ANSWER TO CHAPTER 26 HOMEWORK

1.  We find the internal resistance from

               V = e/r;

               9.8V = [12.0V - (60A)r], which gives r = 0.037W.

         Because the terminal voltage is the voltage across the starter, we have

               V = IR;

            9.8V = ()(60A)R, which gives R =  0.16W.

2. When the resistors are connected in series, as shown in A, we have

                RA = SRi = 3R = 3(1.20kW) = 3.60kW.

         When the resistors are connected in parallel, as shown in B, we have

                1/RB = S(1/Ri) = 3/R = 3/(1.20kW), so RB =  0.40kW.) 

          In circuit C, we find the equivalent resistance of the two resistors in parallel:

                1/R1 = S(1/Ri) = 2/R 2/(1.20kW), so R1 = 0.60kW.

          This resistance is in series with the third resistor, so we have

                 RC = R1 + R = 0.60kW + 1.20kW =  1.80kW.

           In circuit D, we find the equivalent resistance of the two resistors in series:

                 R2 = R + R = 1.20kW + 1.20kW + 2.40kW.

           This resistance is in parallel with the third resistor, so we have

                 1/RD = (1/R2) + (1/R) = (1/2.40kW) + (1/1.20kW), so RD =   0.80kW.

  
3.  (a)  When the switch is closed the addition of R2 to the parallel  set will decrease the equivalent resistance, so the current from the battery will increase. This causes an increase in the voltage across R1, and a corresponding decrease across R3 and R4. The voltage across R2 increases from zero. Thus we have  V1 and V2 increase; V3 and V4 decrease.

       (b)  The current through R1 has increased. This current is now split into three, so currents through R3 and R4 decrease. Thus we have

               I1 (=I) and I2 increase; I3 and I4 decrease.

       (c)  The current through the battery has increased, so the power output of the battery  increases.

       (d)  Before the switch is closed,  I2 = 0.   We find the resistance for R3 and R4 in parallel from

               1/RA = S(1/Ri) = 2/R3 = 2/(100W),

              which gives RA = 50W.    

              For the single loop, we have

                I = I1 = V/(R1 + RA)

                  = (45.0V)/(100W + 50W) =  0.300A.  

              This current will split evenly through R3 and R4:

                I3 = I4 =1/2I = 1/2(0.300A) =  0.150A.

              After the switch is closed, we find the resistance  for R2, R3, and R4 in parallel from

                1/RB = S(1/Ri) = 3/R3 = 3/(100W),

              which gives RB = 33.3W.

              For the single loop, we have

                  I = I1 = V/(R1 + RB)

                    = (45.0V)(100W + 33.3W) =  0.338A.

               This current will split evenly through R2, R3, and R4:

                   I2 = I3 = I4 = 1/3I = 1/3(0.338A) =  0.113A.

             

  
4. For  the conservation of current at point C, we have

                  Iin = Iout;

                  I1 = I2 + I3.

        For the two loops indicated on the diagram, we have

                  loop 1:  V1 - I2R2 - I1R1 = 0;

                               + 9.0V - I2(15W) - I1(22W) = 0;

                  loop 2:  V3 + I2R2 = 0;

                               +6.0V + I2(15W) = 0.

          When we solve these equations, we get

                   I1 = 0.68A,  I2 = -0.40A,  I3 = 1.08A.

           Note that I2 is opposite to the direction shown.         

            

  
5. The given current is I2  = -0.30A. For  the conservation of current at point a, we have

                     I2 + I3 = I1, or  -0.30A + I3 = I1.

        For the top loop indicated on the diagram, we have 

                     loop 1;  e1 - I1r1 -I1R3 + e2 - I2r2 - I2R2 - I1R1 = 0;

                       +12.0V - I1(1.0W) - I1(8.0W) + 12.0V - (-0.30A)(1.0W) - (-0.30A)(10W) - I1(12W) = 0,

        which gives  I1 = 1.30A.

        Thus we have

                                 I3 = I1 - I2 = 1.30A - (-0.30A) = 1.60A.

         For the bottom loop indicated on the diagram, we have

                      loop2:  e - I3r - I3R5 + I2R2 - e2 + I2r2 - I3R4 = 0;

                e - (1.60A)(1.0W) - (1.60A)(18W) + (-0.30A)(10W) - 12.0V + (-0.30A)(1.0W)-                         (1.60A)(15W) = 0,

         which gives  e =  70V.

                

  

6.   The time constant of the circuit is

                  t = RC = (6.7 x 103W)(6.0 x 10-6F) = 0.0402s.

          The capacitor voltage will decrease exponentially;

 

                  0.01Vo = Voe-t/(00402s),  or t/(0.0402s) = ln(100) = 4.61,

          which gives  t =  0.19s.

7*.  We find the resistance for R1 and R2 in parallel from 

                 1/Rp = (1/R1) + (1/R2) = [1/(3.8kW) + (1/2.1kW)],

which gives Rp = 1.35kW.

Because the same current passes through Rp and R3, the higher resistor will have the higher power dissipation, so the limiting resistor is R3, which will have a power dissipation of 1/2W. We find the current from 

                 P3max = I3max2R3;

                 0.50W = I3max2(1.8 x 103W), which gives I3max = 0.0167A.

The maximum voltage for the network is

                 Vmax  = I3max(Rp + R3)

                          = (0.0167A)(1.35 x 103W + 1.8 x 103W) = 53V.

 
8*.  (a) On the diagram, we show the potential difference applied between points a and c. Because all of the resistors are the same, symmetry means that the three currents leaving point a must be the same three currents entering point c. This means that there is no current in the resistor between points b and d, which can be removed without changing the currents. When we redraw the circuit, we see that we have three parallel branches between points a and c.

We find the equivalent resistance from

              1/Rac = (1/R) + (1/2R) + (1/2R),

which gives Rac = R/2.

(b) Ifr we apply a potential difference between points a and b, the same symmetry exists, so all currents leaving a must enter b. Consequently there is no current in R', which can be removed. When we redraw the circuit, we get one identical to that in part (a), so the equivalent resistance is R/2.

9*. (a) For the conservation of current at point a, we have

                    I1 = I2 + I3.

  For the loop on the right, we have

                   +I2R2  - Q/C = 0, or I2R2 = Q/C.

For the outside loop, we have

                  e - I1R1 - Q/C = 0, or I1R1 = e - Q/C.

The current I3 is charging the capacitor: I3 = dQ/dt.

When we use these results in the junction equation, we get

                 (e - Q/C)/R1 = (Q/R2C) + dQ/dt, which becomes

                 e = R1dQ/dt + (R1 + R2)Q/R2C.

This has the same form as the simple RC circuit:

                 e  = RdQ/dt + Q/C,

if we replace R with R1, and C with R2C/(R1 + R2).

Thus the time constant is 

                  t = R1R2C/(R1 + R2).

(b) After a long time, the current through the capacitor wwill be zero and it will have its maximum charge. The current through the resistors will be

                  I = e/(R1 + R2).

The voltage across the capacitor will be the voltage across R2:  V = IR2. Thus the charge on the capacitor will be

                 Qmax = IR2C = eR2C/(R1 + R2).

 

 

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