Problem — same voltage, different resistances

Problem

Four resistors are supplied separately at the same voltage $V = 12\ \text{V}$: (A) $R = 1\ \Omega$; (B) $R = 10\ \Omega$; (C) $R = 100\ \Omega$; (D) $R = 1000\ \Omega$. Rank the four resistors by power dissipated, from greatest to smallest.

Solution

Principle — at fixed voltage, the power dissipated by a resistor is $P = \frac{V^2}{R}$ that is, inversely proportional to the resistance: the smaller $R$, the greater the power. Do not confuse this with $P = R\,i^2$, valid when the current is fixed; here the voltage is fixed, so the correct form is $V^2/R$.

Calculation (with $V^2 = 144\ \text{V}^2$): $P_A = \frac{144}{1} = 144\ \text{W}, \quad P_B = \frac{144}{10} = 14{,}4\ \text{W}$ $P_C = \frac{144}{100} = 1{,}44\ \text{W}, \quad P_D = \frac{144}{1000} = 0{,}144\ \text{W}$

Ranking — the smallest resistance dissipates the most: $P_A > P_B > P_C > P_D$

$\ev{A > B > C > D}$

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Topics: Circuiti elettrici Concepts: Potenza Skills: Ragionamento di ranking