Question

An electric iron of $1\,kW$ is operated at $220\,V$. Find which of the following fuses that respectively rated at $1\,A$. $3\,A$ and $5\,A$ can be used in it.

Answer 1

Hint:Electric Power is the rate at which work is done or energy is transformed into an electrical circuit. Power is denoted by ‘P’. Its SI unit is Watt, joule per second. It is a scalar quantity. Fuse is the safety device or protector, which is used to protect our home appliances with damage by high voltage.

Formula Used:
$P=VI$
$VI=\dfrac{{{V}^{2}}}{R}$
$P$=Power (in Watt), $V$=Potential difference in the circuit (in Volt), $I$ =Electric current (in Ampere) and $R$=Resistance (in Ohm).

Complete step by step answer:
Electric Power: It is the rate of doing work by an energy source or the rate at which the electrical energy is transferred (dissipated or consumed) per unit time in the electric circuit.
So, Power ($P$) = Work done (w)/Time(t)
$P$=Electrical energy dissipated / Time(t)
$\Rightarrow P$=$VI=\dfrac{{{V}^{2}}}{R}$
Given that Power P = $1$ $1KW$
$P= 1000\,W$
And voltage drawn $I=\dfrac{P}{V}$
$I=\dfrac{1000}{220}$
$\Rightarrow I=\dfrac{50}{11}$
$\therefore I=4.54\,A$

Hence, to run an electric iron of $1\,KW$,rated fuse of $5$ Ampere should be used.

Additional Information:
Power can also be written as
$P={{I}^{2}}R$
$\Rightarrow P=\dfrac{{{V}^{2}}}{R}$ by ohm's law $(V=IR)$
As we know Power is the rate of energy consumption so energy can directly be calculated using
$P=\dfrac{E}{t}$
where, $E$= energy consumption (in joules) and $T$=time in seconds.

Now we have two formulas to calculate power
1. $P=VI$ (in terms of V and I)
When the electrical instruments are connected in series (current flowing in each instrument is the same) then we should use this formula.

2. $P=\dfrac{{{V}^{2}}}{R}$(in terms of V and R)
When the electrical instruments are connected in parallel
(voltage drop is same) then use this formula

Note:To choose the fuse we need to check the fuse rating which is
$\text{Fuse rating}= \dfrac{\text{Watts}}{\text{Volts}} \times 1.25$
The principle on which the fuses work is the Heating Effect of the Current. The fuse melts down when the excessive current or heat is generated. So in the above question we took the slightly higher capacity fuse.