Question

Find the value of x
1/484 + 1/1210 = x

Answer 1

Answer:

Answer

Power of 1st lamp, P₁ = 40W , voltage of 1st lamp, V₁ = 220V

power of 2nd lamp , P₂ = 60W , voltage of 2nd lamp ,V₂ = 220V

we know, one things ,

Power = V²/R [ when potential difference is same then consider P = V²/R ]

R = V²/P

so, resistance of 1st lamp , R₁ = V₁²/P₁ = (220)²/60 = 4840/6 = 2420/3Ω

resistance of 2nd lamp , R₂ = V₂²/P₂ = (220)²/40 = 48400/40 = 1210Ω

Now, question said ,

(a) Both the lamps are in parallel ,

So, 1/Req = 1/R₁ + 1/R₂

1/Req = 3/2420 + 1/1210 = 5/2420 = 1/484

Req = 484Ω

Now, Current drawn from electrical supply ,i = potential difference/Req

= 220V/484 = 20/44 = 5/11 A

(b) energy consumed by lamps in one hour = Energy consumed by 1st lamps in one hour + energy consumed by 2nd lamps in one hour

= P₁ × 1hour + P₂ × 1 hour

=(P₁ + P₂) × 1 hour

= (60W + 40W) × 1 hour

= 100 Wh

= 100/1000 KWh [ ∵ 1 KWh = 10²Wh ]

= 0.1 KWh

Answer 2

Answer:

can you send the question one more time