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Given that 1 unit of electricity costs 80 frs. Determine the cost of using 5 living lamps, each rated 120 W, switched on for 4 hours each day throughout the month of June.

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Answer

5760 frs5760 \, \text{frs}

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Problem: Given that 1 unit of electricity costs 80 frs. Determine the cost of using 5 living lamps, each rated 120 W, switched on for 4 hours each day throughout the month of June.

  1. Determine the total power of all lamps.

    There are 5 lamps, and each lamp has a power rating of 120 W.

    Total Power(P)=Number of lamps×Power per lamp\text{Total Power} (P) = \text{Number of lamps} \times \text{Power per lamp} P=5×120 WP = 5 \times 120 \, \text{W} P=600 WP = 600 \, \text{W}

  2. Convert the total power from watts to kilowatts.

    P=600 W1000 W/kWP = \frac{600 \, \text{W}}{1000 \, \text{W/kW}} P=0.6 kWP = 0.6 \, \text{kW}

  3. Calculate the total operating hours for the month of June.

    The lamps are switched on for 4 hours each day. The month of June has 30 days.

    Total Time(t)=Hours per day×Number of days\text{Total Time} (t) = \text{Hours per day} \times \text{Number of days} t=4 h/day×30 dayst = 4 \, \text{h/day} \times 30 \, \text{days} t=120 ht = 120 \, \text{h}

  4. Calculate the total energy consumed in kilowatt-hours (kWh).

    Energy(E)=Power(P)×Time(t)\text{Energy} (E) = \text{Power} (P) \times \text{Time} (t) E=0.6 kW×120 hE = 0.6 \, \text{kW} \times 120 \, \text{h} E=72 kWhE = 72 \, \text{kWh}

  5. Calculate the total cost.

    The cost of 1 kWh (1 unit) is 80 frs.

    Total Cost=Total Energy×Cost per unit\text{Total Cost} = \text{Total Energy} \times \text{Cost per unit} Total Cost=72 kWh×80 frs/kWh\text{Total Cost} = 72 \, \text{kWh} \times 80 \, \text{frs/kWh} Total Cost=5760 frs\text{Total Cost} = \boxed{5760 \, \text{frs}}
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