0195136047.pdf

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