Problem 7.90 PSpice|Multisim - Computer Action...
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Problem 7.90 PSpice|Multisim The energy stored in the capacitor in the circuit is zero at the instant the switch is closed. The ideal operational amplifier reaches saturation in 18 . Part A What is the numerical value of in kiloohms, if = 310 ? ANSWER: =
Transcript of Problem 7.90 PSpice|Multisim - Computer Action...
Problem 7.90 PSpice|Multisim
The energy stored in the capacitor in the circuit is zero at the instant the switch is closed. The ideal operationalamplifier reaches saturation in 18 .
Part A
What is the numerical value of in kiloohms, if = 310 ?
ANSWER:
ms
R C nF
= R kΩ
Problem 8.2 PSpice|Multisim
The circuit elements in the circuit in are = 100 , = 5 , and = 200 . The initial inductor current is
0.3 , and the initial capacitor voltage is 25 .
Part A
Find the expression for for .
ANSWER:
R ΩC μF L mH
A V
v(t) t ≥ 0
70, 000t + 50 Ve−1000t e−1000t
35, 000t − 50 Ve−1000t e−1000t
70, 000t + 25 Ve−1000t e−1000t
70, 000t − 50 Ve−1000t e−1000t
35, 000t + 25 Ve−1000t e−1000t
35, 000t + 50 Ve−1000t e−1000t
70, 000t − 25 Ve−1000t e−1000t
35, 000t − 25 Ve−1000t e−1000t
Problem 8.7
The voltage response for the circuit in is known to be ,
where is in seconds. The initial current in the inductor is 25 , and the initial voltage on the capacitor is 5 . The resistor has a value of 50 .
Part A
Find the value of .
Express your answer to three significant figures and include the appropriate units.
ANSWER:
Part B
Find the value of .
Express your answer to three significant figures and include the appropriate units.
ANSWER:
Part C
Find the value of .
Express your answer using three significant figures.
ANSWER:
v(t) = t + , t ≥ 0D1 e−80t D2 e−80t
t( )I0 mA( )V0 V Ω
L
= L
C
= C
D1
= D1 V/s
Part D
Find the value of .
Express your answer using three significant figures.
ANSWER:
Part E
Find for .
Express your answer in terms of , where is in milliseconds.
ANSWER:
D2
= D2 V
(t)iC t ≥ 0+
t t
= iC mA
Problem 8.24 PSpice|Multisim
The initial energy stored in the circuit in the figure is zero. At , a dc current source of 24 is applied to thecircuit. The value of the resistor is 400 .
Part A
Find for ;
ANSWER:
Part B
Find for ;
ANSWER:
t = 0 mAΩ
v(t) t ≥ 0
32( + ) Ve−20,000t e−40,000t
16( − ) Ve−20,000t e−80,000t
16( − ) Ve−20,000t e−40,000t
32( − ) Ve−20,000t e−80,000t
16( + ) Ve−20,000t e−80,000t
16( + ) Ve−20,000t e−40,000t
32( + ) Ve−20,000t e−80,000t
32( − ) Ve−20,000t e−40,000t
(t)iR t ≥ 0
Part C
Find for ;
ANSWER:
40( − 2 ) mAe−20,000t e−80,000t
40( + 2 ) mAe−20,000t e−80,000t
20( − ) mAe−20,000t e−80,000t
20( − 2 ) mAe−20,000t e−80,000t
20( + 2 ) mAe−20,000t e−80,000t
20( + ) mAe−20,000t e−80,000t
40( − ) mAe−20,000t e−80,000t
40( + ) mAe−20,000t e−80,000t
(t)iC t ≥ 0
−8 + 32 mAe−20,000t e−80,000t
−8 − 32 mAe−20,000t e−80,000t
−8 + 16 mAe−20,000t e−80,000t
−4 − 32 mAe−20,000t e−80,000t
−4 + 16 mAe−20,000t e−80,000t
−8 − 16 mAe−20,000t e−80,000t
−4 + 32 mAe−20,000t e−80,000t
−4 − 16 mAe−20,000t e−80,000t
Problem 8.31 PSpice|Multisim
The switch in the circuit in has been open for a long timebefore closing at = 0.
Part A
Find for .
ANSWER:
t
(t)io t ≥ 0
3 − 1200t + 6 Ae−800t e−800t
3 − 1200t − 6 Ae−800t e−800t
6 − 1200t − 3 Ae−800t e−800t
3 − 1200t + 3 Ae−800t e−800t
6 − 1200t + 6 Ae−800t e−800t
3 − 1200t − 3 Ae−800t e−800t
6 − 1200t + 3 Ae−800t e−800t
6 − 1200t − 6 Ae−800t e−800t
Problem 8.39
The current in the circuit in the followin figure is known to be .
The capacitor has a value of 80 ; the initial value of the current is 6 ; and the initial voltage on the capacitoris 21 .
Part A
Find the value of .
Express your answer with the appropriate units.
ANSWER:
Part B
Find the value of .
Express your answer with the appropriate units.
ANSWER:
Part C
Find the value of .
Express your answer with the appropriate units.
ANSWER:
i = cos1500t + sin1500t, t ≥ 0B1 e−2000t B2 e−2000t
nF mAV
R
= R
L
= L
B1
Part D
Find the value of .
Express your answer with the appropriate units.
ANSWER:
= B1
B2
= B2
