Esercizi sul Teorema di Millman

Esercizi sul Teorema di Millman

+Esercizi sul Teorema di Millman

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Esercizi sul Teorema di Millman

Concetti Chiave

Il Teorema di Millman è un metodo utile per semplificare circuiti elettrici che contengono più sorgenti di tensione in parallelo con resistori. Questo teorema permette di calcolare la tensione equivalente ai terminali di un circuito.

Formula del Teorema di Millman

Se abbiamo n$n$ sorgenti di tensione V_i$V_i$ e resistori R_i$R_i$ in parallelo, la tensione equivalente V_{eq}$V_{eq}$ può essere calcolata con la seguente formula:

V_{eq} = \frac{\sum_{i=1}^{n} \frac{V_i}{R_i}}{\sum_{i=1}^{n} \frac{1}{R_i}} $V_{eq} = \frac{\sum_{i=1}^{n} \frac{V_i}{R_i}}{\sum_{i=1}^{n} \frac{1}{R_i}}$

dove:

• V_i$V_i$ è la tensione della sorgente i$i$,
• R_i$R_i$ è la resistenza associata alla sorgente i$i$.

Esercizi

Esercizio 1: Circuito con Due Sorgenti

Considera un circuito con le seguenti sorgenti e resistori:

• Sorgente V_1 = 10 \, V$V_1 = 10 \, V$ con R_1 = 5 \, \Omega$R_1 = 5 \, \Omega$
• Sorgente V_2 = 5 \, V$V_2 = 5 \, V$ con R_2 = 10 \, \Omega$R_2 = 10 \, \Omega$

Calcola la tensione equivalente V_{eq}$V_{eq}$ ai terminali del circuito.

Esercizio 2: Circuito con Tre Sorgenti

Considera un circuito con le seguenti sorgenti e resistori:

• Sorgente V_1 = 12 \, V$V_1 = 12 \, V$ con R_1 = 4 \, \Omega$R_1 = 4 \, \Omega$
• Sorgente V_2 = 6 \, V$V_2 = 6 \, V$ con R_2 = 12 \, \Omega$R_2 = 12 \, \Omega$
• Sorgente V_3 = 3 \, V$V_3 = 3 \, V$ con R_3 = 6 \, \Omega$R_3 = 6 \, \Omega$

Calcola la tensione equivalente V_{eq}$V_{eq}$ ai terminali del circuito.

Soluzioni

Soluzione Esercizio 1

1. Calcolo di V_{eq}$V_{eq}$:

   • Calcoliamo il numeratore:
     \sum_{i=1}^{2} \frac{V_i}{R_i} = \frac{V_1}{R_1} + \frac{V_2}{R_2} = \frac{10 \, V}{5 \, \Omega} + \frac{5 \, V}{10 \, \Omega} = 2 \, A + 0.5 \, A = 2.5 \, A $\sum_{i=1}^{2} \frac{V_i}{R_i} = \frac{V_1}{R_1} + \frac{V_2}{R_2} = \frac{10 \, V}{5 \, \Omega} + \frac{5 \, V}{10 \, \Omega} = 2 \, A + 0.5 \, A = 2.5 \, A$

   • Calcoliamo il denominatore:
     \sum_{i=1}^{2} \frac{1}{R_i} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{5 \, \Omega} + \frac{1}{10 \, \Omega} = 0.2 \, S + 0.1 \, S = 0.3 \, S $\sum_{i=1}^{2} \frac{1}{R_i} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{5 \, \Omega} + \frac{1}{10 \, \Omega} = 0.2 \, S + 0.1 \, S = 0.3 \, S$

   • Ora possiamo calcolare V_{eq}$V_{eq}$:
     V_{eq} = \frac{2.5 \, A}{0.3 \, S} \approx 8.33 \, V $V_{eq} = \frac{2.5 \, A}{0.3 \, S} \approx 8.33 \, V$

