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A 220 V and 800 W electric kettle and three 220 V nd 100 W bulbs are connected in parallel . on connecting this combination with 220 V supply .the total current in the circuit will be

A 220 V and 800 W electric kettle and three 220 V nd 100 W bulbs are connected in parallel . on connecting this combination with 220 V supply .the total current in the circuit will be

Grade:12

3 Answers

Arun
25763 Points
3 years ago
 resistance of kettle = V^2 / P = 220 x 220 / 800 = 484 / 8  ohm
Resistance of each bulb = V^2 / P = 220 x 220 / 100 = 484 ohm
Now 484/8 , 484, 484, 484 are connected in parallel to a supply of 200 V
Effective resistance = 484/11 ohm
So current in the circuit = 200/ (484/11) = 2200/484 = 4.55 A
Raghulal Prajapati
76 Points
3 years ago
according to ur question all the resistances are in parallel to each other .
so the solution is :
 as we know    W= V2/R    ------>   R= V2/W  
so resistance for kettle is  :  220*220/800= 121/2 ohm
 resistance for (1)  bulb is :   220*220/100= 484 ohm
  as all are in parallel so net resistance is 1/resistance of kettle  +3/resistance of bul = 1/ equivalent resistance 
so equivalent resistance= 44 ohm
so current in circuit = v/R = 220/44= 5 ampere.
thank u.
 
Arun
25763 Points
3 years ago
Sorry
In my answer I have typed a wrong value 
Effective resistance = 484/11
So current in the circuit
= 220/ (484/11)
= 2420/484
= 5 amp

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