Solutions to Week 3 Assignment

1. The rate at which electrons leave the battery is the current:

                I = V/R = [(9.0V)/(1.6W)](60s/min)/(1.60 x 10^-19C/electron) = 2.1 x 10²¹electrons/min.

2. We find the temperature from

               r_0,T = r_cu = r_0,cu(1 + a_cuDT);

              5.6 x 10^-8W·m = (1.68 x 10^-8W·m){1 + [0.0068(C°)^-1](T - 20°C), which gives T = 363°C.

3.. We find the operating resistance from

              P = V²/R;

              60W = (240V)²/R, which gives R = 9.6 x 10²W.

      If we assume that the resistance stays the same, for the lower voltage we have

              P = V²/R = (120V)²/(9.6 x 10²W) = 15W.

      At one-quarter the power, the bulb will be much dimmer.

4. The required current to deliver the power is I = P/V, and the wasted power (thermal losses in the wires) is P_loss = I²R. For the two conditions we have

              I₁ = (520kW)/(12kV) = 43.3A; P_loss,1 = (43.3A)²(3.0W)(10^-3kW/W) = 5.63kW;

              I₂ = (520kW)/(50kV) = 10.4A; P_loss,2 = (10.4A)²(3.0W)(10^-3kW/W) = 0.324kW;

     Thus the decrease in power loss is 

              DP_loss = P_loss,1  - P_loss,2  = 5.633kW - 0.324kW = 5.3kW.

5. The peak voltage is 

               V₀ = Ö2V_max = Ö2(450V) = 636V.

   We find the peak current from

               P = I_rmsV_rms = (I₀/Ö2)V_rms;

               1800W = (I₀/Ö2)(450V), which gives I₀ =5.66A.              

6. The equivalent resistance of the two resistors connected in series is

             R_s = R₁+ R₂.

    We find the equivalent resistance of the two resistors connected in parallel from

            1/R_p = 1/R₁ + 1/R₂, or R_p = R₁R₂/(R₁ + R₂).

   The power dissipated in a resistor is P = I²R, so the ratio of the tow powers is

            P_p/P_s = R_p/R_s = (R₁ + R₂)²/R₁R₂= 4.

   when we we expand the square, we get

           R₁² + 2R₁R₂ + R₂² = 4R₁R₂, or (R₁ - R₂)² = 0, which gives R₁ = R₂ = 2.20kW.