Please give me the below content same as it is without any changes in rendered equations
Got it — I'll use a transfer-function (impedance) approach and give (v(t)) in the time domain.
**Image (your uploaded file):** `/mnt/data/44b8c09b-588f-458d-8eb0-dfa14e2164fb.png`
---
### 1) Write impedances (s-domain / phasor)
Angular frequency: (\omega = 100\ \text{rad/s}).
Inductance (L=40\ \mu\text{H}=40\times10^{-6},\text{H}). Capacitor (C=1\ \mu\text{F}=1\times10^{-6},\text{F}).
Branches across the source (all in parallel):
* (Z_{R1}=10\ \Omega).
* Series R-L branch: (Z_{RL}(s)=6 + sL).
* Capacitor branch: (Z_C(s)=\dfrac{1}{sC}).
Admittance (total):
[
Y(s)=\frac{1}{Z_{R1}}+\frac{1}{Z_{RL}(s)}+\frac{1}{Z_C(s)}
=\frac{1}{10}+\frac{1}{6+sL}+sC.
]
Equivalent impedance:
[
Z_{eq}(s)=\frac{1}{Y(s)}.
]
Transfer (voltage across the parallel network) for a current source:
[
V(s)=I(s),Z_{eq}(s).
]
---
### 2) Evaluate at steady-state frequency (s=j\omega)
Compute numerical impedances at (\omega=100):
[
Z_{RL}=6+j\omega L=6 + j(100)(40\times10^{-6}) = 6 + j0.004\ \Omega,
]
[
Z_C=\frac{1}{j\omega C}=\frac{1}{j(100)(1\times10^{-6})} = -j10000\ \Omega.
]
Total admittance:
[
Y = \frac{1}{10}+\frac{1}{6+j0.004}+\frac{1}{-j10000}.
]
Computing this numerically gives
[
Z_{eq} \approx 3.75000104 + j,0.00015625\ \Omega.
]
(You can see the imaginary part is essentially zero — the capacitor's reactance is huge at this low frequency, and the inductor's reactance is tiny.)
---
### 3) Phasor voltage and back to time domain
Current source in time domain: (i_s(t)=4\cos(100t+80^\circ)).
Phasor (using amplitude convention): (I=4\angle 80^\circ).
Phasor voltage:
[
V = I;Z_{eq} \approx 4\angle 80^\circ \times (3.75000104 + j0.00015625).
]
Numerically:
[
|V|\approx 15.000004\ \text{V},\qquad \angle V \approx 80.0024^\circ.
]
Thus the time-domain voltage (real part) is
[
\boxed{,v(t) \approx 15.00\cos\big(100t + 80.00^\circ\big)\ \text{V},}
]
(rounded—the small phase shift and tiny imaginary part are negligible at this precision).
Question:
Please give me the below content same as it is without any changes in rendered equations
Got it — I'll use a transfer-function (impedance) approach and give (v(t)) in the time domain.
**Image (your uploaded file):** `/mnt/data/44b8c09b-588f-458d-8eb0-dfa14e2164fb.png`
---
### 1) Write impedances (s-domain / phasor)
Angular frequency: (\omega = 100\ \text{rad/s}).
Inductance (L=40\ \mu\text{H}=40\times10^{-6},\text{H}). Capacitor (C=1\ \mu\text{F}=1\times10^{-6},\text{F}).
Branches across the source (all in parallel):
* (Z_{R1}=10\ \Omega).
* Series R-L branch: (Z_{RL}(s)=6 + sL).
* Capacitor branch: (Z_C(s)=\dfrac{1}{sC}).
Admittance (total):
[
Y(s)=\frac{1}{Z_{R1}}+\frac{1}{Z_{RL}(s)}+\frac{1}{Z_C(s)}
=\frac{1}{10}+\frac{1}{6+sL}+sC.
]
Equivalent impedance:
[
Z_{eq}(s)=\frac{1}{Y(s)}.
]
Transfer (voltage across the parallel network) for a current source:
[
V(s)=I(s),Z_{eq}(s).
]
---
### 2) Evaluate at steady-state frequency (s=j\omega)
Compute numerical impedances at (\omega=100):
[
Z_{RL}=6+j\omega L=6 + j(100)(40\times10^{-6}) = 6 + j0.004\ \Omega,
]
[
Z_C=\frac{1}{j\omega C}=\frac{1}{j(100)(1\times10^{-6})} = -j10000\ \Omega.
]
Total admittance:
[
Y = \frac{1}{10}+\frac{1}{6+j0.004}+\frac{1}{-j10000}.
]
Computing this numerically gives
[
Z_{eq} \approx 3.75000104 + j,0.00015625\ \Omega.
]
(You can see the imaginary part is essentially zero — the capacitor's reactance is huge at this low frequency, and the inductor's reactance is tiny.)
---
### 3) Phasor voltage and back to time domain
Current source in time domain: (i_s(t)=4\cos(100t+80^\circ)).
Phasor (using amplitude convention): (I=4\angle 80^\circ).
Phasor voltage:
[
V = I;Z_{eq} \approx 4\angle 80^\circ \times (3.75000104 + j0.00015625).
]
Numerically:
[
|V|\approx 15.000004\ \text{V},\qquad \angle V \approx 80.0024^\circ.
]
Thus the time-domain voltage (real part) is
[
\boxed{,v(t) \approx 15.00\cos\big(100t + 80.00^\circ\big)\ \text{V},}
]
(rounded—the small phase shift and tiny imaginary part are negligible at this precision).
This is AI generated Response.
Asked by: Mohin
Created at: 2025-11-20 08:32:37
Final Answer:
RenderedRaw Text
Did you like this answer?
