Worked exercise — assigned powers for three resistors in parallel

Problem

Three resistors $R_1$, $R_2$, $R_3$ are connected in parallel to a generator $\mathcal{E} = 20\;\text{V}$. It is required that the total power be $P_{\text{tot}} = 10\;\text{W}$ and that $P_2 = 2P_1$ and $P_3 = 3P_1$. Find $R_1$, $R_2$, $R_3$.

Three resistors in parallel: each sees the same voltage $\mathcal{E}$.

Solution

Step 1 — main current. From the energy balance $P_{\text{tot}} = \mathcal{E}\,i_0$: $i_0 = \frac{P_{\text{tot}}}{\mathcal{E}} = \frac{10}{20} = 0{,}5\;\text{A}$

Step 2 — individual powers. The constraint $P_1 + P_2 + P_3 = P_1 + 2P_1 + 3P_1 = 6P_1 = 10\;\text{W}$ gives: $P_1 = \frac{5}{3}\;\text{W} \qquad P_2 = \frac{10}{3}\;\text{W} \qquad P_3 = 5\;\text{W}$

Step 3 — currents and resistances. In a parallel connection with fixed $\mathcal{E}$, each branch has $\Delta V = \mathcal{E}$, so $P_k = \mathcal{E}\,i_k$: $i_1 = \frac{5/3}{20} = \frac{1}{12}\;\text{A} \quad i_2 = \frac{10/3}{20} = \frac{1}{6}\;\text{A} \quad i_3 = \frac{5}{20} = \frac{1}{4}\;\text{A}$ The resistances are obtained from $P_k = R_k\,i_k^2$ (equivalently $R_k = \mathcal{E}^2/P_k$): $R_1 = \frac{5/3}{1/144} = 240\;\Omega \qquad R_2 = 120\;\Omega \qquad R_3 = 80\;\Omega$

Check: $i_1 + i_2 + i_3 = \tfrac{1}{12} + \tfrac{1}{6} + \tfrac{1}{4} = \tfrac{1+2+3}{12} = \tfrac{1}{2}\;\text{A} = i_0$. Correct.

$\ev{R_1 = 240\;\Omega,\quad R_2 = 120\;\Omega,\quad R_3 = 80\;\Omega}$

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Topics: Circuiti elettrici Concepts: Potenza