Question.
An electric oven of $2 \mathrm{~kW}$ power rating is operated in a domestic electric circuit $(220 \mathrm{~V})$ that has a current rating of $5 \mathrm{~A}$. What result do you expect? Explain.
An electric oven of $2 \mathrm{~kW}$ power rating is operated in a domestic electric circuit $(220 \mathrm{~V})$ that has a current rating of $5 \mathrm{~A}$. What result do you expect? Explain.
solution:
Power, $\mathrm{P}=2 \mathrm{~kW}=2000 \mathrm{~W} ; \mathrm{V}=220 \mathrm{~V}$
Now, $\mathrm{P}=\mathrm{V} \times \mathrm{I} \quad$ or $\quad \mathrm{I}=\mathrm{P} / \mathrm{V}$
or $\mathrm{I}=(2000) / 220=9.09 \mathrm{~A} .$
This shows that current flowing through the oven is more than the current rating (5 A). Hence, the fuse in the circuit melts i.e., the circuit breaks preventing the oven from damage.
Power, $\mathrm{P}=2 \mathrm{~kW}=2000 \mathrm{~W} ; \mathrm{V}=220 \mathrm{~V}$
Now, $\mathrm{P}=\mathrm{V} \times \mathrm{I} \quad$ or $\quad \mathrm{I}=\mathrm{P} / \mathrm{V}$
or $\mathrm{I}=(2000) / 220=9.09 \mathrm{~A} .$
This shows that current flowing through the oven is more than the current rating (5 A). Hence, the fuse in the circuit melts i.e., the circuit breaks preventing the oven from damage.
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