Got it — I'll use a transfer-function (impedance) approach and give (v(t)) in the time domain.
**Imageyour uploaded file):** `/mnt/44b8c09b-588f-458d-8ebfa14e2164fb.png`
---
### 1) Write impedances (s-domain / phasor)
Angular frequency: (\omega = 100\ \text{rad/s}).
Inductance (L=40\ \mu\text{H}=40\times10^{-6},\text{H}). Capacitor (C=1\ \mu\text{F}=1\times10^{-6},\text{F}).
Branches across the source (all in parallel):
* (Z_{R1}=10\ \Omega).
* Series R-L branch: (Z_{RL}(s)=6 + sL).
* Capacitor branch: (Z_C(s)=\dfrac{1}{sC}).
Admittance (total):
[
Y(s)=\frac{1}{Z_{R1}}+\frac{1}{Z_{RL}(s)}+\frac{1}{Z_C(s)}
=\frac{1}{10}+\frac{1}{6+sL}+sC.
]
Equivalent impedance:
[
Z_{eq}(s)=\frac{1}{Y(s)}.
]
Transfer (voltage across the parallel network) for a current source:
[
V(s)=I(s),Z_{eq}(s).
]
---
### 2) Evaluate at steady-state frequency (s=j\omega)
Compute numerical impedances at (\omega=100):
[
Z_{RL}=6+j\omega L=6 + j(100)(40\times10^{-6}) = 6 + j.004\ \Omega,
]
[
Z_C=\frac{1}{j\omega C}=\frac{1}{j(100)(1\times10^{-6})} = -j10000\ \Omega.
]
Total admittance:
[
Y = \frac{1}{10}+\frac{1}{6+j.004}+\frac{1}{-j10000}.
]
Computing this numerically gives
[
Z_{eq} \approx 3.75000104 + j,.00015625\ \Omega.
]
(You can see the imaginary part is essentially zero — the capacitor's reactance is huge at this low frequency, and the inductor's reactance is tiny.)
---
### 3) Phasor voltage and back to time domain
Current source in time domain: (i_s(t)=4\cos(100t+80^\circ)).
Phasor (using amplitude convention): (I=4\angle 80^\circ).
Phasor voltage:
[
V = I;Z_{eq} \approx 4\angle 80^\circ \times (3.75000104 + j.00015625).
]
Numerically:
[
|V|\approx 15.000004\ \text{V},\qquad \angle V \approx 80.0024^\circ.
]
Thus the time-domain voltage (real part) is
[
\boxed{,v(t) \approx 15.00\cos\big(100t + 80.00^\circ\big)\ \text{V},}
]
(rounded—the small phase shift and tiny imaginary part are negligible at this precision).
Question:
Please give me the solution in rendered equation form
This is AI generated Response.
Asked by: Mohin
Created at: 2025-11-20 08:33:57
Final Answer:
RenderedRaw Text
Did you like this answer?
# Transfer Function Approach for Voltage in Time Domain
**Image:** 
---
## 1) Impedances in s-domain / Phasor
**Angular frequency:** \( \omega = 100\ \text{rad/s} \)
**Inductance:** \( L = 40\ \mu\text{H} = 40 \times 10^{-6}\ \text{H} \)
**Capacitance:** \( C = 1\ \mu\text{F} = 1 \times 10^{-6}\ \text{F} \)
### Parallel Branches
- \( Z_{R1} = 10\ \Omega \)
- Series R-L branch: \( Z_{RL}(s) = 6 + sL \)
- Capacitor branch: \( Z_C(s) = \frac{1}{sC} \)
### Total Admittance
\[
Y(s) = \frac{1}{Z_{R1}} + \frac{1}{Z_{RL}(s)} + \frac{1}{Z_C(s)} = \frac{1}{10} + \frac{1}{6+sL} + sC
\]
### Equivalent Impedance
\[
Z_{eq}(s) = \frac{1}{Y(s)}
\]
### Transfer Function
For a current source:
\[
V(s) = I(s) \cdot Z_{eq}(s)
\]
---
## 2) Evaluate at Steady-State Frequency \( (s = j\omega) \)
### Compute Numerical Impedances at \( \omega = 100 \)
\[
Z_{RL} = 6 + j\omega L = 6 + j(100)(40 \times 10^{-6}) = 6 + j0.004\ \Omega
\]
\[
Z_C = \frac{1}{j\omega C} = \frac{1}{j(100)(1 \times 10^{-6})} = -j10000\ \Omega
\]
### Total Admittance
\[
Y = \frac{1}{10} + \frac{1}{6+j0.004} + \frac{1}{-j10000}
\]
### Numerical Result
Computing gives:
\[
Z_{eq} \approx 3.75000104 + j0.00015625\ \Omega
\]
*Note: The imaginary part is negligible at this precision due to the capacitor's high reactance and the inductor's low reactance.*
---
## 3) Phasor Voltage and Time Domain Conversion
**Time Domain Current Source:**
\[
i_s(t) = 4\cos(100t + 80^\circ)
\]
**Phasor:**
\[
I = 4\angle 80^\circ
\]
### Phasor Voltage Calculation
\[
V = I \cdot Z_{eq} \approx 4\angle 80^\circ \times (3.75000104 + j0.00015625)
\]
### Numerical Result
\[
|V| \approx 15.000004\ \text{V}, \quad \angle V \approx 80.0024^\circ
\]
### Time-Domain Voltage
The real part yields:
\[
\boxed{v(t) \approx 15.00\cos(100t + 80.00^\circ)\ \text{V}}
\]
*Note: The small phase shift and tiny imaginary part are negligible at this precision.*
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