Solutions to Week 4 Assignment

1. We can reduce the circuit to a single loop be successively combining parallel and series combinations.

    We combine R1 and R2, which are in series:

              R7 = R1 + R2 = 2.20kW + 2.20kW = 4.40kW.

   We combine R3 and R7, which are in parallel:

             1/R8 = (1/R3) + (1/R7) = [1/(2.20kW)] + [1/(4.40kW)],

   which gives R8 = 1.47kW.

   We combine R4 and R8, which are in series:

             R9 = R4 + R8 = 2.20kW + 1.47kW = 3.67kW.

   We combine R5 and R9, which are in parallel:

             1/R10 = (1/R5) + (1/R8) = [1/(2.20kW)] + [1/(3.67kW)],      

   which gives R10 = 1.38kW.     

  We combine R6 and R10, which are in series:

            Req = R10 + R6 = 1.38kW + 2.20kW = 3.58kW.

  The current in the single loop is the current through R6:

            I6 = I = e/Req  = (12V)/(3.58kW) = 3.36mA.

   For VAC we have 

            VAC = IR10 = (3.36mA)(1.38kW) = 4.63V.

   This allow us to find I5 and I4;

            I5 = VAC/R5 = (4.63V)/(2.20kW) = 2.11mA;

           I4 = VAC/R9 = (4.63V)/(4.67kW) = 1.26mA.

   For VBC we have 

           VAB = I4R8 = (1.26mA)(1.47kW) = 1.85V.

   This allow us to find I3 and I2;

            I3 = VAB/R3 = (1.85V)/(2.20kW) = 0.84mA;

           I1 =  I2 = VAB/R7 = (1.85V)/(4.40kW) = 0.42mA.

  For above, we have VAB = 1.85V.

  

2. We find the resistance of a bulb from the nominal rating:

            Rbulb = Vnominal2/Pnominal = (12.0V)2/(3.0W) = 48W.

    We find the current through each bulb when connected to the battery from

             Ibulb = V/Rbulb = (11.8V)/(48W) = 0.246A.

    Because the bulbs are in parallel, the current through the battery is

           I = 2Ibulb = 2(0.246A) =  0.492A.

   We find the internal resistance from

           V = e - Ir,

          11.8V = [12.0V - (0.492A)r], which gives r = 0.4W.

3. From the circuit, we wee that C2 and C3 are in series and find their equivalent capacitance from

                 1/C4 = (1/C2) + (1/C3), which gives C4 = C2C3/(C2 + C3).

    From the new circuit, we see that C1 and C4 are in parallel, with an equivalent capacitance

                 Ceq = C1+ C4 = C1+ [C2C3(C2 + C3)]

                        =(C1C2 + C1C3 + C2C3)/(C2 + C3)

                        =3C1/2 = 3(8.8mF)/2 = 13.2mF.

   Because this capacitance is equivalent to the three, the energy stored in it is the energy stored in the network:

                 U = 1/2 CeqV2 = 1/2(13.2 x 10-6F)(90V)2 = 0.053J.

 

  

 4. If we think of a layer of equal and opposite charges on the interface between the two dielectric, we see that they are in series. For the equivalent capacitance, we have      

                  1/C = (1/C1) + (1/C2) = (1/2d/K1e0A) + (1/2d/K2e0A)

                        = (d/2e0A)[(1/K1) + (1/K2)

                        = (1/2C0)[(K1 + K2)/K1K2],

     which gives C = 2C0K1K2/(K1 + K2).