Chapter 28
Problem 46
†
Given
C = 0.025 µF
L = 340 µH
Solution
a) Find the peak current if the peak voltage on the capacitor is 190 V .
When the capacitor is at peak voltage, there is no current flowing in the circuit. Therefore, all of the
energy is in the capacitor and none is in the inductor. Energy in the capacitor is then
UE = 12 CV 2 = 21 (2.5 × 10−8 F )(190 V )2
UE = 4.51 × 10−4 J
When the current is at a maximum, there is no energy stored in the capacitor and it is all in the inductor
in the form of a magnetic field. The relationship for magnetic energy in an inductor is
UM = 21 LI 2
Solving for current gives
r
2UM
I=
L
Due to conservation of energy the electric energy is converted to magnetic energy and the current must
then be
s
2(4.51 × 10−4 J)
= 1.63 A
I=
(3.4 × 10−4 H)
b) Find the time for the peak current.
Since the peak current occurs when the voltage across the capacitor is zero, the time will be 1/4 of an
oscillation. The time period is given by
T =
2π
ω
The angular frequency for the oscillatory circuit is
ω0 = √
1
1
=p
−4
LC
(3.4 × 10 H)(2.5 × 10−8 H)
ω0 = 3.43 × 105 rad/s
The time period is then
T =
2π
= 1.83 × 10−5 s
(3.43 × 105 rad/s)
The time to reach maximum current is then
t = 14 T = 41 (1.83 × 10−5 s) = 4.58 × 10−6 s
t = 4.58 µs
†
Problem from Essential University Physics, Wolfson