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Find an expression for 𝑖𝑙 in the laplace domain for the circuit below after the switch is opened.

Assume that:

𝐼=0.7 A
𝑅=10.0Ω
𝐿=5.0 mH
𝐶=20 mF

The expression should be in the format of 𝑖𝐿 = A/(s(s^2 + B*s +C)). Please express coefficients A, B, and C as whole numbers.

As the question asks for a Laplace equation, we'll need to do a Laplace transform.
1. Transform impedances and sources into the Laplace domain using: Zc = 1/sC, Zl = sL, Zr = R, and I = i/s.
2. Form a current divider using the s-domain impedances: I_l = I_t * 1/Zl / (1/Zl + 1/Zc + 1/Zr)
3. solve and factorise out some s's and constants to get it in the desired form!

To find the expression for \(i_L\) in the Laplace domain for the given circuit after the switch is opened, we need to follow a series of steps to transform the circuit elements and solve for the desired current.

### Step 1: Transform impedances and sources into the Laplace domain

Given:
- \( I = 0.7 \text{ A} \)
- \( R = 10.0 \, \Omega \)
- \( L = 5.0 \text{ mH} = 5.0 \times 10^{-3} \text{ H} \)
- \( C = 20 \text{ mF} = 20 \times 10^{-3} \text{ F} \)

The impedances in the Laplace domain are:
- \( Z_R = R = 10 \, \Omega \)
- \( Z_L = sL = s \cdot 5.0 \times 10^{-3} = 5.0 \times 10^{-3} s \, \Omega \)
- \( Z_C = \frac{1}{sC} = \frac{1}{s \cdot 20 \times 10^{-3}} = \frac{1}{20 \times 10^{-3} s} = \frac{50}{s} \, \Omega \)

### Step 2: Form a current divider using the s-domain impedances

The total current \( I \) splits between the resistor \( R \), the inductor \( L \), and the capacitor \( C \). We are interested in the current through the inductor \( L \).

The total impedance \( Z_{total} \) is:
\[ Z_{total} = Z_R + Z_L + Z_C = 10 + 5.0 \times 10^{-3} s + \frac{50}{s} \]

Using the current divider rule, the current through the inductor \( I_L(s) \) is:
\[ I_L(s) = I \cdot \frac{1/Z_L}{1/Z_L + 1/Z_C + 1/Z_R} \]

Simplifying the impedance terms:
\[ \frac{1}{Z_L} = \frac{1}{5.0 \times 10^{-3} s} = \frac{200}{s} \]
\[ \frac{1}{Z_C} = \frac{s}{50} \]
\[ \frac{1}{Z_R} = \frac{1}{10} \]

Therefore, the total admittance is:
\[ Y_{total} = \frac{1}{Z_L} + \frac{1}{Z_C} + \frac{1}{Z_R} = \frac{200}{s} + \frac{s}{50} + \frac{1}{10} \]

### Step 3: Solve for \( I_L(s) \)

Let's combine the admittances:
\[ Y_{total} = \frac{200}{s} + \frac{s}{50} + \frac{1}{10} \]
\[ Y_{total} = \frac{2000 + s^2 + 5s}{50s} \]
\[ Y_{total} = \frac{s^2 + 5s + 2000}{50s} \]

Thus, the total impedance \( Z_{total} \) is:
\[ Z_{total} = \frac{50s}{s^2 + 5s + 2000} \]

Now, the current \( I_L(s) \) is given by:
\[ I_L(s) = I \cdot \frac{1/Z_L}{Y_{total}} \]
\[ I_L(s) = 0.7 \cdot \frac{200}{s} \cdot \frac{50s}{s^2 + 5s + 2000} \]
\[ I_L(s) = 0.7 \cdot \frac{200 \cdot 50}{s^2 + 5s + 2000} \]
\[ I_L(s) = 0.7 \cdot \frac{10000}{s^2 + 5s + 2000} \]

Therefore, the expression for \( I_L(s) \) is:
\[ I_L(s) = \frac{7000}{s^2 + 5s + 2000} \]

In the form \( i_L = \frac{A}{s(s^2 + Bs + C)} \), the coefficients are:
- \( A = 7000 \)
- \( B = 5 \)
- \( C = 2000 \)

Would you like more details or have any questions about this solution? Here are some related questions:

1. How do we derive the Laplace transform for an inductor and a capacitor?
2. What is the significance of the impedance in the Laplace domain?
3. How does the current divider rule work in the s-domain?
4. What are the steps to solve circuits with multiple components in the Laplace domain?
5. How do initial conditions affect the Laplace transform of a circuit?
6. What are some common mistakes to avoid when applying Laplace transforms in circuit analysis?
7. Can we use the same approach for circuits with dependent sources?
8. How can we verify the solution obtained using the Laplace transform?

**Tip:** Always double-check the units of your components when transforming to the Laplace domain to avoid errors in your calculations.

Learn how to derive the Laplace domain expression for the current through an inductor in a circuit after the switch is opened. This detailed solution covers the use of Laplace transforms, impedance calculations, and the current divider rule. Suitable for advanced college-level students and engineers.

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