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Problem set #8, EE 221, 10/24/2002 – 10/31/2002
Chapter 6, Problem 16.
Compute vx for the multiple op
amp circuit of Fig. 6.36.
Chapter 6, Solution 16.
The 3 mA source, 1 kΩ resistor and 20 kΩ resistor may be replaced with a –3 V source (“+” reference up) in
series with a 21 kΩ resistor. No current flows through either 1 MΩ resistor, so that the voltage at each of the
four input terminals is identically zero. Considering each op amp circuit separately,
vout
LEFTOPAMP
vout
RIGH OPAMP
100
= 14.29 V
21
100
= - (5)
= - 50 V
10
= - (-3)
vx = vout
LEFTOPAMP
- vout
RIGH OPAMP
= 14.29 + 50 = 64.29 V.
Chapter 6, Problem 22.
Compute vout for the circuit of
Fig. 6.38.
Chapter 6, Solution 22.
vout of stage 1 is (1)(-20/ 2) = -10 V.
vout of stage 2 is (-10)(-1000/ 10) = 1000 V
Note: in reality, the output voltage will be limited to a value less than that by the sources used to power
the op amps, e.g., ±15V.
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Chapter 7, Problem 1.
A capacitor is fabricated from two thin aluminum disks 1 cm in diameter which are separated by a distance of
100µm (0.1 mm). (a) Compute the capacitance, assuming only air between the metal plates. (b) Determine the
voltage that must be applied to store a measly 1 mJ of energy in the capacitor. (c)If the capacitor is needed to
store 2.5µJ of energy in an application that can supply up to 100 V, what value of relative permittivity εr=ε/ε0
would be required for the region between the plates?
Chapter 7, Solution 1.
(a) C =
εA
d
=
8.854 × 10 −12 (78.54 × 10 −6 )
= 6.954 pF
100 × 10 − 6
1
2E
2(1 × 10 − 3 )
2
=
= 16.959kV
(b) Energy , E = CV ∴V =
C
2
6.954 × 10 −12
(c) E =
C=
εA
d
1
2 E 2(2.5 × 10 −6 )
CV 2 ∴ C = 2 =
= 500 pF
2
(100 2 )
V
∴ε =
Cd (500 × 10 −12 )(100 × 10 −6 )
=
= 636.62 pF .m −1
−6
A
(78.54 × 10 )
ε 636.62 × 10 −12
∴ Re lative _ permittivity, =
= 71.9
ε 0 8.854 × 10 −12
Chapter 7, Problem 4.
A voltage v(t)
is applied to a 300-µF capacitor.
(a) Compute the energy stored in the capacitor at t = 2 ms.
(b) At what time has the energy stored in the capacitor dropped to 37 percent of its maximum value? (Round to
the nearest second.)
(c) Determine the current flowing through the capacitor at t = 1.2 s.
(d) Calculate the power delivered by the capacitor to the external circuit at t = 2 s.
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Chapter 7, Solution 4.
(a)
(b)
t
t
−  3 − 
2×10 − 3
dv

5
5
= C∫
3e . − e dt = −1.080µJ
change in stored energy = ∫ v.C
 5

0
t0
dt


t
(absolute) energy stored at t ≥ 0+ with respect to t0 ≤ 0−: w(t) = C/2 v2(t) = 1.35 e−2t/5 mJ
Vmax = 3V
Max. energy at t=0, =
1
CV 2 = 1.35mJ ∴ 37% E max = 499.5µJ
2
V at 37% Emax = 1.825V
v (t ) = 1.825 = 3e
−
t
5
∴ t = 2.486s ⇒≈ 2 s
(c)
1.2 

−
dv
−6  3
i =C
= 300 × 10
− e 5  = −141.593µA


dt
5


(d)
P = − v i = − 2.011(−120.658*10-6) = 242.6µW
Chapter 7, Problem 11.
With reference to Fig. 7.43:
(a) sketch vL as a function of time, 0 < t < 60 ms;
(b) find the value of time at which the inductor is
absorbing a maximum power;
(c) find the value of time at which it is supplying a
maximum power; (d) find the energy stored in the
inductor at t = 40 ms.
Chapter 7, Solution 11.
(a) (note: 20ms should read 10ms)
(b)
PL = vL iL = (−100)(−5) = 500 W at t = 40− ms
(c)
PL min = 100(−5) = −500W at t = 20+ and 40+ ms
(d)
WL =
1 2
1
Li L ∴ WL (40ms) = × 0.2(−5) 2 = 2.5J
2
2
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Chapter 7, Problem 18.
With reference to the circuit shown in Fig.
7.47, find
(a) wL ;
(b) wC ;
(c) the voltage across each circuit element;
(d) the current in each circuit element.
Chapter 7, Solution 18.
(a)
1
wL = × 5 × 1.62 = 6.4J
2
(b)
1
wc = × 20 × 10−6 ×1002 = 0.1J
2
(c)
Left to right (magnitudes): 100, 0, 100, 116, 16, 16, 0 (V)
(d)
Left to right (magnitudes): 0, 0, 2, 2, 0.4, 1.6, 1.6 (A)
Chapter 7, Problem 23.
For the circuit of Fig. 7.52,
(a) reduce the circuit to the fewest
possible components using
series/parallel combinations;
(b) determine vx if all resistors are
10 kΩ, all capacitors are 50 µF,
and all inductors are 1 mH.
Chapter 7, Solution 23.
(a) Assuming all resistors have value R, all inductors have value L, and all capacitors have value C, (note: 3.5R should read
1.5R)
(b) At dc, 20µF is open circuit; 500µH is short circuit.
Using voltage division, V x =
10k
(9 ) = 3.6V
10k + 15k
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Chapter 7, Problem 36.
In the circuit shown in Fig. 7.61,
let is = 60e-200t mA with i1(0) = 20
mA.
(a) Find v(t) for all t .
(b) Find i1(t) for t ≥ 0.
(c) Find i2(t) for t ≥ 0.
Chapter 7, Solution 36.
is = 60e −200t mA, i1 (0) = 20mA
(a)
6 4 = 2.4H ∴ v = Leq is′ = 2.4 × 0.06(−200)e −200t
or v = −28.8e −200t V
(b)
(c)
1 t
4.8 −200t
−28.8e−200t dt + 0.02 =
− 1) + 0.02
(e
∫
o
6
200
= 24e−200t − 4mA(t > 0)
i1 =
i2 = is − i1 = 60e−200t − 24e−200t + 4 = 36e−200t + 4mA(t > 0)
Chapter 7, Problem 44.
A velocity sensor is attached to a rotating wheel. Design a circuit to provide a positive voltage whose magnitude
is equal to the acceleration (revolutions per minute per minute) of the wheel. Assume the velocity sensor’s
output is 1 mV/rpm, and the wheel rotates at less than 3500 rpm.
Chapter 7, Solution 44.
Create a op-amp based differentiator using an ideal op amp with input capacitor C1 and feedback resistor Rf
followed by inverter stage with unity gain.
Vout = +
R
dvs
1mV
R f C1
= 60 ×
/ min
R
dt
rpm
RfC1=60 so choose Rf = 6 MΩ and C1 = 10 µF.