
An electric kettle consumes $1\,kW$ of electric power when operated at $220\,V$. A fuse wire of what rating must be used for it?
A) 1A
B) 2A
C) 4A
D) 5A
Answer
511.5k+ views
Hint: Electric power depends upon the current flowing through the circuit and the potential difference across its terminals. We will use the mathematical formula to get the answer of the suitable fuse wire rating.
Complete step by step answer:
Initially, we will convert Electric power (P) into S.I. unit i.e. W.
Given from the question, $P\, = \,1kW\, = \,1000W$
Now from the formula of electric power (P) we know that $P\, = \,V\, \times \,I$ where V is the voltage and I is the current. Rearranging the above equation, we can write
$I\, = \dfrac{P}{V}$ ----------(1)
On substituting the given values of P and V in the formula, we get
$\Rightarrow I\, = \,\dfrac{{1000}}{{220}}\,A$
So, Current $I\, = \,4.545\,A$
This is the permissible amount of current to flow from the circuit. Thus, rating of fuse wire must be higher than this value ( $4.545\,A$ ), so the most appropriate answer should be $5\,A$.
Hence, Option D is the correct answer.
Additional Information:
When current passes through the conductor there are collisions between charge carriers and the sub atomic particles which leads to some energy dissipation of the electrons in the form of heat energy. Expression for heat produced is given by Joule’s law of heating.
According to Joule’s law of heating, the heat produced depends upon the square of the current passing through the circuit, the resistance offered by the circuit and the time for which current is flowing through the circuit.
Note:
A fuse is used in the circuit which contains a fuse wire of very low melting point hence only current within permissible limit can flow through the circuit. If current above a permissible limit enters the fuse, the current produces heating effect and the wire melts and the fuse breaks, thereby breaking the circuit and not allowing current to flow through it.
This is done to ensure that excessive current does not flow through the electrical circuit and the electrical appliances are safe.
Complete step by step answer:
Initially, we will convert Electric power (P) into S.I. unit i.e. W.
Given from the question, $P\, = \,1kW\, = \,1000W$
Now from the formula of electric power (P) we know that $P\, = \,V\, \times \,I$ where V is the voltage and I is the current. Rearranging the above equation, we can write
$I\, = \dfrac{P}{V}$ ----------(1)
On substituting the given values of P and V in the formula, we get
$\Rightarrow I\, = \,\dfrac{{1000}}{{220}}\,A$
So, Current $I\, = \,4.545\,A$
This is the permissible amount of current to flow from the circuit. Thus, rating of fuse wire must be higher than this value ( $4.545\,A$ ), so the most appropriate answer should be $5\,A$.
Hence, Option D is the correct answer.
Additional Information:
When current passes through the conductor there are collisions between charge carriers and the sub atomic particles which leads to some energy dissipation of the electrons in the form of heat energy. Expression for heat produced is given by Joule’s law of heating.
According to Joule’s law of heating, the heat produced depends upon the square of the current passing through the circuit, the resistance offered by the circuit and the time for which current is flowing through the circuit.
Note:
A fuse is used in the circuit which contains a fuse wire of very low melting point hence only current within permissible limit can flow through the circuit. If current above a permissible limit enters the fuse, the current produces heating effect and the wire melts and the fuse breaks, thereby breaking the circuit and not allowing current to flow through it.
This is done to ensure that excessive current does not flow through the electrical circuit and the electrical appliances are safe.
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