Question

Solve the following example.

Two tungsten bulbs of wattage 100 W and 60 W power work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor? 

Answer 1

 Power of first bulb, P₁ = 100 W
Power of second bulb, P₂ = 60 W
Now,
Resistance of first bulb,  `R_1 = (V^2)/(P_1)=(220xx220)/100 = 484 Ω`

Resistance of second bulb, \[R_2 = \frac{V^2}{P_2} = \frac{220 \times 220}{60} = 806 . 7 \Omega\]

When the bulbs are connected in parallel, their equivalent  resistance is

\[R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{484 \times 806 . 7}{484 + 806 . 7} = 302 . 5 \Omega\]

\[I = \frac{V}{R_{eq}} = \frac{220}{302 . 5} = 0 . 72 A\]

Answer 2

Potential remains same in the resistors when they are connected in the parallel. 
∴ Potential in both the bulbs will be same.  

Now, 
For First Tungsten Bulb, 
Power = 100 W.
Potential = 220 V.

∵ Power = Potential × Current. 
∴ Current = 100/220 
⇒ Current = 0.45 A.

For second Tungsten bulbs, 
Power = 60 W.
Potential = 220 V.

∴ Current = 60/220
   = 0.27 A.

Now, Total current flowing = Current in 1st bulb + Current in 2nd bulb.
     = 0.45 + 0.27
     =  0.72 A.
Hence, the total current flowing in the main conductor is 0.72 A.