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Question: An electric iron of heating element of resistance \(88\Omega \) is used at 220 volts for 2 hours. Th...

An electric iron of heating element of resistance 88Ω88\Omega is used at 220 volts for 2 hours. The electric energy spent, in unit, will be:
A. 0.8
B. 1.1
C. 2.2
D. 8.8

Explanation

Solution

To calculate the energy we need to first calculate the power produced by the electrical system. We will use the formula, P=VIP = VI for the same. Then we will substitute this in the formula, E=PtE = Pt to calculate the energy spent.

Complete step by step answer:
We know the formula to calculate the electric power produced in a circuit is,
P=VIP = VI (Where, p is the electric power; Vis the voltage used; I is the current in amperes)
From ohm’s law, we know that, I=VRI = \dfrac{V}{R}
On substituting the value if I in the power formula, we get,
P=V2RP = \dfrac{{{V^2}}}{R}
(Now, substitute the value of V and R in the above relation)
P=220288P = \dfrac{{{{220}^2}}}{{88}}
P=550W\Rightarrow P = 550W
(Now, for simpler calculations, we will convert watts to kilowatt-hours. This is the commercial unit for calculating the energy consumed by an electrical circuit)
P=0.55P = 0.55kW
So, the formula to calculate the energy consumed is,
E=PtE = Pt (where E is the energy consumed; P is the electric power; t is the time)
On substituting the required values in the above formula, we get,
E=0.55kW×2hrsE = 0.55kW \times 2hrs
E=1.1\Rightarrow E = 1.1 units
Hence, option (B) is the correct answer.

Note: Kilowatt-hour is the commercial unit of electricity. Therefore, one should calculate the answers in the same unit.
-If the resistances are connected in series, then using P=I2RP = {I^2}R, the power developed will be higher in the resistor of higher value as current will be the same in all resistors.
-If the resistances are connected in parallel, then using P=V2RP = \dfrac{{{V^2}}}{R}, the power developed will be higher in the resistor of a lower value as potential will be the same across all resistors.