Question

An electric bulb is designed to draw power \[{{P}_{O}}\] at voltage \[{{V}_{O}}\]. If the voltage is \[V\] it draws a power \[P\], then:
A.\[P=\left( \dfrac{V}{{{V}_{O}}} \right){{P}_{O}}\]
B.\[P=\left( \dfrac{{{V}_{O}}}{V} \right){{P}_{O}}\]
C.\[P={{\left( \dfrac{{{V}_{O}}}{V} \right)}^{2}}{{P}_{O}}\]
D.\[P={{\left( \dfrac{V}{{{V}_{O}}} \right)}^{2}}{{P}_{O}}\]

Answer 1

Hint: The bulb consumes power based on the Joule’s law of heating which states that the power consumed in the resistor is directly proportional to the square of the electric current passing through the resistance and the resistance of the resistor.
The bulb is designed to draw power which is rated at a particular voltage called voltage rating of the bulb.

Complete step-by-step solution:
The rated power of the bulb is ${{P}_{o}}$
The rated voltage of the bulb is ${{V}_{o}}$
Using the formula for the power consumed,
$\text{Power}=\dfrac{{{\left( \text{Voltage} \right)}^{\text{2}}}}{\left( \text{Resistance} \right)}$
Therefore the resistance of the bulb can be calculated as,
$R=\dfrac{{{\left( {{V}_{o}} \right)}^{2}}}{{{P}_{0}}}$
When the resistance is connect across a voltage source of the voltage $V$ then the electric current passing through the resistor can be calculated using Ohm’s law,
$i=\dfrac{V}{R}$
Putting the value of the resistance $R=\dfrac{{{\left( {{V}_{o}} \right)}^{2}}}{{{P}_{0}}}$
$\begin{align}
  & i=\dfrac{V}{\dfrac{{{\left( {{V}_{o}} \right)}^{2}}}{{{P}_{0}}}} \\
 & =\dfrac{V{{P}_{o}}}{{{\left( {{V}_{o}} \right)}^{2}}}
\end{align}$
Using Joule’s law of heating,
Power consumed$={{i}^{2}}R$
$\begin{align}
  & P={{\left( \dfrac{V{{P}_{o}}}{V_{o}^{2}} \right)}^{2}}\times \dfrac{V_{o}^{2}}{{{P}_{o}}} \\
 & ={{\left( \dfrac{V}{{{V}_{o}}} \right)}^{2}}{{P}_{o}}
\end{align}$
Hence, the power drawn by the bulb is${{\left( \dfrac{V}{{{V}_{o}}} \right)}^{2}}{{P}_{o}}$.
So, option B is the right answer.

Note:Make sure to note that
->The bulb draws the electric power and emits the light.
->The electric power is drawn by the resistor using Joule’s law.