Caso A)

Si riconosce che:

     \[ R_{AB} = (R_1 + R_2) + \frac{(R_3 + R_5) \cdot (R_4 + R_6)}{(R_3 + R_5) + (R_4 + R_6)} \]

     \[ R_{AB} = (50 + 50) + \frac{300 \cdot 300}{300 + 300} = 100 + 150 = 250\,\Omega \] [latex] \bigskip \textbf{Caso B)} [latex] \[ R_{AB} = R_1 + \left\{ R_6 \parallel \left[ (R_2 \parallel R_4) + R_3 + R_5 \right] \right\} \]

     \[ R_{AB} = 50 + 100 \parallel \left[ \left( \frac{50 \cdot 200}{250} \right) + 200 + 100 \right] = 50 + 100 \parallel 340 = 50 + \frac{100 \cdot 340}{100 + 340} = 50 + \frac{34000}{440} = 50 + 77{,}27 = 127{,}27\,\Omega \]

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\textbf{Caso C)}

     \[ R_{AB} = R_1 + R_2 + R_3 \parallel R_4 + R_5 \parallel R_6 = 50 + 50 + 200 \parallel 200 + 100 \parallel 100 \]

     \[ R_{AB} = 100 + 100 + 50 = 250\,\Omega \]

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\textbf{Caso D – Deviatori entrambi chiusi:}

     \[ R_{AB} = R_1 + R_{56} = 50 + \left( \frac{100 \cdot 100}{100 + 100} \right) = 50 + 50 = 100\,\Omega \]