Calculating Power Consumption of a Bulb

This problem involves understanding how the power consumed by an electrical device, like a bulb, changes when it is connected to a voltage different from its rated voltage. The key principle here is that the resistance of the bulb remains constant regardless of the applied voltage (assuming a standard resistive load like an incandescent bulb). We are given the bulb's rated specifications (rated voltage and rated power) and the new operating voltage. We need to find the new power consumption.

Step 1: Find the Resistance of the Bulb

The rated voltage ($V_{rated}$) and rated power ($P_{rated}$) are given for the bulb. The relationship between power, voltage, and resistance ($R$) is given by the formula:

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

We can use this formula to find the resistance of the bulb using its rated values. Rearranging the formula to solve for resistance:

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

Given:

• Rated voltage, \(V_{rated} = 220 \text{ volts}\)
• Rated power, \(P_{rated} = 100 \text{ watts}\)

Calculating the resistance:

\(R = \frac{(220 \text{ V})^2}{100 \text{ W}}\)

\(R = \frac{48400 \text{ V}^2}{100 \text{ W}}\)

\(R = 484 \text{ } \Omega\)

So, the resistance of the bulb is 484 ohms.

Step 2: Calculate Power Consumption at the New Voltage

Now that we know the resistance of the bulb and the new voltage ($V_{new}$) it is connected to, we can calculate the new power consumption ($P_{new}$). The new voltage is 110 volts.

Using the same power formula \(P = \frac{V^2}{R}\), but with the new voltage and the calculated resistance:

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

Given:

• New voltage, \(V_{new} = 110 \text{ volts}\)
• Resistance, \(R = 484 \text{ } \Omega\)

Calculating the new power consumption:

\(P_{new} = \frac{(110 \text{ V})^2}{484 \text{ } \Omega}\)

\(P_{new} = \frac{12100 \text{ V}^2}{484 \text{ } \Omega}\)

\(P_{new} = 25 \text{ watts}\)

Conclusion

When the 220 volt, 100 watt bulb is connected to a 110 volt source, the power consumption by the bulb is 25 watts. This is significantly less than the rated power because the voltage is halved, and power is proportional to the square of the voltage when resistance is constant.

Revision Table: Bulb Power Consumption Calculation

Reviewing the steps involved in calculating the actual power consumption of a bulb under different voltage conditions.

• Identify rated voltage and power to determine intrinsic property (resistance).
• Use resistance and new voltage to find actual power consumption.
• Remember power is proportional to the square of voltage (\(P \propto V^2\)) when resistance is constant. Halving the voltage quarters the power.

Additional Information: Electrical Power and Ohm's Law

This problem relies on fundamental concepts of electrical circuits:

• Power (P): The rate at which electrical energy is transferred. Measured in watts (W).
• Voltage (V): The electric potential difference across two points. Measured in volts (V).
• Resistance (R): Opposition to the flow of electric current. Measured in ohms (\(\Omega\)).
• Ohm's Law: Relates voltage, current (\(I\)), and resistance: \(V = IR\).
• Power Formulas: There are different ways to express power based on Ohm's Law:
  • \(P = VI\)
  • \(P = I^2R\)
  • \(P = \frac{V^2}{R}\) (Used in this solution)

When analyzing a device like a bulb under different operating conditions, assume its resistance is constant unless stated otherwise. Then use the appropriate power formula with the actual voltage or current it experiences.