Question

two electric bulbs rated 100W;220V and 60W;220V are connected in parallel to an electric mains of 220V. Find the current drawn by the bulbs from the mains

Answer 1

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Power = P = 100 W

Voltage = V = 220v

Resistance = R

P = V²/R

100 = 220 ×220/R

R  = 220 × 220/100

     = 484 Ω

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Power = P = 60 W

Voltage = V = 220v

Resistance = R

P = V²/R

60 = 220 × 220/R

R = 220× 220/60

    = 806.7 Ω

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As the resistors are connected in parallel , 

total resistance = 1/R

1/R = 1/484 + 1/806.7

      =  806.7 + 484/484×806.7

      = 1290.7/390442.8

R = 390442.8/ 1290.7 = 302.5 ohms

Total resistance = 302.5 Ω

I = current

V = IR

220 = I × 302.5

I = 220/302.5

   = 0.73 A

The current drawn is 0.73 A

Answer 2

Given,

In case of Bulb 1

Potential difference,V1= 220v

Power,P1 = 100w

In case of Bulb 2

Potential Difffernce,D2 =220v

Power,P2 = 60w

Now, since they are connected in parallel combination hence they have same potential difffernce and the total Current in the circuit is given the sum of currents flowing in both the resistors.

Power = Potential difference×Current

=>P = V×I

=>I= P/V.

Now,In case of Bulb 1

I1 = 100/220

=> I1 = 5/11 watts ------------(i)

Now,in case if bulb 2

I2 = 60/220

=>I2= 3/11watts ---------------(ii)

From (i) and (ii) we get=>

Total Current in circuit,(I)n=>

(I)1 + (I)2

=>(3/11)+(5/11)

=>8/11 watts.

Hence, the current in the circuit is 8/11 watts or 0.73 watts.

Remember

1)Ohm's law of resistance

V=I×R

2) Power = Work/Time

or

Power = V×I = I^2×R