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**Hint:**Think about the formula for the power dissipation across a resistor(Bulb has a certain resistance). With the help of ohm’s law we can find the resistance.Also remember the relation or formula of the terms you can derive the definition from there.

**Complete step by step answer:**

(a) Let’s define power first, Power is the rate at which a work is done or power is the amount of energy transferred or converted per unit time. As per definition, we can define power as,

\[P=\dfrac{W}{T}=\dfrac{E}{T}\]

The S.I unit of power is Watt or joule per second. Power dissipated across the resistor \[=VI\].

Here, $V$ is the voltage drop across the resistor, $I$ is the current flowing through that resistor and for the resistor we can use ohm’s law and write \[V=IR\] and substitute the value of $V$ in the power dissipation formula to get a different variation of the formula.

(b) (i)The rating of the bulb is \[5\,V\] and \[500\,mA\] which means the bulb will glow when a voltage of \[5\,V\] is applied and will produce a \[500\,mA\] of current. Let’s solve the numericals now, the power dissipated across the resistor is \[1\,W=1000\,mW\].

\[P=VI \\

\Rightarrow P=5\,V\times 500\,mA \\

\therefore P=2500\,mW=2.5\,W \\ \]

(ii) Resistance across the resistor can be found out by ohm’s law

\[1\,A=1000\,mA\]

\[\Rightarrow V=IR \\

\Rightarrow R=\frac{V}{I} \\

\Rightarrow R=\frac{5}{0.5} \\

\therefore R=10\Omega \\ \]

(iii) From the formula of power we can say,

\[1\,h=3600\,s\]

\[\Rightarrow E=PT \\

\Rightarrow E=2.5\times 2.5\times 3600 \\

\therefore E =22500J=22.5\,kJ \]

Hence, the energy consumed is \[22.5\,kJ\].

**Note:**Other useful formulae for power dissipation are: \[P=VI={{I}^{2}}R=\dfrac{{{V}^{2}}}{R}\]. These are just the rearrangement of the basic formula \[(P=VI)\]. We can define average power as the total energy consumed divided by the total time taken. In simple language, we can say that average power is the average amount of work done or energy converted per unit of time.

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