14.3 Solutions
663
14.29 For parallel resonance circuit
ω = ω 0
1 −
CR
2
L
ω 0 =
1
√
LC
=
1
√
10 −3 × 5 × 10 −6
= 1.414214 × 10
6 rad/s
f =
ω
2π
=
ω 0
2π
1 −
CR
2
L
=
1.414214 × 10 6
2π
1 −
5 × 10 −10 × 10 2
10 −3
= 2.250736 × 10
5
/s
Quality factor
Q =
ωL
R
=
1.414 × 10 6 × 10 −3
10
= 141.4
14.30 V = V 0 e
−t/RC
ln V = ln V 0 −
t
RC
V
V
= 0 −
t
C
d
dR
1
R
R −
t
R
d
dC
1
C
C
=
t
C R
R
R
+
C
C
=
50 × 10 −6
10 −8 × 5 × 10 4
5
100
+
10
100
= 0.015
14.31 (a) V = ξ e −t/RC
The voltage on the condensor will fall to l/e of its initial value when the
time
t = RC = 10
4
× 10
−5
= 0.1 s
(b) Error on t will result from error on R.
t = CR
∴ t/t = R/R
Power P = i 2 R =
ξ 2
e 2 R
=
(3000) 2
(2.718) 2 × 10 4 = 121.8 W
Energy U = P.t = 121.8 × 0.1 = 12.18 J
Heat H = 12.18/4.18 = 2.914 cal