Animated Solution for Physics - Current Electricity: An electric bulb is rated 220 V-100 W. The power consumed by it when operated on 110 V will be

Have you ever noticed that when the voltage in your house drops, the incandescent bulbs become noticeably dimmer? This isn't just a random occurrence; it's a direct consequence of the fundamental laws of electricity. In this problem, we are going to explore exactly how the power consumed by a bulb changes when the voltage supplied to it is altered.

When you buy a light bulb from the store, it comes with a specific rating printed on the box. In our case, the bulb is rated at 220 V and 100 W. But what does this actually mean?

The rated voltage ($V_R=220\text{ V}$) is the optimal potential difference the manufacturer designed the bulb to operate at. The rated power ($P_R=100\text{ W}$) is the amount of electrical energy the bulb will convert into light and heat every second, provided it is connected to exactly that rated voltage.

If you connect it to a lower voltage, it won't consume 100 W. It will consume less. To figure out exactly how much less, we need to find the one property of the bulb that remains constant regardless of the voltage: its resistance.

The filament inside the bulb is a physical piece of tungsten wire. Its resistance ($R$) is a built-in property. While resistance can change slightly with temperature, in standard physics problems like this one, we assume it remains perfectly constant.

We can link power, voltage, and resistance using the famous electrical power formula:

By rearranging this formula, we can solve for the resistance of our bulb using its rated parameters:

Let's plug in the numbers. The rated voltage is 220 V, and the rated power is 100 W:

We could calculate this out to be $484\Omega$, but as a pro-tip for competitive exams like JEE, it's often better to leave it in its fractional form. This prevents messy calculations and allows for elegant cancellations later on!

Operating at 110 V

Now, the problem states that we take this exact same bulb and connect it to a new voltage source. Let's call this new operating voltage $V^{\prime }$, which is given as 110 V.

Because the voltage has changed, the power consumed will also change. Let's call this new consumed power $P^{\prime }$. We use the exact same power formula, but this time with our new voltage and our constant resistance:

This is where the magic happens. Let's substitute our new voltage ($110\text{ V}$) and the expression we found for resistance ($R=\frac{220^2}{100}$) into our power equation:

When you divide by a fraction, you multiply by its reciprocal. Let's rearrange the terms to group the voltages together:

Look at how beautifully that simplifies! The new voltage (110 V) is exactly half of the rated voltage (220 V).

Squaring the fraction gives us one-fourth:

And finally, one-fourth of 100 is simply 25.

Our final answer is 25 W. But there's a powerful shortcut hidden in this math. Because the resistance is constant, the power consumed is directly proportional to the square of the voltage ($P\propto V^2$).

If you halve the voltage (from 220 V to 110 V), the power doesn't just halve; it drops by a factor of two squared, which is four! So, the new power is simply the old power divided by four ($100/4=25\text{ W}$). Remembering this proportional relationship can save you precious minutes in an exam!