Question

When a 60w,220v) bulband af 100w 220v bulb are connected in series, then which



bulb will glow brighter?

Answer 1

Let resistances of the given bulbs be r1 and r2 resp.
A/q
R1=220 ×220/60 (since, v^2/R = p)
=220×22/6 ohm
R2=220×220/100
=22×22 ohm
Eq. Resistance(in series) =(22)^2 {5/3 + 1}
=484×8/3=3872/3=1290.66 ohm
Total voltage (in series voltages add up) =440v
Total current = 440×3/3872 =40×3/352=30/88
=15/44ampere
Now, in series current remains constant so
Power dissipated in resistor1=(Ir1^2) 15/44×15/44×220×22/6=5/2×15/2×10/2=75×5/4=375/4 =93.75 watts
Power in resistor 2 = 15/44×15/44×22×22
=225/4=56.25 watt
So power dissipated in resistor 1, i.e. First bulb is more.
So we can also alternatively say that as resistance of first bulb is more... And when connected in series... The resistances add up,and the current remains same. So more resistance will give more power. Which is jst the opposite case when the bulbs are connected in parallel. Anyways hope this helps