ANSWER TO CHAPTER 25 HOMEWORK

1.  For the device we have V = IR.

         (a)  If we assume constant resistance and divide expressions for the two conditions, we get

                     V2/V1 = I2/I1;

                     0.90 = I2/(5.50A), which gives I2 = 4.95A.

         (b)  With the voltage constant, if we divide expressions for the two conditions, we get

                     I2/I1 = R1/R2;

                     I2/(5.50A) = 1/0.90, which gives I2 = 6.11A.

2.  (a)  We find the resistance from

                     V= IR;

                     12V = (7.5A)R, which gives R =  16W.

          (b) The charge that passes through the hair dryer is

                     DQ = IDt = (7.5A)(15min)(60s/min) =  6.8 x 103C.

3.  We find the temperature change from 

                     R = Ro(1 + aDT), or DR = RoaDT

                     0.20Ro = Ro[0.0068(C°)-1]DT, which gives DT = 29°C.

4.   90A.h is the total charge that passed through the battery when it was charged. We find the energy from

                     Energy = Pt = VIt = VQ = (12V)(90A.h)(10-3kW/W) =  1.1kWh =  3.9 x 106 J.

5.   The peak voltage is

                     Vo = Ö2Vrms = Ö2(450V) =  636V.

          We find the peak current from

                     P = IrmsVrms  = (IoÖ2)Vrms;

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

6.  From Example 25-12 we know that the density of free electrons in copper is

                       n = 8.4 x 1028m-3.

         (a)  We find the drift speed from

                    I = neAvd = ne(1/4pD2)vd; 

                    2.5 x 10-6A = (8.4 x 1028m-3)(1.60 x 10-19C)[1/4p(0.55 x10-3m)2]vd,

                     which gives vd =  7.8 x10-10m/s.

          (b)  The current density is

                     j = I/A = (2.5 x10-6A)/1/4p(0.55 x 10-3m)2 =  10.5A/m2 along the wire.

          (c) We find the electric field from

                    j = E/r;

                   10.5A/m2 = E/(1.68 x 10-8W.m), which gives E = 1.8 x 10 -7V/m.

7*. WE choose a spherical shell of radius r and thickness dr as a differential element. The area f the element is 4pr2. We add (integrate) the resistances of all the shells:

                   R = ò r1r2(1/s4pr2)dr = -(1/s4p)(1/r2 - 1/r1)

                      = (r2 - r1)/4psr1r2.

8* For the water to remove the thermal energy produced, we have

                  P = IV = (m/t)cDT;

                  (14.5A)(240V) = (m/t)(4186J/kg·C°)(6.50C°), which gives m/t = 0.128kg/s.

 

9*. If we take north as the positive direction, for the current density we have

                  I/A = n+(+2e)vd+ + n-(-e)vd-

                        = (2.8 x 1012m-3)(2)(1.6 x 10-19C)(2.0 x 106m/s) + (8.0 x 1011m-3)(-1.6 x 10-19C)(-7.2 x 106m/s)

                        = 2.7A/m2 north.

 

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