The voltage V o across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1.

Question

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Answer 1

Textbook Question

Chapter 10, Problem 1RQ

The voltage V_o across the capacitor in Fig. 10.43 is:

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 1

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 2

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 3

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 4

Figure 10.43

For Review Question 10.1.

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 5

Expert Solution & Answer

To determine

Choose the correct option to find the voltage Vo across the capacitor in Figure 10.43.

Answer to Problem 1RQ

The correct option from the given choices is (c) 7.071∠−45° V.

Explanation of Solution

Given data:

Refer to Figure 10.43 in the textbook for circuit analysis.

Calculation:

From Figure 10.43, write the expression for voltage Vo using voltage division rule.

Vo=(10∠0° V)(−j1) Ω(1−j1) Ω=(10∠0° V)1∠−90° Ω1.41421 ∠−45° Ω=(10∠0° V)[11.41421 ∠(−90°−(−45°))]=(10)11.41421 ∠−45° V

Reduce the equation as follows.

Vo=7.07108∠−45° V≅7.071∠−45° V

Hence, the voltage across the capacitor is 7.071∠−45° V. Therefore, option (c) is correct and options (a), (b), and (d) are incorrect.

Conclusion:

Thus, the correct option from the given choices is (c) 7.071∠−45° V.

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A plane wave propagating in the +z direction in medium 1 is normally incident to medium 2 located at the z=0 plane as below. Both mediums are general, characterized by ( ε i, Mi, Ơi ). tot = [ ει μη σ] [ε, μη σε ] Ex Ex tot E₁₂ (z) = Ee Ex z=0 From conservation of energy: P₁AV'(z=0) + Piav'(z=0) = P2av²(z=0). Using the above show for lossless media that: ( 1 - ||²) = (1/M2 )|T|² .

A plane wave propagating in the +z direction in medium 1 is normally incident to medium 2 located at the z=0 plane as below. Both mediums are general, characterized by ( ε i, Hi, σ¡ ). [ ει μη σ] Ex [ ει μη ση ] Ex tot E₁₂ (z) = E'₁e¹² -122 E(z) = Ee+ E₁₁₁² E₁x z=0 1. Specify the electric field reflection coefficient г and transmission coefficient T: E ΓΔ E E TA EL 2. Show that T=1+г. Can the transmitted electric field amplitude in region 2 be LARGER than the incident electric field amplitude? 3. Determine expressions for P₁AV'(z), PIAV'(z) and P2AV'(z) (note the sign for the reflected power direction should be (-z).

Answer 2

Textbook Question

Chapter 10, Problem 1RQ

The voltage V_o across the capacitor in Fig. 10.43 is:

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 1

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 2

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 3

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 4

Figure 10.43

For Review Question 10.1.

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 5

Expert Solution & Answer

To determine

Choose the correct option to find the voltage Vo across the capacitor in Figure 10.43.

Answer to Problem 1RQ

The correct option from the given choices is (c) 7.071∠−45° V.

Explanation of Solution

Given data:

Refer to Figure 10.43 in the textbook for circuit analysis.

Calculation:

From Figure 10.43, write the expression for voltage Vo using voltage division rule.

Vo=(10∠0° V)(−j1) Ω(1−j1) Ω=(10∠0° V)1∠−90° Ω1.41421 ∠−45° Ω=(10∠0° V)[11.41421 ∠(−90°−(−45°))]=(10)11.41421 ∠−45° V

Reduce the equation as follows.

Vo=7.07108∠−45° V≅7.071∠−45° V

Hence, the voltage across the capacitor is 7.071∠−45° V. Therefore, option (c) is correct and options (a), (b), and (d) are incorrect.

Conclusion:

Thus, the correct option from the given choices is (c) 7.071∠−45° V.

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Answer 3

Textbook Question

Chapter 10, Problem 1RQ

The voltage V_o across the capacitor in Fig. 10.43 is:

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 1

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 2

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 3

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 4

Figure 10.43

For Review Question 10.1.

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 5

Expert Solution & Answer

To determine

Choose the correct option to find the voltage Vo across the capacitor in Figure 10.43.

Answer to Problem 1RQ

The correct option from the given choices is (c) 7.071∠−45° V.

Explanation of Solution

Given data:

Refer to Figure 10.43 in the textbook for circuit analysis.

Calculation:

From Figure 10.43, write the expression for voltage Vo using voltage division rule.

Vo=(10∠0° V)(−j1) Ω(1−j1) Ω=(10∠0° V)1∠−90° Ω1.41421 ∠−45° Ω=(10∠0° V)[11.41421 ∠(−90°−(−45°))]=(10)11.41421 ∠−45° V

Reduce the equation as follows.

Vo=7.07108∠−45° V≅7.071∠−45° V

Hence, the voltage across the capacitor is 7.071∠−45° V. Therefore, option (c) is correct and options (a), (b), and (d) are incorrect.

Conclusion:

Thus, the correct option from the given choices is (c) 7.071∠−45° V.

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Answer 4

Textbook Question

Chapter 10, Problem 1RQ

The voltage V_o across the capacitor in Fig. 10.43 is:

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 1

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 2

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 3

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 4

Figure 10.43

For Review Question 10.1.

