = 1322.5/(1322.5+529) * 230 = 164.28 Volts

P = V * I = 164.28 *0.124 = 20.37 Watts

Therefore, the brigher bulb will be the 40W bulb.