Please give me the below content same as it is without any changes in rendered equations
**Given (both problems):**
(\omega = 377\ \text{rad/s}). Phasor notation: (V_s(t)=12\cos(\omega t+\phi_V)\Rightarrow \mathbf{V}=12\angle\phi_V).
---
### Problem 1 — Inductor (L=10\ \text{mH}), (v_s(t)=12\cos(377t+40^\circ))
1. Convert source to phasor: (\mathbf{V}=12\angle 40^\circ).
2. Inductor impedance:
[
Z_L = j\omega L = j(377)(10\times10^{-3}) = j(3.77)\ \Omega
]
so (|Z_L|=3.77;\Omega,;\angle Z_L=+90^\circ.)
3. Phasor current: (\mathbf{I}=\dfrac{\mathbf{V}}{Z_L}=\dfrac{12\angle40^\circ}{3.77\angle90^\circ}=\dfrac{12}{3.77}\angle(40^\circ-90^\circ)).
4. Magnitude and angle: (\dfrac{12}{3.77}=3.183) and angle (40^\circ-90^\circ=-50^\circ).
5. Back to time domain:
[
\boxed{i(t)=3.183\cos(377t-50^\circ)\ \text{A}}
]
(Interpretation: current **lags** the voltage by (50^\circ) because of the inductor.)
---
### Problem 2 — Capacitor (C=500\ \mu\text{F}), (v_s(t)=12\cos(377t-60^\circ))
1. Phasor: (\mathbf{V}=12\angle(-60^\circ)).
2. Capacitor impedance:
[
Z_C=\frac{1}{j\omega C}=-j\frac{1}{\omega C}.
]
Compute (\omega C = 377\times500\times10^{-6}=0.1885), so (\dfrac{1}{\omega C}=5.305\ \Omega). Thus (Z_C=-j5.305\ \Omega) (magnitude (5.305), angle (-90^\circ)).
3. Phasor current:
[
\mathbf{I}=\dfrac{\mathbf{V}}{Z_C}=\dfrac{12\angle(-60^\circ)}{5.305\angle(-90^\circ)}=\dfrac{12}{5.305}\angle(-60^\circ-(-90^\circ)).
]
4. Magnitude and angle: (\dfrac{12}{5.305}=2.262) and angle (-60^\circ+90^\circ=30^\circ).
5. Time domain:
[
\boxed{i(t)=2.262\cos(377t+30^\circ)\ \text{A}}
]
(Interpretation: capacitor current **leads** the voltage by (90^\circ); here net result is current leads the given source by (+30^\circ).)
---
**Final answers (rounded):**
* For the 10 mH inductor: (i(t)=3.183\cos(377t-50^\circ)\ \mathrm{A}.)
* For the 500 µF capacitor: (i(t)=2.262\cos(377t+30^\circ)\ \mathrm{A}.)
Question:
Please give me the below content same as it is without any changes in rendered equations
**Given (both problems):**
(\omega = 377\ \text{rad/s}). Phasor notation: (V_s(t)=12\cos(\omega t+\phi_V)\Rightarrow \mathbf{V}=12\angle\phi_V).
---
### Problem 1 — Inductor (L=10\ \text{mH}), (v_s(t)=12\cos(377t+40^\circ))
1. Convert source to phasor: (\mathbf{V}=12\angle 40^\circ).
2. Inductor impedance:
[
Z_L = j\omega L = j(377)(10\times10^{-3}) = j(3.77)\ \Omega
]
so (|Z_L|=3.77;\Omega,;\angle Z_L=+90^\circ.)
3. Phasor current: (\mathbf{I}=\dfrac{\mathbf{V}}{Z_L}=\dfrac{12\angle40^\circ}{3.77\angle90^\circ}=\dfrac{12}{3.77}\angle(40^\circ-90^\circ)).
4. Magnitude and angle: (\dfrac{12}{3.77}=3.183) and angle (40^\circ-90^\circ=-50^\circ).
5. Back to time domain:
[
\boxed{i(t)=3.183\cos(377t-50^\circ)\ \text{A}}
]
(Interpretation: current **lags** the voltage by (50^\circ) because of the inductor.)
---
### Problem 2 — Capacitor (C=500\ \mu\text{F}), (v_s(t)=12\cos(377t-60^\circ))
1. Phasor: (\mathbf{V}=12\angle(-60^\circ)).
2. Capacitor impedance:
[
Z_C=\frac{1}{j\omega C}=-j\frac{1}{\omega C}.
]
Compute (\omega C = 377\times500\times10^{-6}=0.1885), so (\dfrac{1}{\omega C}=5.305\ \Omega). Thus (Z_C=-j5.305\ \Omega) (magnitude (5.305), angle (-90^\circ)).
3. Phasor current:
[
\mathbf{I}=\dfrac{\mathbf{V}}{Z_C}=\dfrac{12\angle(-60^\circ)}{5.305\angle(-90^\circ)}=\dfrac{12}{5.305}\angle(-60^\circ-(-90^\circ)).
