Dear studentfor bulb1P=V2/R60=2202/RR=806 6net resistance=bulb1+bulb 2=806 6×2 (identical)=1613 333Pnet=Vnet×Inet=220×2201613 33=30 W

Sum no 9

D e a r s t u d e n t f o r b u l b 1 P = V^{2} / R 60 = 220^{2} / R R = 806.6 n e t r e s i s \tan c e = b u l b 1 + b u l b 2 = 806.6 \times 2 (i d e n t i c a l) = 1613.333 P_{n e t} = V_{n e t} \times I_{n e t} = 220 \times \frac{220}{1613.33} = 30 W