The Z-10/45° ohms and Y=0.4-j0.2 siemens are connected in series. What is the phas difference between v(t) and i(t)? i(t) J + +1 v(t) Z Y
The Z-10/45° ohms and Y=0.4-j0.2 siemens are connected in series. What is the phas difference between v(t) and i(t)? i(t) J + +1 v(t) Z Y
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.61P
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Question
PHASE DIFFERENCE FIND( NEED NEAT HANDWRITTEN SOLUTION ONLY OTHERWISE DOWNVOTE).
![**Problem Statement:**
The \( Z = 10 / \angle 45^\circ \) ohms and \( Y = 0.4 - j0.2 \) siemens are connected in series. What is the phase difference between \( v(t) \) and \( i(t) \)?
**Explanation and Diagram:**
The given complex impedance \( Z = 10 / \angle 45^\circ \) ohms and complex admittance \( Y = 0.4 - j0.2 \) siemens need to be analyzed to determine the phase relationship between the voltage \( v(t) \) and the current \( i(t) \).
Here's the provided diagram:
```
i(t)
↓
+ ----( Z )-----+
| |
v(t) ( Y )
| |
+---------------+
```
The diagram shows that the current \( i(t) \) flows through the impedance \( Z \) and then through the admittance \( Y \) arranged in series, while the voltage \( v(t) \) is applied across the entire series combination.
To solve for the phase difference, we'll use the properties of impedance and admittance in AC circuits:
1. **Convert Y to Z:**
Since \( Y \) is given in admittance:
\[
Y = 0.4 - j0.2 \ \text{siemens}
\]
The corresponding impedance \( Z_Y \) is:
\[
Z_Y = \frac{1}{Y} = \frac{1}{0.4 - j0.2}
\]
This can be simplified using complex conjugates:
\[
Z_Y = \frac{0.4 + j0.2}{(0.4)^2 + (0.2)^2} = \frac{0.4 + j0.2}{0.16 + 0.04} = \frac{0.4 + j0.2}{0.2} = 2 + j1 = \sqrt{5} \angle \tan^{-1}(0.5)
\]
2. **Calculate the total impedance:**
Total series impedance \( Z_{total} \) is:
\[
Z_{total} = Z + Z_Y = 10 / \angle 45^\circ + \sqrt{5} \angle](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0aee0e3d-a57d-4304-8930-413dab086e20%2F65682100-5e25-4fff-a208-612f29ad63cc%2Fhwbgk4o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
The \( Z = 10 / \angle 45^\circ \) ohms and \( Y = 0.4 - j0.2 \) siemens are connected in series. What is the phase difference between \( v(t) \) and \( i(t) \)?
**Explanation and Diagram:**
The given complex impedance \( Z = 10 / \angle 45^\circ \) ohms and complex admittance \( Y = 0.4 - j0.2 \) siemens need to be analyzed to determine the phase relationship between the voltage \( v(t) \) and the current \( i(t) \).
Here's the provided diagram:
```
i(t)
↓
+ ----( Z )-----+
| |
v(t) ( Y )
| |
+---------------+
```
The diagram shows that the current \( i(t) \) flows through the impedance \( Z \) and then through the admittance \( Y \) arranged in series, while the voltage \( v(t) \) is applied across the entire series combination.
To solve for the phase difference, we'll use the properties of impedance and admittance in AC circuits:
1. **Convert Y to Z:**
Since \( Y \) is given in admittance:
\[
Y = 0.4 - j0.2 \ \text{siemens}
\]
The corresponding impedance \( Z_Y \) is:
\[
Z_Y = \frac{1}{Y} = \frac{1}{0.4 - j0.2}
\]
This can be simplified using complex conjugates:
\[
Z_Y = \frac{0.4 + j0.2}{(0.4)^2 + (0.2)^2} = \frac{0.4 + j0.2}{0.16 + 0.04} = \frac{0.4 + j0.2}{0.2} = 2 + j1 = \sqrt{5} \angle \tan^{-1}(0.5)
\]
2. **Calculate the total impedance:**
Total series impedance \( Z_{total} \) is:
\[
Z_{total} = Z + Z_Y = 10 / \angle 45^\circ + \sqrt{5} \angle
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