]
4. Magnitude and angle: (\dfrac{12}{5.305}=2.262) and angle (-60^\circ+90^\circ=30^\circ).
5. Time domain:
[
\boxed{i(t)=2.262\cos(377t+30^\circ)\ \text{A}}
]
(Interpretation: capacitor current **leads** the voltage by (90^\circ); here net result is current leads the given source by (+30^\circ).)
---
**Final answers (rounded):**
* For the 10 mH inductor: (i(t)=3.183\cos(377t-50^\circ)\ \mathrm{A}.)
* For the 500 µF capacitor: (i(t)=2.262\cos(377t+30^\circ)\ \mathrm{A}.)
This is AI generated Response.
Asked by: jack
Created at: 2025-11-20 22:28:00
Final Answer:
RenderedRaw Text
Did you like this answer?
```markdown
**Given (both problems):**
(\omega = 377\ \text{rad/s}). Phasor notation: (V_s(t)=12\cos(\omega t+\phi_V)\Rightarrow \mathbf{V}=12\angle\phi_V).
---
### Problem 1 — Inductor (L=10\ \text{mH}), (v_s(t)=12\cos(377t+40^\circ))
1. Convert source to phasor: (\mathbf{V}=12\angle 40^\circ).
2. Inductor impedance:
[
Z_L = j\omega L = j(377)(10\times10^{-3}) = j(3.77)\ \Omega
]
so (|Z_L|=3.77;\Omega,;\angle Z_L=+90^\circ.)
3. Phasor current: (\mathbf{I}=\dfrac{\mathbf{V}}{Z_L}=\dfrac{12\angle40^\circ}{3.77\angle90^\circ}=\dfrac{12}{3.77}\angle(40^\circ-90^\circ)).
4. Magnitude and angle: (\dfrac{12}{3.77}=3.183) and angle (40^\circ-90^\circ=-50^\circ).
5. Back to time domain:
[
\boxed{i(t)=3.183\cos(377t-50^\circ)\ \text{A}}
]
(Interpretation: current **lags** the voltage by (50^\circ) because of the inductor.)
---
### Problem 2 — Capacitor (C=500\ \mu\text{F}), (v_s(t)=12\cos(377t-60^\circ))
1. Phasor: (\mathbf{V}=12\angle(-60^\circ)).
2. Capacitor impedance:
[
Z_C=\frac{1}{j\omega C}=-j\frac{1}{\omega C}.
]
Compute (\omega C = 377\times500\times10^{-6}=.1885), so (\dfrac{1}{\omega C}=5.305\ \Omega). Thus (Z_C=-j5.305\ \Omega) (magnitude (5.305), angle (-90^\circ)).
3. Phasor current:
[
\mathbf{I}=\dfrac{\mathbf{V}}{Z_C}=\dfrac{12\angle(-60^\circ)}{5.305\angle(-90^\circ)}=\dfrac{12}{5.305}\angle(-60^\circ-(-90^\circ)).
]
4. Magnitude and angle: (\dfrac{12}{5.305}=2.262) and angle (-60^\circ+90^\circ=30^\circ).
5. Time domain:
[
\boxed{i(t)=2.262\cos(377t+30^\circ)\ \text{A}}
]
(Interpretation: capacitor current **leads** the voltage by (90^\circ); here net result is current leads the given source by (+30^\circ).)
---
**Final answers (rounded):**
* For the 10 mH inductor: (i(t)=3.183\cos(377t-50^\circ)\ \mathrm{A}.)
* For the 500 µF capacitor: (i(t)=2.262\cos(377t+30^\circ)\ \mathrm{A}.)
```
Question:
please provide the solution in rendered equation form
This is AI generated Response.
Asked by: jack
Created at: 2025-11-20 22:29:05
Final Answer:
RenderedRaw Text
Did you like this answer?
```markdown
**Given (both problems):**
\(\omega = 377\ \text{rad/s}\). Phasor notation: \(V_s(t)=12\cos(\omega t+\phi_V)\Rightarrow \mathbf{V}=12\angle\phi_V\).
---
### Problem 1 — Inductor (L=10\ \text{mH}), (v_s(t)=12\cos(377t+40^\circ))
1. Convert source to phasor: \(\mathbf{V}=12\angle 40^\circ\).
2. Inductor impedance:
\[
Z_L = j\omega L = j(377)(10\times10^{-3}) = j(3.77)\ \Omega
\]
so \(|Z_L|=3.77\;\Omega,\;\angle Z_L=+90^\circ\).
3. Phasor current: \(\mathbf{I}=\dfrac{\mathbf{V}}{Z_L}=\dfrac{12\angle40^\circ}{3.77\angle90^\circ}=\dfrac{12}{3.77}\angle(40^\circ-90^\circ)\).
