If an electric iron of $1200 W$ is used for $30 m i n u t e s$ everyday, find electric energy consumed in the month of April.

12 k W h

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18 k W h

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36 k W h

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6 k W h

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Solution

The correct option is B $18 k W h$
$P o w e r P = 1200 W = \frac{1200}{1000} = 1.2 k W$
$t i m e t = 30 m i n = \frac{30}{60} = 0.5 h$

Number of days in the month of april = $30$
Hence, Total time for which iron is used is given by (in hrs) = $0.5 \times 30$
$E n e r g y E = P o w e r \times T o t a l t i m e$
$= 1.2 \times 0.5 \times 30$
$= 18 k W h$

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