0195136047.pdf

(Joyce) #1

28 CIRCUIT CONCEPTS


Since

i(t)=C

dv
dt

=( 5 × 10 −^6 )

dv
dt
it follows that
i(t)= 0 ,t≤− 1 μs
=25 mA, − 1 ≤t≤ 1 μs
= 0 , 1 ≤t≤ 3 μs
=−50 mA, 3 ≤t≤ 4 μs
= 0 , 4 ≤tμs

which is sketched in the center of Figure E1.2.3(a).
Since the energy stored at any instant is

w(t)=

1
2

Cv^2 (t)=

1
2

( 5 × 10 −^6 )v^2 (t)

it follows that:
w(t)= 0 ,t≤− 1 μs
= 62. 5 (t^2 + 2 t+ 1 )pJ, − 1 ≤t≤ 1 μs
=250 pJ, 1 ≤t≤ 3 μs
= 250 (t^2 − 8 t+ 16 )pJ, 3 ≤t≤ 4 μs
= 0 , 4 ≤tμs

which is sketched at the bottom of Figure E1.2.3(a).
(b) From Figure E1.2.3(b) it follows that
i(t)= 0 ,t≤− 1 μs
= 5 (t+ 1 )mA, − 1 ≤t≤ 1 μs
=10 mA, 1 ≤t≤ 3 μs
=− 10 (t− 4 )mA, 3 ≤t≤ 4 μs
= 0 , 4 ≤tμs

Since

v(t)=

1
C

∫t

−∞

i(τ) dτ =

1
5 × 10 −^6

∫t

−∞

i(τ) dτ

it follows that
v(t)= 0 ,t≤− 1 μs

=

(
t^2
2

+t+

1
2

)
mV, − 1 ≤t≤ 1 μs
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