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Chapter 29
Alternating-Current Circuits
*1 •
Determine the Concept Because the rms current through the resistor is given by
I rms   rms R and both rms and R are independent of frequency, (b) is correct.
*44 ••
Picture the Problem We can apply Kirchhoff’s loop rule to obtain expressions for IR and
IL and then use trigonometric identities to show that I = IR + IL = Imax cos (t  ), where
tan  = R/XL and Imax = max/Z with Z 2  R 2  X L2 .
(a) Apply Kirchhoff’s loop rule to a
clockwise loop that includes the
source and the resistor:
Solve for IR:
(b) Apply Kirchhoff’s loop rule to a
clockwise loop that includes the
source and the inductor:
Solve for IL:
(c) Express the current drawn from
the source in terms of Imax and the
phase constant :
Use a trigonometric identity to
expand cos(t  ):
From our results in (a):
 max cos t  I R R  0
 max cos t
IR 
R
 max cost  90  I L X L  0
because the current lags the potential
difference across the inductor by 90.
 max cost  90
IL 
XL
I  I R  I L  I max cost   
I  I max cos t cos   sin t sin  
 I max cos t cos   I max sin t sin 
I  IR  IL 


 max cos t
R
 max cost  90
XL
 max cos t   max sin t
R
XL
A useful trigonometric identity is:
A cos t  B sin t
 A2  B 2 cost   
where
  tan 1
B
A
  
I   max    max
 R   XL
Apply this identity to obtain:
2
2

 cost    (1)

and
  max

1  X L
  tan
  max
 R

Simplify equation (1) and rewrite
equation (2) to obtain:


  tan 1  R 
X 

 L


  
I   max    max
 R   XL
2
(2)
2

 cost   

2
  max
1  1 
 cost   
   
 R  XL 
  max
1
  cost   
Z

2
2
 max cost   
Z
where
tan  
1
1
1
R
and 2  2  2
Z
R
XL
XL
*115 ••
Picture the Problem The average of any quantity over a time interval T is the integral
of the quantity over the interval divided by T. We can use this definition to find both the
average of the voltage squared, V 2 av and then use the definition of the rms voltage.
 
V 
(a) From the definition of Vrms we
have:
Vrms 
Noting that  V02  V02 , evaluate
Vrms  V02  V0  12.0 V
2
0 av
Vrms:
(b) Noting that the voltage during
the second half of each cycle is now
zero, express the voltage during the
first half cycle of the time interval
1
2 T :
V  V0
Express the square of the voltage
during this half cycle:
Calculate V 2 av by integrating V2
V 2  V02
 
from t = 0 to t =
1
2
T and dividing
V 
2
av

V02
T

1
T
2
0
dt 
V02 12 T 1 2
t 0  2 V0
T
by T:
Substitute to obtain:
Vrms 
1
2
V02 
V0 12 V

 8.49 V
2
2
117 ••
Picture the Problem We can use the definition of capacitance to find the charge on each
capacitor and the definition of current to find the steady-state current in the circuit. We
can find the maximum and minimum energy stored in the capacitors using U  12 CeqV 2 ,
where V is either the maximum or the minimum potential difference across the
capacitors.
(a) Use the definition of capacitance
to express the charge on each
capacitor:
Substitute to obtain:
Q1  C1V1 and Q2  C2V2
or, because the capacitors are in parallel,
Q1  C1V and Q2  C2V
where V    24 V
Q1  C1   24 V 
 3 F20 V cos120t   24 V

