Calculating Electric Bulb Power Rating at Reduced Voltage

This problem asks us to find the power rating of an electric bulb when it is operated at a voltage different from its rated voltage. The key principle here is that the resistance of the bulb filament remains constant regardless of the operating voltage (assuming the temperature effect is negligible for this calculation).

Step 1: Determine the Resistance of the Bulb

The bulb is rated at a voltage \(V_{rated} = 220 \, V\) and power \(P_{rated} = 80 \, W\). We can use the formula relating power, voltage, and resistance: \(P = \frac{V^2}{R}\). From this, we can find the resistance \(R\) of the bulb filament:

\(R = \frac{V_{rated}^2}{P_{rated}}\)

Let's calculate the resistance:

\(R = \frac{(220 \, V)^2}{80 \, W}\)

\(R = \frac{48400 \, V^2}{80 \, W}\)

\(R = 605 \, \Omega\)

The resistance of the bulb is \(605 \, \Omega\).

Step 2: Calculate the Power at the New Operating Voltage

Now, the bulb is operated on a new voltage \(V_{new} = 110 \, V\). Since the resistance \(R\) remains constant (\(605 \, \Omega\)), we can use the same formula to find the new power \(P_{new}\) at this voltage:

\(P_{new} = \frac{V_{new}^2}{R}\)

Let's plug in the values:

\(P_{new} = \frac{(110 \, V)^2}{605 \, \Omega}\)

\(P_{new} = \frac{12100 \, V^2}{605 \, \Omega}\)

Performing the division:

\(P_{new} = 20 \, W\)

Conclusion

When the electric bulb rated 220 V and 80 W is operated on 110 V, its power rating would be 20 W.

This shows that when the operating voltage is halved (from 220V to 110V), the power rating becomes one-fourth of the original power rating (from 80W to 20W), which is consistent with the \(P \propto V^2\) relationship when resistance is constant.