Numerical

An electric iron of 1100 W is operated for 2 hrs daily. What will be the electrical consumption expenses for that in the month of April? (The electric company charges Rs 5 per unit of energy).

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#### Solution

Electric power required for working of iron, P = 1100 W

Duration for which the iron is operated daily = 2 h = \[2 \times 60 \times 60 = 7200 s \]

Electric energy consumed by iron in 7200 s is

\[E = 1100 \times 7200 = 7920000 J\]

Thus, total energy consumed in the month of April,

\[E' = E \times 30 = 7920000 \times 30 = 237600000 J\]

We know,

1 unit =\[ 3 . 6 \times {10}^6 J\]

or, \[1 J = \frac{1}{3 . 6 \times {10}^6} \] unit

Thus,

\[237600000 J = \frac{237600000}{3 . 6 \times 1000000} = 66\] units

or, \[1 J = \frac{1}{3 . 6 \times {10}^6} \] unit

Thus,

\[237600000 J = \frac{237600000}{3 . 6 \times 1000000} = 66\] units

Cost of 1 unit of energy = Rs 5

Thus, total electrical consumption expenses for the month of April = 66 \[\times\]5 =

**Rs 330** Concept: Effects of Electric Current - Heating Effect of Electric Current

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