ECEU210 Solutions to HW #6
P 3.3
P 3.17
i (0) = -10mA i ( ) = 0 A
=
L 1 .5 = = 1.5ms R 1000
-3 -t 1.5 ms
i (t ) = -10 10 (1 - e
-t
)A
di v(t ) = L = -10 e1.5 ms V dt
P 3.23
P 3.25
P 3.40
P 3.41
L 5 10 -3 = = = 50 10 -6 s = 50 s R 100
ECEU210 Solutions to HW #5
P 3.6
P 3.7
P 3.19
P 3.22
P 3.29
First we see, from v() = 10R/(100+R) that for R = 8.1 v() = 0.75, so the transient will never reach that value in finite time; therefore R > 8.1. The inequality is too complex for an al
ECEU210 Solutions to HW #4
P 9.37
P 9.38
P 9.39
P 9.40
P 9.49
b a
3 k
2V 2 k
va = 0 (because + terminal is grounded) i1 = 2/2000 = 1 mA i2 = - 1 mA (KCL) vb = -0.001 * 3000 = - 3 V
ECEU210 Solutions to HW #1
P 1.6
!
P 1.14
P 1.20
Comment for P 1.20 (c) -> The current and voltage references for the 4-volt battery are a "load set", so the power output of the battery is p=-iv. The numerical result of the power output is negativ
ECEU210 Solutions to HW #3
P 10.1
P 10.2
P 10.3
P 10.4
Refer to Figure P10.4 on page 562. The Thevenin equivalent circuit for the potentiometer is VT = Voc = xV Req = xRp | (1-x)Rp =
x(1-x) Rp2 / [xRp + (1-x) Rp] = x(1-x)Rp
A load RL will reduce