Two lamps, one rated 100w at 220v and the other 200w at 220v are connected in parallel to electric supply of 220v. find the current drawn.

We have given two lamps in such a way :

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

= 0.1 KWh

Step by step:

We have given two lamps in such a way :

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

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explanation: