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Physics

An electric iron rated 1100 W, 220 V is operated for 5 hours.

Calculate:

(a) the minimum rating of the fuse required.

(b) the energy consumed in kWh.

(c) the cost of the energy consumed, if the rate is ₹ 10 per unit.

Household Circuits

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Answer

Given,

  • Power rating of the iron (P\text P) = 1100 W
  • Voltage rating of the iron (V\text V) = 220 V
  • Operating time = 5 hours

(a) Power consumed by an appliance is given by,

P=VI⇒I=PV=1100220=5 A\text P = \text {VI} \\[1em] \Rightarrow \text I = \dfrac{\text P}{\text V} \\[1em] = \dfrac{1100}{220} \\[1em] = 5\ \text A

Here, I\text I is the operating current flowing through the appliance.

Since the fuse rating should be slightly higher than the operating current, so the minimum suitable fuse rating is 6 A.

Hence, 6 A is the minimum rating of the fuse required.

(b) Energy consumed by the iron = Power consumed x Operating time

= 1100 x 5

= 5500 Wh

= 5.5 x 103 Wh

= 5.5 kWh

Hence, the energy consumed by the iron is 5.5 kWh.

(c) Given,

  • Electricity rate = ₹ 10 per unit

As, 1 unit = 1 kWh

Then,

The cost of energy consumed by the iron = Energy consumed x Electricity rate

= 5.5 x 10

= ₹ 55

Hence, the cost of the energy consumed is ₹ 55.

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