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**Hint:**Power is the electrical work done per unit time. Therefore, substituting the given values in the formula for power, we can calculate the work done in joules. Kilo-watt hour is the work done in kilowatt in an hour. Therefore, to calculate the work done in kilo-watt hour, we multiply by the number of hours and divide by 1000. Convert the units as required.

**Formula used:**

$P=\dfrac{W}{t}$

**Complete answer:**

The electrical work done is defined as the work done to move a unit charge from a point of low potential to a point of high potential. Its SI unit is joules ($J$).

Power is defined as the electrical work done in unit time. Its SI unit is watt ($W$). It is given by-

$P=\dfrac{W}{t}$

Here, $P$ is the power

$W$ is the work done

$t$ is the time taken

Given, an oven has power $1500W$ and the time taken is $20\min =20\times 60s=1200s$

Substituting the given values in above equation, we get

$\begin{align}

& 1500=\dfrac{W}{1200} \\

& \Rightarrow 180\times {{10}^{4}}joules=W \\

\end{align}$

Therefore, the work consumed by the oven in 20 minutes in joules is $180\times {{10}^{4}}joules$.

To convert $W$ to $kW\,hr$ we multiply by $\dfrac{1\,hr}{60\min }$. Therefore, to convert $1500W$ into $kW\,hr$, we get,

$\begin{align}

& 1500W=1500\times 20\min \times \dfrac{1\,hr}{60\,\min } \\

& \Rightarrow 1500W=500W\,hr\times \dfrac{1kW}{100W} \\

& \therefore 1500W=0.5kW\,hr \\

\end{align}$

Therefore, $0.5kW\,hr$ is the power in $kW\,hr$.

Therefore, the electrical energy consumed in joules is $180\times {{10}^{4}}joules$ and in $kW\,hr$, it is. $0.5kW\,hr$

**Note:**

The electrical energy in a circuit depends on the square of current, resistance and the time taken. Electrical energy is also dissipated as heat. According to the Ohm’s law, the resistance is the ratio of potential difference to the current in the circuit. The power consumed in everyday life is calculated in units.

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