Question

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –

• A. 100 W
• B. 75 W
• C. 50 W
• D. 25 W

Answer 1

The Correct Option is D

Energy consumed by bulb \(𝑃=\frac {𝑉^2}{R}\)
⟹\(𝑅 = \frac {𝑉^2}{P}\)
Here, \(V = 220 \ V\) and \(P = 100\  W\)
\(𝑅 = \frac {(220)^2}{100}\)
\(R = 484\  Ω\)
The resistance of the bulb remains constant if the supply voltage is reduced to \(110 V\). If the bulb is operated on \(110 V\), then the energy consumed by it is given by the expression for power
\(𝑃 = \frac {𝑉^2}{R}\)

\(P= \frac {(110)^2}{484}\)

\(P= \frac {12100}{484}\)

\(P= 25 \ 𝑊\)

Hence, the correct option is (D): \(25\ W\)

Answer 2

Given:
The electric bulb is rated 220 V and 100 W.
To find the power consumed by the electric bulb when it is operated on 110 V, we can use the formula:
\(P = \frac{V^2}{R}\)
Where:
- P is the power consumed
- V is the voltage
- R is the resistance

We can find the resistance R using the formula:
\(R = \frac{V^2}{P}\)
Using the rated values:
\(R = \frac{(220)^2}{100} = 484 \, \Omega\)

Now, when the bulb is operated on 110 V, we can use the same formula to find the power consumed:
\(P = \frac{(110)^2}{484}\)

\(P = \frac{12100}{484}\)

\(P =25 \, \text{W}\)

So, the correct option is (D): \(25\ W\)