1. **Problem Statement:** Find the input impedance $Z_{in}$ of the given circuit at angular frequency $\omega = 10\,000$ rad/s (10 krad/s). 2. **Circuit Elements and Values:** - Resistor $R = 50\ \Omega$ - Inductor $L = 2$ mH $= 2 \times 10^{-3}$ H - Capacitor $C = 1$ $\mu$F $= 1 \times 10^{-6}$ F - Dependent voltage source with voltage $2v$ where $v$ is the voltage across the resistor. 3. **Angular Frequency:** $$\omega = 10,000\ \text{rad/s}$$ 4. **Impedances of Inductor and Capacitor:** - Inductive reactance: $$Z_L = j\omega L = j(10,000)(2 \times 10^{-3}) = j20\ \Omega$$ - Capacitive reactance: $$Z_C = \frac{1}{j\omega C} = \frac{1}{j(10,000)(1 \times 10^{-6})} = \frac{1}{j0.01} = -j100\ \Omega$$ 5. **Define Variables:** - Let $v$ be the voltage across the resistor $R$. - The dependent voltage source voltage is $2v$. 6. **Write KVL (Kirchhoff's Voltage Law) around the loop:** Starting from the input terminal and moving clockwise: $$v + Z_L i + (-2v) = 0$$ where $i$ is the current through the series elements. 7. **Express $v$ in terms of $i$:** Since $v$ is voltage across resistor $R$, $$v = R i = 50 i$$ 8. **Substitute $v$ into KVL:** $$50 i + j20 i - 2(50 i) = 0$$ Simplify: $$50 i + j20 i - 100 i = 0$$ $$(-50 + j20) i = 0$$ 9. **This implies $i=0$ or the equation is balanced. But we need to consider the capacitor branch connected from the node between inductor and dependent source to ground.** 10. **Node voltage at the node between inductor and dependent source:** Let this node voltage be $V_x$. 11. **Current through capacitor:** $$i_C = \frac{V_x}{Z_C} = \frac{V_x}{-j100} = j0.01 V_x$$ 12. **Current through inductor and dependent source branch is $i$. The current $i$ splits at node $V_x$ into $i_C$ and current through dependent source branch.** 13. **Apply KCL (Kirchhoff's Current Law) at node $V_x$:** Current into node from inductor is $i$. Current out through capacitor is $i_C$. Current out through dependent voltage source branch is $i_{dep}$. Since dependent voltage source is in series with resistor and inductor, the current $i$ flows through it, so no separate branch current. 14. **Voltage at node $V_x$ is:** $$V_x = v + Z_L i = 50 i + j20 i = (50 + j20) i$$ 15. **Current through capacitor:** $$i_C = j0.01 V_x = j0.01 (50 + j20) i = j0.01 \times 50 i + j0.01 \times j20 i = j0.5 i + j^2 0.2 i = j0.5 i - 0.2 i = (-0.2 + j0.5) i$$ 16. **Total input current $i_{in}$ is current through resistor and inductor branch, which is $i$, plus current through capacitor $i_C$:** $$i_{in} = i + i_C = i + (-0.2 + j0.5) i = (1 - 0.2 + j0.5) i = (0.8 + j0.5) i$$ 17. **Input voltage $V_{in}$ is voltage across resistor $R$ plus inductor $L$ plus dependent source voltage:** $$V_{in} = v + Z_L i + (-2v) = 50 i + j20 i - 2(50 i) = (50 + j20 - 100) i = (-50 + j20) i$$ 18. **Input impedance $Z_{in}$ is:** $$Z_{in} = \frac{V_{in}}{i_{in}} = \frac{(-50 + j20) i}{(0.8 + j0.5) i} = \frac{-50 + j20}{0.8 + j0.5}$$ 19. **Calculate $Z_{in}$ by multiplying numerator and denominator by complex conjugate of denominator:** $$Z_{in} = \frac{(-50 + j20)(0.8 - j0.5)}{(0.8 + j0.5)(0.8 - j0.5)}$$ Denominator: $$0.8^2 + 0.5^2 = 0.64 + 0.25 = 0.89$$ Numerator: $$(-50)(0.8) + (-50)(-j0.5) + j20(0.8) + j20(-j0.5) = -40 + j25 + j16 - 10$$ Simplify numerator: $$(-40 - 10) + j(25 + 16) = -50 + j41$$ 20. **Final $Z_{in}$:** $$Z_{in} = \frac{-50 + j41}{0.89} = -56.18 + j46.07\ \Omega$$ **Answer:** $$\boxed{Z_{in} = -56.18 + j46.07\ \Omega}$$ This is the input impedance at $10$ krad/s for the given circuit.