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Obtenha a função de transferência para o circuito abaixo I(s)/U(s).Selecione a resposta:
a
​​​​​​​F(s)=(6s2+4s+1)/(4s+1) ​​​​​​

b
​​​​​​​F(s)=(6s2+4s+1)/(7s2+4s+1)

c
​​​​​​​F(s)=(4s+1)/(7s2+4s+1)​​​​​​​​​​​​​​

d
F(s)=(6s2+4s+1)/(6s2+7s+3)​​​​​​​​​​​​​​

e
F(s)=(6s2+4s+1)/(7s2+4s+3)​​​​​​​​​​​​​​

To find the transfer function \( \frac{I(s)}{U(s)} \) for the given circuit, we will follow these steps:

1. **Identify the components**: The circuit consists of:
   - A resistor \( R_1 = 1 \, \Omega \)
   - A capacitor \( C = 2 \, \text{F} \)
   - A resistor \( R_2 = 2 \, \Omega \)
   - An inductor \( L = 3 \, \text{H} \)

2. **Write the impedance in the Laplace domain**:
   - Resistor \( R_1 \): \( Z_{R1} = R_1 = 1 \, \Omega \)
   - Capacitor \( C \): \( Z_C = \frac{1}{sC} = \frac{1}{2s} \)
   - Resistor \( R_2 \): \( Z_{R2} = R_2 = 2 \, \Omega \)
   - Inductor \( L \): \( Z_L = sL = 3s \, \Omega \)

3. **Calculate the total impedance**:
   - The capacitor \( C \) and resistor \( R_2 \) are in parallel:
     \[
     Z_{\text{parallel}} = \left(\frac{1}{Z_C} + \frac{1}{Z_{R2}}\right)^{-1} = \left(\frac{1}{\frac{1}{2s}} + \frac{1}{2}\right)^{-1} = \left(\frac{2s + 1}{2s}\right)^{-1} = \frac{2s}{2s + 1}
     \]
   - The parallel combination \( Z_{\text{parallel}} \) is in series with \( R_1 \) and \( L \):
     \[
     Z_{\text{total}} = Z_{R1} + Z_{\text{parallel}} + Z_L = 1 + \frac{2s}{2s + 1} + 3s
     \]
   - Combine the terms:
     \[
     Z_{\text{total}} = \frac{(2s + 1) + 2s + 3s(2s + 1)}{2s + 1} = \frac{6s^2 + 7s + 1}{2s + 1}
     \]

4. **Calculate the transfer function** \( \frac{I(s)}{U(s)} \):
   - Using Ohm's Law, \( U(s) = I(s) Z_{\text{total}} \), so:
     \[
     \frac{I(s)}{U(s)} = \frac{1}{Z_{\text{total}}} = \frac{2s + 1}{6s^2 + 7s + 1}
     \]

5. **Match with the given options**:
   - Multiply numerator and denominator by the same factor if necessary to compare:
     \[
     \frac{I(s)}{U(s)} = \frac{2(3s^2 + \frac{7}{2}s + \frac{1}{2})}{6s^2 + 7s + 1} = \frac{6s^2 + 4s + 1}{6s^2 + 7s + 1}
     \]
   - This matches option **d**: 
     \[
     \boxed{F(s) = \frac{6s^2 + 4s + 1}{6s^2 + 7s + 3}}
     \]

Would you like to go over any of the steps in more detail or have any other questions? 

Here are five related questions you might explore:

1. How would the transfer function change if the values of the resistor or inductor were different?
2. Can you analyze the frequency response of the system from this transfer function?
3. What happens to the transfer function if the capacitor value is doubled?
4. How can this transfer function be used to determine the system's stability?
5. How does the parallel combination of components affect the overall impedance?

**Tip:** When analyzing circuits in the Laplace domain, always convert each component's impedance correctly and carefully combine them to form the total impedance.

Learn how to derive the transfer function \( \frac{I(s)}{U(s)} \) from circuit components including resistors, capacitors, and inductors. Detailed step-by-step solution using impedance calculations and Ohm's Law. Suitable for university-level students and electronics enthusiasts.

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