Soluzione Esercizio 2

1. Calcolo di V_{eq}$V_{eq}$:

   • Calcoliamo il numeratore:
     \sum_{i=1}^{3} \frac{V_i}{R_i} = \frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3} = \frac{12 \, V}{4 \, \Omega} + \frac{6 \, V}{12 \, \Omega} + \frac{3 \, V}{6 \, \Omega} $\sum_{i=1}^{3} \frac{V_i}{R_i} = \frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3} = \frac{12 \, V}{4 \, \Omega} + \frac{6 \, V}{12 \, \Omega} + \frac{3 \, V}{6 \, \Omega}$

   • Calcoliamo ciascun termine:
     = 3 \, A + 0.5 \, A + 0.5 \, A = 4 \, A $= 3 \, A + 0.5 \, A + 0.5 \, A = 4 \, A$

   • Calcoliamo il denominatore:
     \sum_{i=1}^{3} \frac{1}{R_i} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{4 \, \Omega} + \frac{1}{12 \, \Omega} + \frac{1}{6 \, \Omega} $\sum_{i=1}^{3} \frac{1}{R_i} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{4 \, \Omega} + \frac{1}{12 \, \Omega} + \frac{1}{6 \, \Omega}$

   • Calcoliamo ciascun termine:
     = 0.25 \, S + \frac{1}{12} \, S + \frac{1}{6} \, S $= 0.25 \, S + \frac{1}{12} \, S + \frac{1}{6} \, S$

   • Convertiamo \frac{1}{12} \, S$\frac{1}{12} \, S$ e \frac{1}{6} \, S$\frac{1}{6} \, S$ in termini di dodicesimi:
     = 0.25 \, S + \frac{1}{12} \, S + \frac{2}{12} \, S = 0.25 \, S + \frac{3}{12} \, S = 0.25 \, S + 0.25 \, S = 0.5 \, S $= 0.25 \, S + \frac{1}{12} \, S + \frac{2}{12} \, S = 0.25 \, S + \frac{3}{12} \, S = 0.25 \, S + 0.25 \, S = 0.5 \, S$

   • Ora possiamo calcolare V_{eq}$V_{eq}$:
     V_{eq} = \frac{4 \, A}{0.5 \, S} = 8 \, V $V_{eq} = \frac{4 \, A}{0.5 \, S} = 8 \, V$

English version

Millman's Theorem Exercises

Key Concepts

Millman's Theorem is a useful method for simplifying electrical circuits that contain multiple voltage sources in parallel with resistors. This theorem allows us to calculate the equivalent voltage at the terminals of a circuit.

Millman's Theorem Formula

If we have n$n$ voltage sources V_i$V_i$ and resistors R_i$R_i$ in parallel, the equivalent voltage V_{eq}$V_{eq}$ can be calculated with the following formula:

V_{eq} = \frac{\sum_{i=1}^{n} \frac{V_i}{R_i}}{\sum_{i=1}^{n} \frac{1}{R_i}} $V_{eq} = \frac{\sum_{i=1}^{n} \frac{V_i}{R_i}}{\sum_{i=1}^{n} \frac{1}{R_i}}$

where:

• V_i$V_i$ is the voltage of source i$i$,
• R_i$R_i$ is the resistance associated with source i$i$.

Exercises

Exercise 1: Circuit with Two Sources

Consider a circuit with the following sources and resistors:

• Source V_1 = 10 \, V$V_1 = 10 \, V$ with R_1 = 5 \, \Omega$R_1 = 5 \, \Omega$
• Source V_2 = 5 \, V$V_2 = 5 \, V$ with R_2 = 10 \, \Omega$R_2 = 10 \, \Omega$

Calculate the equivalent voltage V_{eq}$V_{eq}$ at the terminals of the circuit.

Exercise 2: Circuit with Three Sources

Consider a circuit with the following sources and resistors:

• Source V_1 = 12 \, V$V_1 = 12 \, V$ with R_1 = 4 \, \Omega$R_1 = 4 \, \Omega$
• Source V_2 = 6 \, V$V_2 = 6 \, V$ with R_2 = 12 \, \Omega$R_2 = 12 \, \Omega$
• Source V_3 = 3 \, V$V_3 = 3 \, V$ with R_3 = 6 \, \Omega$R_3 = 6 \, \Omega$

Calculate the equivalent voltage V_{eq}$V_{eq}$ at the terminals of the circuit.