Chapter 10, Problem 1RQ, The voltage Vo across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1. , example 5

Expert Solution & Answer

To determine

Choose the correct option to find the voltage Vo across the capacitor in Figure 10.43.

Answer to Problem 1RQ

The correct option from the given choices is (c) 7.071∠−45° V.

Explanation of Solution

Given data:

Refer to Figure 10.43 in the textbook for circuit analysis.

Calculation:

From Figure 10.43, write the expression for voltage Vo using voltage division rule.

Vo=(10∠0° V)(−j1) Ω(1−j1) Ω=(10∠0° V)1∠−90° Ω1.41421 ∠−45° Ω=(10∠0° V)[11.41421 ∠(−90°−(−45°))]=(10)11.41421 ∠−45° V

Reduce the equation as follows.

Vo=7.07108∠−45° V≅7.071∠−45° V

Hence, the voltage across the capacitor is 7.071∠−45° V. Therefore, option (c) is correct and options (a), (b), and (d) are incorrect.

Conclusion:

Thus, the correct option from the given choices is (c) 7.071∠−45° V.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Choose the correct option to find the voltage Vo across the capacitor in Figure 10.43.

Answer to Problem 1RQ

The correct option from the given choices is (c) 7.071∠−45° V.

Explanation of Solution

Given data:

Refer to Figure 10.43 in the textbook for circuit analysis.

Calculation:

From Figure 10.43, write the expression for voltage Vo using voltage division rule.

Vo=(10∠0° V)(−j1) Ω(1−j1) Ω=(10∠0° V)1∠−90° Ω1.41421 ∠−45° Ω=(10∠0° V)[11.41421 ∠(−90°−(−45°))]=(10)11.41421 ∠−45° V

Reduce the equation as follows.

Vo=7.07108∠−45° V≅7.071∠−45° V

Hence, the voltage across the capacitor is 7.071∠−45° V. Therefore, option (c) is correct and options (a), (b), and (d) are incorrect.

Conclusion:

Thus, the correct option from the given choices is (c) 7.071∠−45° V.

Answer 8

Expert Solution & Answer

To determine

Choose the correct option to find the voltage Vo across the capacitor in Figure 10.43.

Answer to Problem 1RQ

The correct option from the given choices is (c) 7.071∠−45° V.

Explanation of Solution

Given data:

Refer to Figure 10.43 in the textbook for circuit analysis.

Calculation:

From Figure 10.43, write the expression for voltage Vo using voltage division rule.

Vo=(10∠0° V)(−j1) Ω(1−j1) Ω=(10∠0° V)1∠−90° Ω1.41421 ∠−45° Ω=(10∠0° V)[11.41421 ∠(−90°−(−45°))]=(10)11.41421 ∠−45° V

Reduce the equation as follows.

Vo=7.07108∠−45° V≅7.071∠−45° V

Hence, the voltage across the capacitor is 7.071∠−45° V. Therefore, option (c) is correct and options (a), (b), and (d) are incorrect.

Conclusion:

Thus, the correct option from the given choices is (c) 7.071∠−45° V.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Choose the correct option to find the voltage Vo across the capacitor in Figure 10.43.

Answer 12

Answer to Problem 1RQ

The correct option from the given choices is (c) 7.071∠−45° V.

Answer 13

Explanation of Solution

Given data:

Refer to Figure 10.43 in the textbook for circuit analysis.

Calculation:

From Figure 10.43, write the expression for voltage Vo using voltage division rule.

Vo=(10∠0° V)(−j1) Ω(1−j1) Ω=(10∠0° V)1∠−90° Ω1.41421 ∠−45° Ω=(10∠0° V)[11.41421 ∠(−90°−(−45°))]=(10)11.41421 ∠−45° V

Reduce the equation as follows.

Vo=7.07108∠−45° V≅7.071∠−45° V

Hence, the voltage across the capacitor is 7.071∠−45° V. Therefore, option (c) is correct and options (a), (b), and (d) are incorrect.

Conclusion:

Thus, the correct option from the given choices is (c) 7.071∠−45° V.

Answer 14

Ch. 10.2 - Using nodal analysis, find v1 and v2 is in the...Ch. 10.2 - Calculate V1 and V2 in the circuit shown in Fig....Ch. 10.3 - Find Io in Fig. 10.8 using mesh analysis. Figure...Ch. 10.3 - Figure 10.11 For Practice Prob. 10.4. Calculate...Ch. 10.4 - Find current Io in the circuit of Fig. 10.8 using...Ch. 10.4 - Calculate vo in the circuit of Fig. 10.15 using...Ch. 10.6 - Determine the Norton equivalent of the circuit in...Ch. 10.7 - Find vo and io in the op amp circuit of Fig....Ch. 10.7 - Obtain the closed-loop gain and phase shift for...Ch. 10.8 - Use PSpice to obtain vo and io in the circuit of...

The voltage V o across the capacitor in Fig. 10.43 is: Figure 10.43 For Review Question 10.1.

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