60 Fcos120t   72 F
and
Q2  C2   24 V 
 1.5 F20 V cos120t   24 V

(b) Express the steady-state current
as the rate at which charge is being
delivered to the capacitors:
I
30 Fcos120t   36 F
dQ d
 Q1  Q2 
dt dt
Substitute for Q1 and Q2 and
evaluate I:
I
d
60 F cos120t   72 F
dt
 30 F cos120t   36 F
 120 60 F sin 120t 
 120 30 F sin 120t 
  33.9 mA  sin 120t 
2
U max  12 CeqVmax
(c) Express Umax in terms of the
maximum potential difference
across the capacitors:
U max 
Because Vmax = 44 V and
Ceq = C1 + C2 = 3 F + 1.5 F
= 4.5 F:
1
2
4.5 F44 V2 
4.36 mJ
2
U min  12 CeqVmin
(d) Express Umin in terms of the
minimum potential difference across
the capacitors:
U min 
The minimum energy stored in the
capacitors occurs when
Vmin = 24 V  max = 4 V:
1
2
4.5 F4 V2 
36.0 J
119 ••
Picture the Problem We can apply Kirchhoff’s loop rule to express the current in the
circuit in terms of the emfs of the sources and the resistance of the resistor. We can then
find Imax and Imin by considering the conditions under which the time-dependent factor in I
will be a maximum or a minimum. The average of any quantity over a time interval T is
the integral of the quantity over the interval divided by T. We can use this definition to
find average of the current squared, I 2 av and then Irms.
 
Apply Kirchhoff’s loop rule to
obtain:
Solve for I:
1   2  IR  0
I
1   2
R

20 V cos2 180 s 1 t   18 V
I
Substitute numerical values to
obtain:
36 


 0.5 A  0.556 A cos 1131s 1 t


Express the condition that must be
satisfied if the current is to be a
maximum:
cos 1131s 1 t  1
Evaluate Imax:
I max  0.5 A  0.556 A  1.06 A
Express the condition that must be
satisfied if the current is to be a
minimum:
cos 1131s 1 t  1
Evaluate Imin:
I min  0.5 A  0.556 A   0.0560 A
Because the average value of cost
= 0:
I av  0.500 A
Express and evaluate the average
current delivered by the source
whose emf is 2:
I2 



2
R

18 V
 0.5 A
36 

Because I1  0.556 A cos 1131s 1 t :
I 
2
1 av
1
5.56 ms


5.56 ms
0
Use the trigonometric identity cos 2 x 
I 
2
1 av
1
2
0.556 A 2 cos 2 1131s 1 tdt
1  cos 2 x  to obtain:


0.309 A 2 5.56 ms
1  cos 2 1131s 1 t dt
25.56 ms  0


1
 27.8 A / s t 
sin 2262 s 1
-1
2262
s


2


5.56 ms

t
0

 
Evaluate I12
I 
2
1 av
av :


1
 27.8 A 2 / s 5.56 ms 
sin 2262 s-1 5.56 ms   0.1543 A 2
1
2262 s






 
Express I 2
av
:
I 
2
av
   I 
 I12
2
2 av
av
 0.1543 A 2  0.5 A 
2
 0.4043 A 2
Evaluate Irms:
I rms 
I 
2
av
 0.4043 A 2
 0.636 A
*120 ••
Picture the Problem We can apply Kirchhoff’s loop rule to obtain an expression for
charge on the capacitor as a function of time. Differentiating this expression with respect
to time will give us the current in the circuit. We can then find Imax and Imin by
considering the conditions under which the time-dependent factor in I will be a maximum
or a minimum. We can use the maximum value of the current to find Irms.
Apply Kirchhoff’s loop rule to
obtain:
Substitute numerical values and
solve for q(t):
 1   2  qt   0
C


qt   2 F20 V cos 1131s 1 t
 2 F18 V 


 40 Ccos 1131s 1 t  36 C
Differentiate this expression with
respect to t to obtain the current as a
function of time:
I



dq d
40 C cos 1131s 1 t

dt dt
 36 C


  45.2 mA sin 1131s 1 t
Express the condition that must be
satisfied if the current is to be a
minimum:
Express the condition that must be
satisfied if the current is to be a
maximum:
Because the dc source sees the
capacitor as an open circuit and the
average value of the sine function


sin 1131s 1 t  1
and
I min   45.2 mA


sin 1131s 1 t  1
and
I max  45.2 mA
I av  0
over a period is zero:
Because the peak current is
45.2 mA:
I rms 
I max 45.2 mA

 32.0 mA
2
2