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 are | 0.5 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.
Last revised 1998/03/29 |