A 220 V–1000 W bulb is connected across a 110 V mains supply. The power consumed will be
(a) 750 W
(

Q: Why does the power change if it's a 1000W bulb? Shouldn't it always consume 1000W?

The $1000\text{ W}$ is the *rated* power, not an absolute property. It means the bulb is designed to consume $1000\text{ W}$ *only* when exactly $220\text{ V}$ is applied across it. If you apply less voltage, it draws less current and consumes less power.

Q: Why do we assume the resistance of the bulb is constant?

In reality, the resistance of a tungsten filament increases as it gets hotter. However, in standard physics problems like this, unless a temperature coefficient is given, we assume the resistance remains constant for simplicity.

Q: Can I use the formula $P = VI$ to solve this?

Yes, but it's a longer route. You would first need to find the rated current, then find the resistance, then find the *new* current at $110\text{ V}$, and finally use $P = VI$. Using $P = \frac{V^2}{R}$ is much faster because it directly relates the given voltages to the constant resistance.

Q: Why did the power become one-fourth when the voltage was only halved?

Look at the formula $P = \frac{V^2}{R}$. Power is directly proportional to the *square* of the voltage. So, if voltage becomes $\frac{1}{2}$, the power becomes $(\frac{1}{2})^2 = \frac{1}{4}$ of its original value.

Q: What would happen if we accidentally connected this bulb to a 440V supply?

If the voltage is doubled ($440\text{ V}$), the power consumed would become $2^2 = 4$ times the rated power, which is $4000\text{ W}$. The filament would overheat instantly and the bulb would fuse (burn out).

Question

A 220 V–1000 W bulb is connected across a 110 V mains supply. The power consumed will be

Options:

(A) 750 W

(B) 500 W

(C) 250 W

(D) 1000 W

Sigma Percentile · Accepted Answer

### Decoding Bulb Ratings

Imagine you buy a light bulb from the market. Printed on its glass are two crucial numbers: its **rated voltage** and its **rated power**. In our problem, the bulb is rated at $220	ext{ V}$ and $1000	ext{ W}$.

What does this actually mean? It means that the manufacturer guarantees the bulb will consume exactly $1000	ext{ W}$ of electrical power *only if* you connect it to a potential difference of exactly $220	ext{ V}$.

### The Hidden Constant: Resistance

When you take this bulb and connect it to a different voltage source, the power it consumes will change. However, one physical property of the bulb remains fundamentally constant (assuming temperature variations are negligible): the **resistance** ($R$) of its tungsten filament.

We can extract this hidden constant using the power formula that relates voltage, power, and resistance:

$$P = \frac{V^2}{R}$$

Rearranging this to solve for resistance, we get:

$$R = \frac{V_{	ext{rated}}^2}{P_{	ext{rated}}}$$

Let's plug in our rated values. To make our future calculations elegant, we won't compute the final decimal value just yet. We'll leave it as a raw fraction:

$$R = \frac{220^2}{1000} \ \Omega$$

### The Reality Check: Applied Voltage

Now, the problem states that we are connecting this exact same bulb to a mains supply of $110	ext{ V}$. This is our **applied voltage** ($V_{	ext{applied}}$). Because the applied voltage is lower than the rated voltage, the bulb will draw less current and glow much dimmer.

To find the actual power consumed in this new scenario, we use the power formula again. We use the new applied voltage, but we use the *same* resistance $R$ we found earlier:

$$P_{	ext{consumed}} = \frac{V_{	ext{applied}}^2}{R}$$

### The Elegant Calculation

Let's substitute our values into the equation:

$$P_{	ext{consumed}} = \frac{110^2}{\frac{220^2}{1000}}$$

By rearranging the fraction, we can group the voltage terms together. This reveals a beautiful mathematical shortcut:

$$P_{	ext{consumed}} = \left(\frac{110}{220}ight)^2 	imes 1000$$

Notice the ratio inside the parentheses. $110$ is exactly half of $220$.

$$P_{	ext{consumed}} = \left(\frac{1}{2}ight)^2 	imes 1000$$

Squaring the fraction gives us $\frac{1}{4}$. This tells us a profound physical truth: **because power is proportional to the square of the voltage, halving the voltage reduces the power to one-fourth.**

$$P_{	ext{consumed}} = \frac{1}{4} 	imes 1000 = 250 	ext{ W}$$

The bulb will consume $250	ext{ W}$ of power. Therefore, the correct option is (c).

Animated Solution for Physics - Current Electricity: A 220 V–1000 W bulb is connected across a 110 V mains supply. The power consumed will be

Select Answer:

Visualized Solution

$Understanding Bulb Ratings$

$The Constant Resistance$

$Setting up Resistance$

$The New Applied Voltage$

$Formula for Consumed Power$

$Substituting Values$

$Rearranging the Expression$

$Simplifying the Ratio$

$Final Calculation$

$Food for Thought$

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

Decoding Bulb Ratings

The Hidden Constant

The Reality Check

The Elegant Calculation

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Animated Solution for Physics - Current Electricity: A 220 V–1000 W bulb is connected across a 110 V mains supply. The power consumed will be

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Understanding Bulb Ratings

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Constant Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The New Applied Voltage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formula for Consumed Power

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rearranging the Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought