Calculating Resistance Ratio of Bulbs

This problem asks us to find the ratio of electrical resistance for two light bulbs given their power ratings and the voltage they are designed to operate at. The key is to understand the relationship between power, voltage, and resistance in an electrical circuit.

Relationship Between Power, Voltage, and Resistance

The power ($\text{P}$) dissipated by a resistor (like the filament of a light bulb) is related to the voltage ($\text{V}$) across it and its resistance ($\text{R}$) by the formula:

\( P = \frac{V^2}{R} \)

From this formula, we can rearrange it to find the resistance ($\text{R}$) when the voltage ($\text{V}$) and power ($\text{P}$) are known:

\( R = \frac{V^2}{P} \)

Calculating Resistance for Each Bulb

We are given two bulbs:

• Bulb 1: Power \( P_1 = 500 \) W, Voltage \( V = 250 \) V
• Bulb 2: Power \( P_2 = 200 \) W, Voltage \( V = 250 \) V

Let \( R_1 \) be the resistance of the 500 W bulb and \( R_2 \) be the resistance of the 200 W bulb. Using the formula \( R = \frac{V^2}{P} \):

• Resistance of Bulb 1: \( R_1 = \frac{V^2}{P_1} = \frac{(250)^2}{500} \)
• Resistance of Bulb 2: \( R_2 = \frac{V^2}{P_2} = \frac{(250)^2}{200} \)

Note that both bulbs are rated at the same voltage, 250 V.

Determining the Resistance Ratio

We need to find the ratio \( R_1 : R_2 \). Let's set up the ratio:

\( \frac{R_1}{R_2} = \frac{\frac{(250)^2}{500}}{\frac{(250)^2}{200}} \)

We can see that the \( (250)^2 \) term appears in both the numerator and the denominator, so it cancels out:

\( \frac{R_1}{R_2} = \frac{\frac{1}{500}}{\frac{1}{200}} \)

Dividing by a fraction is the same as multiplying by its reciprocal:

\( \frac{R_1}{R_2} = \frac{1}{500} \times \frac{200}{1} = \frac{200}{500} \)

Now, simplify the fraction:

\( \frac{R_1}{R_2} = \frac{200 \div 100}{500 \div 100} = \frac{2}{5} \)

So, the ratio of resistances \( R_1 : R_2 \) is \( 2 : 5 \).

This shows that for bulbs rated at the same voltage, the resistance is inversely proportional to the power rating. A higher power bulb has lower resistance, and a lower power bulb has higher resistance.

Summary of Steps

1. Identify the given power ratings and voltage for both bulbs.
2. Recall or derive the formula for resistance in terms of power and voltage: \( R = \frac{V^2}{P} \).
3. Set up the ratio of the resistances using this formula.
4. Substitute the given values and simplify the ratio.

Conclusion on Resistance Ratio

Based on our calculation, the resistance ratio of the 500 W bulb to the 200 W bulb, both rated at 250 V, is \( 2 : 5 \).

Revision Table: Key Concepts

Additional Information: Power and Resistance

It's interesting to note the inverse relationship between power and resistance when the voltage is constant. This means that a bulb designed to dissipate more power at a specific voltage must have a lower resistance to allow more current to flow (\(I = V/R\), and \(P = VI\)). Conversely, a lower power bulb at the same voltage will have higher resistance, limiting the current and thus the power dissipated.

This concept is crucial when connecting appliances in parallel to a constant voltage source, like in household wiring. Appliances with lower resistance will draw more current and consume more power.