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Electric Current> A 220 V and 800 W electric kettle and thr...

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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

raghav , 7 Years ago

Grade 12

3 Answers

Arun

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

Last Activity: 7 Years ago

Raghulal Prajapati

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.

Last Activity: 7 Years ago

Arun

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

Last Activity: 7 Years ago

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