Question

Three bulbs of power 10 W, 25 W, and 50 W are connected parallel across a voltage source \( V \). What is the effective power of the combination?

Hint:

When resistors are connected in parallel, the total power is the sum of the individual powers.

Answer 1

Step 1: Use the formula for power in terms of voltage and resistance.
The power \( P \) of a bulb is related to its resistance \( R \) and the voltage \( V \) by the formula: \[ P = \frac{V^2}{R} \] Thus, the resistance of each bulb can be calculated as: \[ R = \frac{V^2}{P} \]
Step 2: Calculate the resistance of each bulb.
Let the voltage across each bulb be \( V \). Then, the resistances of the bulbs are: - For the 10 W bulb: \[ R_1 = \frac{V^2}{10} \] - For the 25 W bulb: \[ R_2 = \frac{V^2}{25} \] - For the 50 W bulb: \[ R_3 = \frac{V^2}{50} \]
Step 3: Use the formula for total resistance in a parallel combination.
For parallel resistors, the total resistance \( R_{\text{total}} \) is given by: \[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \] Substitute the values for \( R_1 \), \( R_2 \), and \( R_3 \): \[ \frac{1}{R_{\text{total}}} = \frac{1}{\frac{V^2}{10}} + \frac{1}{\frac{V^2}{25}} + \frac{1}{\frac{V^2}{50}} \] Simplify the equation: \[ \frac{1}{R_{\text{total}}} = \frac{10}{V^2} + \frac{25}{V^2} + \frac{50}{V^2} = \frac{85}{V^2} \] Thus, the total resistance is: \[ R_{\text{total}} = \frac{V^2}{85} \]
Step 4: Calculate the total power.
The total power \( P_{\text{total}} \) of the combination is given by: \[ P_{\text{total}} = \frac{V^2}{R_{\text{total}}} \] Substitute \( R_{\text{total}} = \frac{V^2}{85} \): \[ P_{\text{total}} = \frac{V^2}{\frac{V^2}{85}} = 85 \, \text{W} \] Thus, the effective power of the combination is: \[ \boxed{85 \, \text{W}} \]