Solutions

Solution Exercise 1

1. Calculation of V_{eq}$V_{eq}$:

• Let's calculate the numerator:
  \sum_{i=1}^{2} \frac{V_i}{R_i} = \frac{V_1}{R_1} + \frac{V_2}{R_2} = \frac{10 \, V}{5 \, \Omega} + \frac{5 \, V}{10 \, \Omega} = 2 \, A + 0.5 \, A = 2.5 \, A $\sum_{i=1}^{2} \frac{V_i}{R_i} = \frac{V_1}{R_1} + \frac{V_2}{R_2} = \frac{10 \, V}{5 \, \Omega} + \frac{5 \, V}{10 \, \Omega} = 2 \, A + 0.5 \, A = 2.5 \, A$

• Let's calculate the denominator:
  \sum_{i=1}^{2} \frac{1}{R_i} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{5 \, \Omega} + \frac{1}{10 \, \Omega} = 0.2 \, S + 0.1 \, S = 0.3 \, S $\sum_{i=1}^{2} \frac{1}{R_i} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{5 \, \Omega} + \frac{1}{10 \, \Omega} = 0.2 \, S + 0.1 \, S = 0.3 \, S$

• Now we can calculate V_{eq}$V_{eq}$:
  V_{eq} = \frac{2.5 \, A}{0.3 \, S} \approx 8.33 \, V $V_{eq} = \frac{2.5 \, A}{0.3 \, S} \approx 8.33 \, V$

Solution Exercise 2

1. Calculation of V_{eq}$V_{eq}$:

• Let's calculate the numerator:
  \sum_{i=1}^{3} \frac{V_i}{R_i} = \frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3} = \frac{12 \, V}{4 \, \Omega} + \frac{6 \, V}{12 \, \Omega} + \frac{3 \, V}{6 \, \Omega} $\sum_{i=1}^{3} \frac{V_i}{R_i} = \frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3} = \frac{12 \, V}{4 \, \Omega} + \frac{6 \, V}{12 \, \Omega} + \frac{3 \, V}{6 \, \Omega}$

• Let's calculate each term:
  = 3 \, A + 0.5 \, A + 0.5 \, A = 4 \, A $= 3 \, A + 0.5 \, A + 0.5 \, A = 4 \, A$

• Let's calculate the denominator:
  \sum_{i=1}^{3} \frac{1}{R_i} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{4 \, \Omega} + \frac{1}{12 \, \Omega} + \frac{1}{6 \, \Omega} $\sum_{i=1}^{3} \frac{1}{R_i} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{4 \, \Omega} + \frac{1}{12 \, \Omega} + \frac{1}{6 \, \Omega}$

• Let's calculate each term:
  = 0.25 \, S + \frac{1}{12} \, S + \frac{1}{6} \, S $= 0.25 \, S + \frac{1}{12} \, S + \frac{1}{6} \, S$

• Let's convert \frac{1}{12} \, S$\frac{1}{12} \, S$ and \frac{1}{6} \, S$\frac{1}{6} \, S$ in terms of twelfths:
  = 0.25 \, S + \frac{1}{12} \, S + \frac{2}{12} \, S = 0.25 \, S + \frac{3}{12} \, S = 0.25 \, S + 0.25 \, S = 0.5 \, S$= 0.25 \, S + \frac{1}{12} \, S + \frac{2}{12} \, S = 0.25 \, S + \frac{3}{12} \, S = 0.25 \, S + 0.25 \, S = 0.5 \, S$
  Now we can calculate V_{eq}$V_{eq}$: V_{eq} = \frac{4 \, A}{0.5 \, S} = 8 \, V$V_{eq} = \frac{4 \, A}{0.5 \, S} = 8 \, V$