Question

A right joules law of heating be two lamps one rated 100 w 220 v and other 60 w 220 v are connected in parallel to electric main supply wire the current drawn by two bulb for the line if the supply voltage is 220 v

Answer 1

Hi there.....

Answer.....

Joule's law of heating = I^2RT

.............

Case of 1st bulb

Now , Power = V^2 / R

Power = 100 W

Potential difference = 220 v

P = V^2 / R

Resistance = V^2 / P

= 220 × 220 / 100

= 484 ohms

Case of 2nd bulb......

P.D = 220 v

Power = 60W

Resistance = 220 × 220 / 60

= 806.67 ohms

Resistance in parallel = 1 / R1 + 1 / R2

1 / Rp = 1 / 484 + 1 / 806.67

= 806.67 + 484 / 484 × 806.67

= 1290.67 / 390428.28

Rp = 390428.28 / 1290.67

Rp = 302.5 ohms

***********************

Now , according to ohms law

IR = V

V / R = I

...............

Resistance = 302.5 ohms

P.D = 220 v

I = 220 / 302 .5

= 0.727 A or 0.73 A

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I hope this helps u.. .

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Answer 2

Hello friend___❤

Here is your answer ✒

For the 1st lamp:

Power P1 = 100 W and

Potential difference V = 220 V

Therefore,

I1= P1/ V = 100/220 = 0.455 A

For the 2nd lamp:

Power P2 = 60 W and

Potential difference V = 220 V

Therefore,

I2= P2/ V =60/ 220 = 0.273 A

So, the net current drawn from the supply is given by

= I1 + I2 = 0.455 + 0.273 = 0.728

I hope, this will help you___❤❤

Thank you

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