Question

Two lamps one rated 60 w at 220v and other 40w at 220 v are connected in parallel to electric supply at 220v calculate the current drawn from electric supply

Answer 1

Hi..

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

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 ...


Hope this helps u!!

Answer 2

Answer:

I_series = 0.30 A

I_parallel = 1.36 A

Explanation:

Given :

Two lamps power as 100 W and 200 W at 220 V.

We have resistance formula :

R₁ = 220 × 220 / 100 = 484 Ω

R₂ = 220 × 220 / 200 = 242 Ω

In series combination :

R_net = 484 + 242 Ω

= > 726 Ω

We have ohm's law :

V = I R

I = 220 / 726 A

I = 0.30 A

( ii ) In parallel combination :

1 / R_net = 1 / 484 + 1 / 242 Ω

R_net = 484 / 3 Ω

Given :  V = 220 V

I' = 220 / 484 × 3 A

I' = 1.36 A