A voltage V is applied across a single resistor of resistance R, and the power dissipated is P. The same voltage V is now applied across a parallel combination of three equal resistors, each of resistance R. What is the total power dissipated in the parallel combination?

Introduction / Context:
This problem examines how electric power dissipation changes when we replace a single resistor with several equal resistors in parallel, while keeping the applied voltage the same. Understanding series and parallel combinations of resistors and how they affect equivalent resistance and power is fundamental in DC circuit analysis and practical electrical design.

Given Data / Assumptions:
• Case 1: A single resistor of resistance R has voltage V across it and power dissipated P. • Case 2: Three equal resistors, each of resistance R, are connected in parallel across the same voltage V. • Power dissipation in both cases is resistive heating, obeying P = V^2 / R_eq. • All resistors are ohmic and have the same resistance value R.

Concept / Approach:
When voltage is fixed, the power dissipated in a resistor or network is P = V^2 / R_eq, where R_eq is the equivalent resistance seen by the source. For a single resistor, R_eq = R. For three equal resistors in parallel, the reciprocal of the equivalent resistance is the sum of reciprocals: 1 / R_eq = 1 / R + 1 / R + 1 / R = 3 / R, which gives R_eq = R / 3. Because the equivalent resistance decreases in parallel, the total current drawn increases and therefore total power increases. By comparing P in the two situations, we can find the new power in terms of P.

Step-by-Step Solution:
Step 1: For the single resistor case, power is P = V^2 / R. Step 2: For three resistors of resistance R in parallel, compute equivalent resistance. Step 3: Use 1 / R_eq = 1 / R + 1 / R + 1 / R = 3 / R. Step 4: Solve for R_eq: R_eq = R / 3. Step 5: In the parallel case, the power dissipated is P_parallel = V^2 / R_eq. Step 6: Substitute R_eq = R / 3 to get P_parallel = V^2 / (R / 3) = 3 * V^2 / R. Step 7: Since P = V^2 / R, we have P_parallel = 3P.

Verification / Alternative check:
We can check with simple numeric values. Let R = 3 ohm and V = 3 V. In the single resistor case, P = V^2 / R = 9 / 3 = 3 W. For three 3 ohm resistors in parallel, each branch gets 3 V. Power in each resistor is P_branch = 9 / 3 = 3 W. There are three such branches, so total power is 3 W + 3 W + 3 W = 9 W. This is exactly 3 times the original 3 W, confirming that the parallel combination dissipates 3P.

Why Other Options Are Wrong:
Option a (P): This would mean the parallel network draws the same power as a single resistor, which is not true because equivalent resistance is smaller. Option c (P/3): This suggests power decreases when adding parallel resistors, which contradicts P = V^2 / R_eq and the reduced resistance. Option d (2P/3): This is an arbitrary fraction and does not follow from the series parallel rules or the calculations. Option e (9P): This would correspond to nine times the original power, which requires either nine resistors or different values; three parallel equal resistors give three times, not nine times, the power.

Common Pitfalls:
A common mistake is to mix up the power formulas P = I^2 * R and P = V^2 / R without paying attention to which quantity (voltage or current) is held constant. In this question, voltage is the same in both cases, so using P = V^2 / R_eq is appropriate. Another error is doing the parallel resistance calculation incorrectly, such as adding resistances instead of their reciprocals. Always remember: series resistances add directly, while parallel resistances use reciprocal addition.

Final Answer:
With the same applied voltage, the parallel combination of three equal resistors dissipates 3P total power.