
An electric oven of $2kW$ power rating is operated in a domestic electric circuit that has a current rating of $5A$. If the supply voltage is $220V$, what do you expect?
A.)Circuit will be broken
B.)Fuse will blow
C.)Both
D.)None
Answer
586.5k+ views
Hint: The power utilized by an electrical device is the product of the voltage across it and the current through it. We can solve this problem, by finding the current that should be flowing through the oven to get the rated power at the given voltage and check whether the current exceeds the current rating of the domestic circuit or not.
Formula used:
Power $P$ expended by an electric device is given by
$P=VI$
where $V$ is the voltage across the device and $I$ is the current flowing through it.
Complete step by step answer:
As explained in the hint, we will find out the current that the oven must draw to get the rated power output when the given voltage is applied across it and check whether it is greater than or lower than the current rating of the domestic circuit.
Hence, let us proceed to do that.
Power $P$ expended by an electric device is given by
$P=VI$ --(1)
where $V$ is the voltage across the device and $I$ is the current flowing through it.
Now, the rated power of the oven is given as $P=2kW=2\times {{10}^{3}}W$ $\left( \because 1kW={{10}^{3}}W \right)$
The voltage across the oven is $V=220Volt$.
Let the current that the oven draws to get the rated power in the given potential difference be $I$.
Hence, using (1), we get,
$2\times {{10}^{3}}W=220\times I$
$\therefore \dfrac{2\times {{10}^{3}}}{220}\approx 9.09A$
Hence, the current drawn by the oven to get the rated power is clearly greater than $5A$ which is the rating of the domestic circuit.
Hence, the fuse will blow and the circuit will be broken.
Hence, the correct option is C) both.
Note: Students might think that the oven will draw only that amount of current as permitted by the rating of the domestic circuit, but that is not the case. Any electrical device when subjected to a voltage draws the current that would help it to obtain or produce a power output equal to its rated power. Hence, this current is also called the rated current. The electrical device will not draw a lower current and compromise on the power output so that the fuse does not blow and the circuit is not overloaded.
Formula used:
Power $P$ expended by an electric device is given by
$P=VI$
where $V$ is the voltage across the device and $I$ is the current flowing through it.
Complete step by step answer:
As explained in the hint, we will find out the current that the oven must draw to get the rated power output when the given voltage is applied across it and check whether it is greater than or lower than the current rating of the domestic circuit.
Hence, let us proceed to do that.
Power $P$ expended by an electric device is given by
$P=VI$ --(1)
where $V$ is the voltage across the device and $I$ is the current flowing through it.
Now, the rated power of the oven is given as $P=2kW=2\times {{10}^{3}}W$ $\left( \because 1kW={{10}^{3}}W \right)$
The voltage across the oven is $V=220Volt$.
Let the current that the oven draws to get the rated power in the given potential difference be $I$.
Hence, using (1), we get,
$2\times {{10}^{3}}W=220\times I$
$\therefore \dfrac{2\times {{10}^{3}}}{220}\approx 9.09A$
Hence, the current drawn by the oven to get the rated power is clearly greater than $5A$ which is the rating of the domestic circuit.
Hence, the fuse will blow and the circuit will be broken.
Hence, the correct option is C) both.
Note: Students might think that the oven will draw only that amount of current as permitted by the rating of the domestic circuit, but that is not the case. Any electrical device when subjected to a voltage draws the current that would help it to obtain or produce a power output equal to its rated power. Hence, this current is also called the rated current. The electrical device will not draw a lower current and compromise on the power output so that the fuse does not blow and the circuit is not overloaded.
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