Five LED bulbs of 8 W each, two fans of 40 W each, electric motor 3kW and a geyser of 5 kW are used for 8 hours per day by a family. Calculate the total energy consumed in 30 days and the cost of electricity used in one day at the rate of Rs.5 per unit.

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Solution

Step 1: Given Data
Power of a LED, $P_{L}$ = $8 W = \frac{8}{1000} k W$
Power of a LED, $P_{L}$ = $5 \times 8 W = 40 W = \frac{40}{1000} k W$
Power of a fan, $P_{f}$ = $40 W = \frac{40}{1000} W$
Power of two fans, $P_{f}$ = $2 \times 40 W = 80 W = \frac{80}{1000} k W$
Power of a motor, $P_{m}$ = $3 k W$
Power of a geyser, $P_{g}$ = $5 k W$
Step 2: Calculation
Formula used,
$W o r k d o n e = P o w e r (P) \times T i m e (T)$ ,
Total energy consumed per hour $P_{h}$ = Sum of power of all devices = $\frac{40}{1000} + \frac{80}{1000} + 3 + 5 k W = \frac{8120}{1000} k W$ .
No of hours per day = $8 h$
Total time period (T) = $H o u r s p e r d a y \times t o t a l d a y s$ = $8 h \times 30 d a y s = 240 h$
Time period (T) = $120 H o u r s$
We need to find the energy consumed in one month.
Calculation of energy:
$W o r k d o n e = \frac{8120}{1000} k W \times 240 h = 1948.8 K J$
Calculation of cost :
Cost for 1 kWh = 5
Cost for 300 kWh = $1948.8 k W h \times 5 = 9744 r u p e e s$
Hence, the total cost is Rs.1500 and the energy consumed is 9744 KJ.

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