Topic: Transmission Lines
Characteristic impedance of a transmission line is 50 $\Omega$. Input impedance of open circuited line is $Zoc=100+j150~\Omega$. The value of short circuited input impedance is
- A. 7.69-j11.54 $\Omega$
- B. 2-j50 $\Omega$
- C. 50 $\Omega$
- D. $100+j150~\Omega$
Answer: A. 7.69-j11.54 $\Omega$
Explanation: The characteristic impedance $Z_0$ of a transmission line is related to its open-circuit impedance $Z_{oc}$ and short-circuit impedance $Z_{sc}$ by the formula: $Z_0 = \sqrt{Z_{oc} Z_{sc}}$.
Given $Z_0 = 50~\Omega$ and $Z_{oc} = 100 + j150~\Omega$.
$Z_{sc} = \frac{Z_0^2}{Z_{oc}} = \frac{50^2}{100 + j150} = \frac{2500}{100 + j150}$.
To simplify, multiply numerator and denominator by the complex conjugate $(100 - j150)$:
$Z_{sc} = \frac{2500(100 - j150)}{100^2 + 150^2} = \frac{2500(100 - j150)}{10000 + 22500} = \frac{2500(100 - j150)}{32500} = \frac{100 - j150}{13} = 7.69 - j11.54~\Omega$.