(a) Given V = 240 V and R = 100 ohms
Therefore,
Power (P) = V2/R = (240)2/100
= 576 W
Energy consumed by bulb A = P × t
E = 576 x 4 x 7 × 60 × 60
E = 58,060.8 kJ
(b) When bulbs A and B are connected in series:
Rnet = R1 + R2
= 100 + 100
Rnet = 200 ohms
Total power consumed by bulb A when connected in series with bulb B
Ptot = V2/Rnet = (240)2/200 = 288 W
PA' = Ptot/2 = 144 W
Power consumed by bulb A when connected without bulb B to 240 V
PA= V2/R
= (240)2/100
= 576 W
As PA' < PA, the brightness of the bulb A decreases when connected in series with bulb B.