a.
Series connection Since in series, the flow of current is constant. We need to compare power with ‘R’.
P = V × I
⇒ P = V2R
RA = 200250 = 800 Ω
RB = 2002100 = 400 Ω
Since, current is constant in P = i2 R. More the resistance more will be the power and thus, the bulb glows more brightly.
Hence, Bulb A glows more.
b.
When connected in parallel voltage will be constant
When voltage is constant then P = V2R
P ∝ 1R so when R↑ power↓ and viceversa
Bulb A: P = V2R
⇒
RA = V2P = 220250 = 968
Bulb B: P = V2R
⇒
RB = V2P = 2202100 = 484
So, RB < RA
⇒ PB > PA
⇒ Bulb B will glow brightly