Question

Calculate the current through a 60 W lamp rated for 250 V. If the line voltage falls to 200 V, how is the power consumed by the bulbs affected?

Answer 1

Hint: Use the relation between current, voltage and power to find the value of current. The expression of power using voltage and resistance get the value of resistance. Use this value to get the second result.

Formula used:
Power consumption expression in terms of voltage and current:
$P=V.I$ ………..(1)
Where,
$P$ is power consumption,
$V$ is the voltage supplied,
$I$ is the current through the bulb.
Power consumption expression in terms of voltage and resistance:
$P={\displaystyle \frac{V^2}{R}}$ ……….(2)
Where,
$R$ is the internal resistance of the bulb.

Complete step by step solution:
Given:
Initially, the power consumed by the bulb is $60W$ at $250V$.
Later the voltage becomes $200V$.
To find:
Current through the bulb at first.
Power consumption later.
Step 1
Use the given values of $P$ and $V$ in eq.(1) to get $I$ as:
$$\begin{gathered} P=V.I \\ \Rightarrow I={\displaystyle \frac{P}{V}}={\displaystyle \frac{60W}{250V}}=0.24A \end{gathered}$$

Step 2
Now, use the values of $P$ and $V$ in eq.(2) to get $R$ as:
$$\begin{gathered} P={\displaystyle \frac{V^2}{R}} \\ \Rightarrow R={\displaystyle \frac{V^2}{P}}={\displaystyle \frac{{(250V)}^2}{60W}}=1041.67\mathrm{\Omega } \end{gathered}$$
Since resistance is an intrinsic property so it will always remain unchanged.

Step 3
Substitute the value of R in eq.(2) again when the voltage is $V^{\prime }=200V$ to get the power $P^{\prime }$:
$P^{\prime }={\displaystyle \frac{{V^{\prime }}^2}{R}}={\displaystyle \frac{{(200V)}^2}{1041.67\mathrm{\Omega }}}=38.4W$

$\therefore$ The current through a 60W rated bulb at 250V is 0.24A. The power consumed by the bulb at 200V is 38.4W. Hence option (B) is the correct answer.

Note:
The second part of the problem can be done in a trickier manner. From eq.(2) notice that power is proportional to the square of the voltage. Hence, the power ratio will be the square of the voltage ratio. Here it’ll be ${\left({\displaystyle \frac{200}{250}}\right)}^2={\displaystyle \frac{16}{25}}$. So, you can get the final power as, $60\times {\displaystyle \frac{16}{25}}W=38.4W$. And this is a shortcut technique.