Chapter 28
Problem 41
†
Given
C = 1.2 µF
v = Vp sin(2πf t)
VP = 22 V
f = 60 Hz
Solution
a) Find the peak current.
First find the angular frequency.
ω = 2πf = 2π(60 Hz) = 120π rad/s
The capacitive reactance is
XC =
1
1
=
ωC
(120π rad/s)(1.2 × 10−6 F )
XC = 2210 Ω
Now with the capacitor attached to the power supply the peak current can be found through Ohm’s law
and replacing the resistance with the capacitive reactance.
IP =
VP
22 V
=
XC
2210 Ω
IP = 0.00995 A = 9.95 mA
b) Find the voltage at t = 6.5 ms.
Substituting the time into the voltage function gives
v = VP sin(2πf t)
v = (22 V ) sin(2π(60 Hz)(6.5 × 10−3 s))
Remember to use the radians mode for this calculation.
v(6.5 ms) = 14.0 V
c) Find the current at t = 6.5 ms.
†
Problem from Essential University Physics, Wolfson
The current function will lead the voltage function by π/2 and, therefore, the function will be
i = IP sin(2πf t + π/2)
Substituting the time into the current function gives
i = (9.95 mA) sin(2π(60 Hz)(6.5 × 10−3 s) + π/2)
Remember to use the radians mode for this calculation.
i(6.5 ms) = −7.67 mA
The magnitude is then 7.67 mA.