082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

478 Chapter 15 Applications of integration
Solution
(a) As the waveform isperiodic, we need only consider the interval 0 t < 10.
I av
= 1 T
∫ T
∫ t2
√
i
t 2 dt
(b) I eff
=
1
t 2
−t 1
Here
0
i(t)dt = 1
10
∫ 10
0
20e −t dt
t 1
=0 t 2
=10 i(t)=20e −t
Firstwe evaluate
So
∫ t2
t 1
i 2 dt=
∫ 10
0
[ e
−2t
= 400
−2
√
200.00
I eff
=
10−0
= 4.4721 A
= 1
10 [−20e−t ] 10
0
= 1
10 (−20e−10 +20e 0 )
= 20
10 (1 −e−10 ) = 2 ×0.99995 = 2.000A
400e −2t dt
] 10
= 400
−2 (e−20 −e 0 )
= 200(1 −e −20 ) = 200.00
0
EXERCISES15.3
1 Calculate the r.m.s.values ofthe functions in
Question1in Exercises15.2.
2 Calculate the r.m.s.values ofthe functions in
Question2in Exercises15.2.
3 Calculate the r.m.s.values ofthe functionsin
Question3in Exercises15.2.
4 Calculate the r.m.s.values ofthe functionsin
Question4in Exercises15.2.
Solutions
1 (a) 2.0817 (b) 1.5275 (c) 0.4472
(d) 1.7889 (e) 6.9666
2 (a) 12.4957 (b) 0.7071 (c) 1
(d) 1.0690 (e) 0.1712
3 (a) 0.7071
(b) 0.7071
√
1
(c)
2 − sinπωcosπω
2πω