Find the input impedance \(Z_{in}(s)\) of the coupled-inductor network shown.
Find the input impedance \(Z_{in}(s)\) of the coupled-inductor network shown.
Show Hint
- \(\dfrac{23s^2 + 46s + 20}{4s + 5}\)
- \(6s + 4\)
- \(7s + 4\)
- \(\dfrac{25s^2 + 46s + 20}{4s + 5}\)
The Correct Option is A
Solution and Explanation
Step 1: Mutual inductance.
Given inductances: \(L_1 = 6H,\; L_2 = 4H,\; M = 1H\). Using dot convention, the impedance matrix is: \[ Z = \begin{bmatrix} 6s & -Ms \\ -Ms & 4s + 5 \end{bmatrix} = \begin{bmatrix} 6s & -s \\ -s & 4s + 5 \end{bmatrix} \] Step 2: Input impedance when secondary has 5Ω load.
Reflect load to primary: \[ Z_{in}(s) = 4 + Z_{11} - \frac{Z_{12}^2}{Z_{22}} \] \[ = 4 + 6s - \frac{(-s)^2}{4s + 5} \] \[ = 4 + 6s - \frac{s^2}{4s + 5} \] Step 3: Combine terms.
\[ Z_{in}(s) = \frac{(4 + 6s)(4s + 5) - s^2}{4s + 5} \] Expand numerator: \[ (4 + 6s)(4s + 5) = 16s + 20 + 24s^2 + 30s = 24s^2 + 46s + 20 \] Therefore: \[ Z_{in}(s)=\frac{24s^2 + 46s + 20 - s^2}{4s + 5} \] \[ Z_{in}(s)=\frac{23s^2 + 46s + 20}{4s + 5} \] Step 4: Final result.
Matches option (A).
Final Answer: \(\boxed{\dfrac{23s^2 + 46s + 20}{4s + 5}}\)
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