Question

Two bulbs are marked as
$(a)$ $100W$ and $220V$,
$(b)$ $40W$ and $220V$.
Which bulb has a larger resistance?

Answer 1

Electric power is the rate at which electric energy is transferred in an electric circuit. While keeping a constant potential difference, power is inversely proportional to the resistance. So, at a constant voltage, the bulb having lesser power will have higher resistance.

Complete step by step solution:
As we know that electric power is the rate at which electric energy is transferred in an electric circuit or the rate at which electric work is done.
$P=\dfrac{W}{t}$
Expressing work done in terms of potential difference as:
$\begin{aligned} & V=\dfrac{W}{q} \\\ & \Rightarrow W=Vq \\\ \end{aligned}$
Therefore we can say that,
Electric power is the measure of electric current having $q$ coulombs of charge passing through a potential difference of $V$ volts in $t$ seconds.
Mathematically, it can be expressed as:
$P=\dfrac{Vq}{t}$
Here,
Electric power ($P$) is measured in watts.
Potential difference ($V$) applied across the end of the conductor is measured in volts.
Electric charge ($q$) is measured in coulombs.
Time ($t$) is measured in seconds.
Now, as we know that electric current is rate of flow of electric charge,
Replacing $\dfrac{q}{t}$ with $I$(electric current):
$\Rightarrow P=VI$ ……(1)
Now, according to the Ohm’s law:
$I=\dfrac{V}{R}$
Applying Ohm’s Law in equation (1), we get:
$P=\dfrac{{{V}^{2}}}{R}$ …...(2)
Rewriting equation (2) as:
$R=\dfrac{{{V}^{2}}}{P}$ ……(3)
Here, we can see that while keeping the potential difference constant, the electric power is inversely proportional to the resistance or we can say that the resistance is inversely proportional to the electric power.
Therefore, the bulb with low electric power will have high resistance and the bulb with high electric power will have low resistance.
Now, by taking the first bulb:
($a$) $100W$ and $220V$
Here, we calculate the resistance by using equation (3):
$\begin{aligned} & R=\dfrac{{{V}^{2}}}{P} \\\ & \Rightarrow R=\dfrac{{{(220)}^{2}}}{100}\Omega \\\ & \Rightarrow R=\dfrac{220\times 220}{100}\Omega \\\ & \Rightarrow R=484\Omega \\\ \end{aligned}$
Now, by taking the second bulb:
($b$) $40W$ and $220V$
Here, we calculate the resistance by using equation (3):
$\begin{aligned} & R=\dfrac{{{V}^{2}}}{P} \\\ & \Rightarrow R=\dfrac{{{(220)}^{2}}}{40}\Omega \\\ & \Rightarrow R=\dfrac{220\times 220}{40}\Omega \\\ & \Rightarrow R=1210\Omega \\\ \end{aligned}$
Therefore, the second bulb ($b$) with power,$P=$ $40W$ and potential difference $V=$ $220V$ has higher resistance.

Note: The power and resistance relation can also be expressed as$P={{I}^{2}}R$.
Here, in this case, when the electric current is constant, the electric power is directly proportional to resistance. Therefore as the resistance increases, the electric power will also increase.