Solutions to Homework Questions 7

Physics 112 Homework 7 (solutions) (2004 Fall)
Chapt21, Problem-16: A 2.40-µF capacitor is connected across an alternating voltage with an rms value of
9.00 V. The rms current in the circuit is 25.0 mA. (a) What is the source frequency (b) If the capacitor is replaced by an
ideal coil with an inductance of 0.160 H, what is the rms current in the coil
Solution:
(a)
I rms = !V rms
X C
= (!V rms )( 2" fC), so
I
f = rms
( )2" C = 25.0 # 10 $3 A
( 9.00 V)2" 2.40 #10 $6 F
!V rms
(b)
I rms = !V rms
X L
( )
= !V rms
2" fL = 9.00 V
2" 184 Hz
= 184 Hz
( )( 0.160 H ) = 4.86 # 10$2 A =
48.6 mA
Chapt21, Problem-20: A series AC circuit contains the following components: R = 150 !, L = 250 mH,
C = 2.00 µF, and a generator with "V max = 210 V operating at 50.0 Hz. Calculate the (a) inductive reactance, (b) capacitive
reactance, (c) impedance, (d) maximum current, and (e) phase angle between the current and the generator voltage.
Solution:
X L = 2! fL= 2! ( 50.0 Hz) ( 0.250 H)= 78.5 !
(a)
1
(b) X C =
2! fC = 1
2! 50.0 Hz
( ) = 1.59 " 103 $ =
( ) 2 .00" 10 #6 F
(c) Z = R 2 + ( X L ! X C ) 2 = ( 150 ") 2 + ( 78.5 "!1.59 #10 3 ") 2 =
(d) I max = !V max
Z = 210 V
138 mA
1.52 "10 3 = 0.138 A =
#
(e) ! = tan "1 # X L
" X C & #
% ( = tan "1 78.5 )"1.59* 103 )&
$ R '
%
( = " 84.3° ,
$ 150 ) '
or
the current leads the voltage by 84.3°
1.59 k!
1.52 k!
Chapt21, Problem-25: A person is working near the
secondary of a transformer, as shown in Figure P21.25. The primary
voltage is 120 V (rms) at 60.0 Hz. The capacitance C s , which is the
stray capacitance between the hand and the secondary winding, is 20.0 pF.
Assuming the person has a body resistance to ground of R b = 50.0 k!,
determine the rms voltage across the body. (Hint: Redraw the circuit with
the secondary of the transformer as a simple AC source.
Solution:
1
X C =
2! fC = 1
2! 60.0 Hz
( ) = 1.33" 108 $
( ) 2 + ( 1.33! 10 8 ") 2 = 1.33 !10 8 "
( ) 20.0 "10 #12 F
Z RC = R 2 + X C 2 = 50.0 ! 10 3 "
( ) rms
20.5 H
and I rms = !V 200 V
secondary 5 000 V
=
Z RC
1.33 "10 8 # = 3.76 "10 100 Hz
$5 A
35.0 !
Therefore, !V b,rms = I rms R b = ( 3.76 "10 #5 A) ( 50.0 "10 3 $ )= 1.88 V
C
2