Solution :
\( P = R \times I^2 = 50 \times 4^2 = 800 \, \text{W} \)

Solution :
\( E = P \times t = 0,8 \, \text{kW} \times 2 \, \text{h} = 1,6 \, \text{kWh} \)

Solution :
\( \eta = \frac{P_{\text{utile}}}{P_{\text{reçue}}} = \frac{400}{500} = 0,8 \) soit 80%

Solution :
\( I = \frac{E}{R + r} = \frac{12}{6} = 2 \, \text{A} \)
Puissance fournie : \( P_{\text{gen}} = E \times I = 12 \times 2 = 24 \, \text{W} \)
Puissance dans R : \( P_R = R \times I^2 = 5 \times 4 = 20 \, \text{W} \)

Solution :
\( R = \rho \frac{L}{S} = 1,7 \times 10^{-8} \times \frac{10}{\pi (10^{-3})^2} \approx 0,054 \, \Omega \)
\( E = R I^2 t = 0,054 \times 25 \times 3600 \approx 4860 \, \text{J} \)

Solution :
\( I = \frac{E – E’}{r + r’} = \frac{3}{3} = 1 \, \text{A} \)
\( \eta = \frac{E’ \times I}{(E’ + r’ I) \times I} = \frac{9 \times 1}{11 \times 1} \approx 81,8\% \)

Solution :
\( I = \frac{P}{U} = \frac{36}{12} = 3 \, \text{A} \)
\( t = \frac{50}{3} \approx 16,7 \, \text{heures} \)

Solution :
\( P = R I^2 = R \left(\frac{E}{R + r}\right)^2 \)
La puissance est maximale quand \( R = r \) (théorème de la puissance maximale)

Solution :
\( E = 2 \times 1,5 \times 30 = 90 \, \text{kWh} \)
Coût = \( 90 \times 0,8 = 72 \, \text{DH} \)

Solution :
1. \( I = \frac{E – E’}{r + r’ + R} = \frac{6}{9} \approx 0,67 \, \text{A} \)
2. \( P_{\text{utile}} = E’ I = 18 \times 0,67 \approx 12 \, \text{W} \)
3. \( \eta = \frac{E’ I}{E I} = \frac{12}{24 \times 0,67} \approx 75\% \)