Answers for Assignment 6

30. d = 2 mm

31. [Agrees with HRW]:

      (a) 3.1 x 1011 electrons

      (b) 25 microamps

      (c) average power 1300 watts [except that 1250 should be rounded to 1200, not 1300] and peak power 25 Megawatts

32. R = rho L / (pi a b)
If a=b, R = rho L A

33.

Reff = (R1 R2 + R1 R3 + R2 R3 ) / (R2 + R3 )

I1 = (R2 + R3 ) V / (R1 R2 + R1 R3 + R2 R3 )

I2 = R3 V / (R1 R2 + R1 R3 + R2 R3 )

I3 = R2 V / (R1 R2 + R1 R3 + R2 R3 )

V1 = R1 (R2 + R3 ) V / (R1 R2 + R1 R3 + R2 R3 )

V2 = V3 = R2 R3 V / (R1 R2 + R1 R3 + R2 R3 )

34.
The currents are0.5 V/R in the left branch,

0.25 V/R in the center branch, and

0.25 V/R in the right branch.
The left-hand battery supplies 1 V2/R watts
the upper battery supplies 0.5 V2/R watts. and
the right-hand battery absorbs 0.25 V2/R watts.

From left to right, the resistors dissipate 0.25 V2/R, 0.5 V2/R, 0.125 V2/R, and 0.375 V2/R watts. All the power dissipated by the resistors comes from the power supplied by the batteries, of course [1.25 V2/R watts total].

35.

Req = (5/3) R
Currents:

                ---- (1/5) (V/R) ---------------------
               |                                      |
        a------ -- (2/5) (V/R) --- --- (1/5) (V/R) --- ---- (2/5) (V/R) - --b
        |                         |                                      |  |
        |                          ---- (1/5) (V/R) ---------------------   |
        |                                                                   |
         ---------------------- (3/5) (V/R) --------------------------------

36.

Using MATHCAD,

MATHCAD lists the solutions from the top down, so the top line is I1, the next line is I2, etc.


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Last revised 1998/03/29