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Concept explainers
(a)
Sketch the voltage waveform of capacitor.
(a)
![Check Mark](/static/check-mark.png)
Explanation of Solution
Given data:
Value of capacitor is
Formula used:
Refer to FIGURE 7.44 in the textbook.
The expression for the voltage across the capacitor is:
Here,
Calculation:
Refer to the FIGURE 7.43 (a).
Substitute the limits
The equation for the current waveform for time period
Substitute
Here,
Substitute
Since initial value of voltage at
Therefore,
Substitute
Substitute
Substitute
Since initial value of voltage at
Therefore,
Substitute
Substitute
Substitute
Substitute
Substitute
Substitute
Since initial value of voltage at
Substitute
Substitute
Substitute
Since initial value of voltage at
Substitute
Substitute
Substitute
Substitute
The expression for the voltage across capacitor for time period
Substitute
The labeled sketch of voltage waveform for
Conclusion:
Thus, the resulting voltage waveform is sketched.
(b)
Find the voltage of capacitor.
(b)
![Check Mark](/static/check-mark.png)
Answer to Problem 11E
The voltage of capacitor for given time duration is
Explanation of Solution
Given Data:
Value of capacitor is
Value of time duration for capacitor voltage is
Formula used:
The expression for the voltage at given point of time is,
Calculation:
Substitute
Substitute
Substitute
Conclusion:
Thus, the voltage of capacitor for given time duration is
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Chapter 7 Solutions
Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf
- A three-phase 20 kV medium-voltage line is 10 km. Resistance is 0.252 2/km and reactance is 0.128 92/km (inductive). Voltage at the beginning of line is 21.0 kV. At the end of the line is loading P = 2.5 MW with power factor 0.92ind. Draw 1-phase equivalent diagram and calculate line voltage at the end the of line, active and reactive power at the beginning of the line and power losses of the line.arrow_forwardA three-phase 20 kV medium-voltage line is 10 km. Resistance is 0.365 2/km and reactance is 0.363 2/km (inductive). Voltage at the beginning of line is 20.5 kV. At the end of the line is loading P= 800 kW with power factor 0.95ind. Draw 1-phase equivalent diagram and calculate load current, line voltage at the end the of line, voltage drop and power losses of the line.arrow_forward6. Answer the following questions. Take help from ChatGPT to answer these questions (if you need). Write the answers briefly using your own words with no more than two sentences, and make sure you check whether ChatGPT is giving you the appropriate answers in our context. A) What is a model in our context? B) What is an LTI system? C) What are the three forms of model we have used in the class so far to represent an LTI system? Among the above three forms, which forms can still be used to represent a nonlinear system?arrow_forward
- 5. Consider the following block diagram of a system in the Figure 4. Y₁(s) G₁ G2. R(s) C(s) Y₂(s) G3 G4 Figure 4 The models of the blocks G1, G2, G3 and G4 are represented by a differential equation, transfer function, state-space form, and impulse response as the followings. dy1 G₁: +2y₁ = 3r(t) dt 1 G2: G₂(s) = S+3 G3: x=2x+r, y2=3x-r G4: h(t)=8(t) + et 1(t) Find the simplified expression of the overall transfer function of the system i.e., G(s) = Note for G3 block, you may need to use the formula H(s) = C (sI - A)-¹ B+ D. C(s) R(s)arrow_forward4. Simplify the block diagram in Figure 3 and find the closed-loop transfer function G(s) = C(s) R(s) G₁ R(s) Figure 3 C(s) G2 H₁ H₂arrow_forward1. Consider a system defined by the following state-space equations. -5 2 N-MAN-G = 3 -1 y = [12] Find the transfer function H(s) = x1 x2. Y(s) U(s)' + 5arrow_forward
- 3. Simplify the block diagram in Figure 2 and find the closed-loop transfer function G(s) = C(s) R(s)' G₁ C(s) R(s) G2 G3 G4 Figure 2arrow_forwardRigid network supplies Feeder 1 through 110/21 kV transformer (Figure 1). Short circuit power of the supplying network is 5000 MVA and voltage is 110 kV. Determine 3-phase short circuit current for the point A. Draw 1-phase equivalent diagram. How big is the current if the 3-phase short circuit occurs in the Busbar? 110/21 kV Busbar Supplying network S = 16MVA 4-10% Figure 1. Feeder 1: 1-5km - r = 0.337 2/km x 0.361 2/km Aarrow_forwardRigid network supplies Feeder 1 through 110/21 kV transformer (Figure 1). Short circuit power of the supplying network is 3000 MVA and voltage is 110 kV. Length of feeder 1 is 5 km. Determine 3-phase short circuit current for the point A. Draw 1-phase equivalent diagram. 110/21 kV Busbar Supplying network S = 16MVA 4-10% Feeder 1: Figure 1. - 1 = 5km r = 0.337 2/km x = 0.361 2/km Aarrow_forward
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