4. Magnitude and angle: \(\dfrac{12}{3.77}=3.183\) and angle \(40^\circ-90^\circ=-50^\circ\).
5. Back to time domain:
\[
\boxed{i(t)=3.183\cos(377t-50^\circ)\ \text{A}}
\]
(Interpretation: current **lags** the voltage by \(50^\circ\) because of the inductor.)
---
### Problem 2 — Capacitor (C=500\ \mu\text{F}), (v_s(t)=12\cos(377t-60^\circ))
1. Phasor: \(\mathbf{V}=12\angle(-60^\circ)\).
2. Capacitor impedance:
\[
Z_C=\frac{1}{j\omega C}=-j\frac{1}{\omega C}.
\]
Compute \(\omega C = 377\times500\times10^{-6}=0.1885\), so \(\dfrac{1}{\omega C}=5.305\ \Omega\). Thus \(Z_C=-j5.305\ \Omega\) (magnitude \(5.305\), angle \(-90^\circ\)).
3. Phasor current:
\[
\mathbf{I}=\dfrac{\mathbf{V}}{Z_C}=\dfrac{12\angle(-60^\circ)}{5.305\angle(-90^\circ)}=\dfrac{12}{5.305}\angle(-60^\circ-(-90^\circ)).
\]
4. Magnitude and angle: \(\dfrac{12}{5.305}=2.262\) and angle \(-60^\circ+90^\circ=30^\circ\).
5. Time domain:
\[
\boxed{i(t)=2.262\cos(377t+30^\circ)\ \text{A}}
\]
(Interpretation: capacitor current **leads** the voltage by \(90^\circ\); here net result is current leads the given source by \(+30^\circ\).)
---
**Final answers (rounded):**
* For the 10 mH inductor: \(i(t)=3.183\cos(377t-50^\circ)\ \mathrm{A}\).
* For the 500 µF capacitor: \(i(t)=2.262\cos(377t+30^\circ)\ \mathrm{A}\).
```
Question:
please provide the content in rendered equations
This is AI generated Response.
Asked by: jack
Created at: 2025-11-20 22:30:32
Final Answer:
RenderedRaw Text
Did you like this answer?
```markdown
**Given (both problems):**
\(\omega = 377\ \text{rad/s}\). Phasor notation: \(V_s(t)=12\cos(\omega t+\phi_V)\Rightarrow \mathbf{V}=12\angle\phi_V\).
---
### Problem 1 — Inductor (L=10\ \text{mH}), (v_s(t)=12\cos(377t+40^\circ))
1. Convert source to phasor: \(\mathbf{V}=12\angle 40^\circ\).
2. Inductor impedance:
\[
Z_L = j\omega L = j(377)(10\times10^{-3}) = j(3.77)\ \Omega
\]
so \(|Z_L|=3.77\;\Omega,\;\angle Z_L=+90^\circ\).
3. Phasor current: \(\mathbf{I}=\dfrac{\mathbf{V}}{Z_L}=\dfrac{12\angle40^\circ}{3.77\angle90^\circ}=\dfrac{12}{3.77}\angle(40^\circ-90^\circ)\).
4. Magnitude and angle: \(\dfrac{12}{3.77}=3.183\) and angle \(40^\circ-90^\circ=-50^\circ\).
5. Back to time domain:
\[
\boxed{i(t)=3.183\cos(377t-50^\circ)\ \text{A}}
\]
(Interpretation: current **lags** the voltage by \(50^\circ\) because of the inductor.)
---
### Problem 2 — Capacitor (C=500\ \mu\text{F}), (v_s(t)=12\cos(377t-60^\circ))
1. Phasor: \(\mathbf{V}=12\angle(-60^\circ)\).
2. Capacitor impedance:
\[
Z_C=\frac{1}{j\omega C}=-j\frac{1}{\omega C}.
\]
Compute \(\omega C = 377\times500\times10^{-6}=0.1885\), so \(\dfrac{1}{\omega C}=5.305\ \Omega\). Thus \(Z_C=-j5.305\ \Omega\) (magnitude \(5.305\), angle \(-90^\circ\)).
3. Phasor current:
\[
\mathbf{I}=\dfrac{\mathbf{V}}{Z_C}=\dfrac{12\angle(-60^\circ)}{5.305\angle(-90^\circ)}=\dfrac{12}{5.305}\angle(-60^\circ-(-90^\circ)).
\]
4. Magnitude and angle: \(\dfrac{12}{5.305}=2.262\) and angle \(-60^\circ+90^\circ=30^\circ\).
5. Time domain:
\[
\boxed{i(t)=2.262\cos(377t+30^\circ)\ \text{A}}
\]
(Interpretation: capacitor current **leads** the voltage by \(90^\circ\); here net result is current leads the given source by \(+30^\circ\).)
---
**Final answers (rounded):**
* For the 10 mH inductor: \(i(t)=3.183\cos(377t-50^\circ)\ \mathrm{A}\).
* For the 500 µF capacitor: \(i(t)=2.262\cos(377t+30^\circ)\ \mathrm{A}\).
```
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