Answer to Question 1

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 a. Remove the 0.8 ohm resistor.  

     Total resistance = 1 + 5 + 4 = 10 ohms

      Current = 12 ÷ 10 = 1.2 A

     Thévenin voltage = 4 × 1.2 = 4.8 V

 

b. Take away the battery and look in.

    Parallel Resistors: 1/R = ¼ + 1/6  = 0.4167 Þ R = 2.4 ohm

 

 c. The circuit is a perfect battery of EMF 4.8 V in series with a

     2.4 ohm internal resistor

     I = 4.8 ÷ (0.8 + 2.4) = 1.5 A