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A shop owner uses two fluorescent tubes of $40 W$ each, a fan of $60 W$ rating and a TV of $160 W$ rating. If he uses all these gadgets for $5$ hours daily, calculate the units of electrical energy consumed by him in the month of April.

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158.5 \times 1 0^{6} J

148.5 \times 1 0^{6} J

138.5 \times 1 0^{6} J

128.5 \times 1 0^{6} J

answer is B.

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Detailed Solution

The total power consumed by the shop owner in the month April is

148.5 \times 1 0^{6} J

.
Given:
Power consumed by the shop:
The total power consumed by two tubes of power

40 W

in one hour;

\Rightarrow P_{1} = 40 \times 2

\Rightarrow P_{1} = 80 W

The power consumed by the fan of power

60 W

in one hour is,

\Rightarrow P_{2} = 60 W

The power consumed by the television of power

160 W

in one hour is;

\Rightarrow P_{3} = 160 W

Total power consumed in one hour is,

P = P_{1} + P_{2} + P_{3}

\Rightarrow P = 80 + 60 + 160

\Rightarrow P = 300 W

Duration of operating the electrical appliances

\Rightarrow Duration, t = 5 \times 30

\Rightarrow Duration, t = 150 hours

So, the power can be calculated as;

\Rightarrow Power = \frac{Energy}{time}

\Rightarrow 300 = \frac{Energy}{150}

By rearranging the above equation,

\Rightarrow Energy = 300 \times 150

\Rightarrow Energy = 45000 Wh

Or,

\Rightarrow Energy = 45 kWh

As we know

1 kwh = 3.6 \times 1 0^{6} Joules

So the the energy in joules will be,

\Rightarrow Energy = 45 \times 3.6 \times 1 0^{6} J

\Rightarrow Energy = 148.5 \times 1 0^{6} J

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