Three lamps $100 W, 60 W$ and $40 W$ respectively are connected in parallel to a power supply of $220 V$ . Calculate (a) total energy consumed (b) total resistance of arrangement (c) cost of keeping them lighted for 6 hours daily for 30 days, if cost of electricity per unit is ₹ 2.80.

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Solution

Step 1: Given data.
Three lamps $100 W, 60 W$ , and $40 W$ respectively are connected in parallel to a power supply of $220 V$ .
Step 2: calculations.
(a) Three lamps in Parallel Hence total consumption
$= 100 + 60 + 40$
$= 200 W$
Energy consumed in 30 days 6 hrs $= 200 \times 6 \times 30 = 36000 Wh$ $= 36 kwh$
$1 kwh = 1$ unit
$= 36$ units
cost of electricity per unit is $R s . 2.80$ .
$\Rightarrow Total Cost = 36 \times 2.8 = Rs 100.8$
(b) Power $= V^{2} / R_{1}$
$100 = 220^{2} / R_{1} \Rightarrow R_{1} = 484 Ω$
$60 = 220^{2} / R_{2} \Rightarrow R_{2} = 2420 / 3 Ω$
$40 = 220^{2} / R_{3} \Rightarrow R_{3} = 1210 Ω$
(c)Total resistance $= R$
$\Rightarrow \frac{1}{R} = \frac{1}{484} + \frac{2420}{3} + \frac{1}{1210}$
$\Rightarrow \frac{1}{R} = \frac{5 + 3 + 2}{2420}$
$\Rightarrow \frac{1}{R} = \frac{10}{2420}$
$\Rightarrow R = 242$
Total resistance $= 242 Ω$

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