Solutions
Chapter 11
Problem 1
Part a
I = Em /Z,
s
R2 + ωd L −
Z=
1
ωd C
2
As there is no resistor, R = 0. As there is no capacitor, C = ∞ or
I=
1
ωd C
= 0. So Z = ωd L, and
Em
= 4.17 × 10−3 A.
ωd L
Part b
i = I sin(ωd t − φ)
where φ is given by:
tan φ =
ωd L −
1
ωd C
R
= ∞.
So, φ = π/2 and
i = I sin(ωd t − π/2) = −I cos(ωd t)
Hence, i is a maximum when ωd t = π, 3π, 5π, . . . . At these times sin(ωd t) = 0. So, at these times,
E = Em sin(ωd t) = 0.
Problem 2
Part a
I = Em /Z,
s
1 2
Z=
+ ωd L −
ωd C
As there is no resistor, R = 0. As there is no inductor, L = 0. So Z = 1/(ωd C), and
R2
I = Em ωd C = 5.00 × 10−2 A.
1
Part b
i = I sin(ωd t − φ)
where φ is given by:
tan φ =
ωd L −
1
ωd C
R
= −∞.
So, φ = −π/2 and
i = I sin(ωd t + π/2) = I cos(ωd t)
Hence, i is a maximum when ωd t = 0, 2π, 4π, . . . . At these times sin(ωd t) = 0. So, at these times,
E = Em sin(ωd t) = 0.
Problem 3
Part a
The resonant frequency is
√
ωd = 1/ LC = 500rad/sec
Part b
The current amplitude is
I=
Em
Em
=p 2
Z
R + (XL − XC )2
At resonance, XL = XC , hence,
I=
Em
10
=
= 2.50 A
R
4
Also at resonance
ωd = √
1
LC
and the voltage amplitude across the inductor is
IL
VL = IXL = Iωd L = √
=I
LC
s
L
= 2500 V
C
Part c
The inductor may have a higher voltage across it than the source because the capacitor voltage
can partially (or completely in case of resonance) cancel the inductor voltage.
2
Problem 4
tan φ =
XL − XC
R
Hence,
R=
2πfd L − 2πf1d C
ωd L − ωd1C
XL − XC
=
=
= 140 Ω
tan φ
tan φ
tan φ
Problem 5
Part a
s
Z=
R2 + ωd L −
1
ωd C
2
= 67 Ω
Part b
I=
Em
= 6.7 A
Z
Part c
The phase constant is given by
tan φ =
ωd L − 1/(ωd C)
= 1.35
R
Hence,
φ = 53◦
Problem 6
...................................................................................................
..
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i
...
..
...
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>
>
r>
>
>
>
>
..>
.
R>
>
.. ..... .........
>
>
E ..... ∼ .....
>
............
...>
..
...
..
...
..
..................................................................................................
3
Power dissipation in R is
2
P = Irms
R
where
Irms =
Erms
Erms
=
Z
r+R
So,
R
2
P = Erms
(r + R)2
The value of R for which P is maximum, must obey the maxima condition from calculus:
This gives:
1
2R
2
Erms
−
= 0.
(r + R)2 (r + R)3
The solution to this equation is:
r = R.
Problem 7
The voltage of the secondary is
Vs = Ns
Vp
= 4.0 V
Np
4
dP
dR
= 0.