Question

Two electrical bulbs marked 40w,220v and 60w,220V, when conneted in series across same voltage supply of 220V,the effective power is p1 and when connected in parallel the effective power is p2. Then (p1/p2) is

Answer 1

First of all find resistance of each bulb
Resistance of {40W, 220V} :
We know, Power , P = V²/R
R₁ = (220)²/40 Ω

Similarly resistance of {60W, 220V} :
R₂ = (220)²/60 Ω

Now, if bulbs are in series with 220V supply :
Then, Req = R₁ + R₂ = (220)²/40 + (220)²/60
= (220)²/24Ω
Then, Power , P₁= V²/Req
= (220)²/{(220)²/24} = 24W

And if bulbs are in parallel with 220V supply :
Then, Req = R₁R₂/(R₁ + R₂) = {(220)²/40}{(220)²/60}/{(220)²/40 + (220)²/60}
= (220)²/100
Now, power, P₂= V²/Req = (220)²/{(220)²/100} = 100W

So, P₁/P₂ = 24/100 = 6/25

Answer 2

Given two bulbs:
First bulb case:
P=40W
V=220V
R1=?
P=V²/R
R1=V²/P
=(220)²/40 ohms

Second bulb :
P=60W
V=220V
R2=V²/P
=(220)²/60

When resistance are connected in series :
R=R1+R2
=(220)²/40 +(220)²/60
=(220)²[1/40+1/60]
=(220)²[3+2]/120
=(220)²5/120
=(220)²/24 ohms

Power when resistance are connected in series:
P1=V²/R
=(220)²/(220)²/24
=24W

Effective resistance when resistance are connected in series:
1/R=1/R1+1/R2
=40/(220)²+60/(220)²
=(1/[220]²)[40+60]
=[1/(220)²]x100
R=(220)²/100
Power in parallel :
P2=V²/R
=[(220)²/(220)²]x100
=100W
∴P1/P2=24/100=12/50=6/25