please solve this problem step by step6.5 Find it) in the crcut in Exgure P55 V(t) = 12 cos (377t + 40°) & 10 mH FIGURE P55 6 Deion 4) he nebcck Eu P50 it V(t) = 12 cos (3771 - 60°) V 500 uF
Question:
please solve this problem step by step
6.5 Find it) in the crcut in Exgure P55
V(t) = 12 cos (377t + 40°) & 10 mH
FIGURE P55
6 Deion 4) he nebcck Eu P50
it
V(t) = 12 cos (3771 - 60°) V 500 uF
This is AI generated Response.
Asked by: jack
Created at: 2025-11-20 22:31:36
Asked by: jack
Created at: 2025-11-20 22:31:36
Final Answer:
Rendered
Raw Text
Did you like this answer?
Let's solve both problems step by step:
---
## **Problem 1: Inductor Circuit**
### **Given:**
- Source: \( v_s(t) = 12 \cos(377t + 40^\circ) \) V
- Inductor: \( L = 10\,\text{mH} = .01\,\text{H} \)
### **Step 1: Phasor Representation**
- Convert \( v_s(t) \) to phasor:
\[
\vec{V}_s = 12\angle40^\circ \text{ V}
\]
### **Step 2: Inductive Reactance**
- Angular frequency: \( \omega = 377\,\text{rad/s} \)
- Inductive reactance:
\[
X_L = \omega L = 377 \times .01 = 3.77\,\Omega
\]
- Impedance of inductor:
\[
Z_L = jX_L = j3.77\,\Omega
\]
### **Step 3: Apply Ohm's Law (Phasor Domain)**
\[
\vec{I} = \frac{\vec{V}_s}{Z_L} = \frac{12\angle40^\circ}{3.77\angle90^\circ}
\]
\[
= \frac{12}{3.77}\angle(40^\circ - 90^\circ)
\]
\[
= 3.18\angle(-50^\circ) \text{ A}
\]
### **Step 4: Convert Back to Time Domain**
\[
i(t) = 3.18\cos(377t - 50^\circ) \ \text{A}
\]
---
## **Problem 2: Capacitor Circuit**
### **Given:**
- Source: \( v_s(t) = 12\cos(377t - 60^\circ) \) V
- Capacitor: \( C = 500\,\mu\text{F} = 500 \times 10^{-6}\,\text{F} \)
### **Step 1: Phasor Representation**
- Convert \( v_s(t) \) to phasor:
\[
\vec{V}_s = 12\angle{-60^\circ} \text{ V}
\]
### **Step 2: Capacitive Reactance**
- Angular frequency: \( \omega = 377\,\text{rad/s} \)
- Capacitive reactance:
\[
X_C = \frac{1}{\omega C} = \frac{1}{377 \times 500 \times 10^{-6}}
\]
\[
= \frac{1}{.1885} = 5.31\,\Omega
\]
- Impedance of capacitor:
\[
Z_C = -jX_C = -j5.31\,\Omega
\]
(or \( 5.31\angle{-90^\circ} \,\Omega \))
### **Step 3: Apply Ohm's Law (Phasor Domain)**
\[
\vec{I} = \frac{\vec{V}_s}{Z_C} = \frac{12\angle{-60^\circ}}{5.31\angle{-90^\circ}}
\]
\[
= \frac{12}{5.31}\angle(-60^\circ - (-90^\circ))
\]
\[
= 2.26\angle(30^\circ)\ \text{A}
\]
### **Step 4: Convert Back to Time Domain**
\[
i(t) = 2.26\cos(377t + 30^\circ) \ \text{A}
\]
---
## **Final Answers**
### **1. Inductor Circuit**
\[
\boxed{i(t) = 3.18\cos(377t - 50^\circ)\ \text{A}}
\]
### **2. Capacitor Circuit**
\[
\boxed{i(t) = 2.26\cos(377t + 30^\circ)\ \text{A}}
\]
Let me know if you'd like a more detailed explanation for any step!
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :
VIPSolutions