Tabii, bu tip anahtarlamalı devre sorularında adım adım çözüm aşağıdaki gibi yapılır: ### **Adım 1: Başlangıç (t < 0) - Anahtar "a" Konumunda** - Anahtar "a" konumunda, devrede sadece 50V kaynağı, 10Ω direnç ve 1H endüktans var. - **Endüktansın kararlı halde davranışı:** Kararlı halde endüktans kısa devre gibi davranır. **Devre:** 50V — 10Ω — 1H (kısa devre) **İndüktör akımı (t = 0⁻):** \[ i_L(0^-) = \frac{50V}{10\Omega} = 5 \text{ A} \] --- ### **Adım 2: Anahtar "b" Konumuna Geçiyor (t = 0 ve sonrası)** - Anahtar "b" konumuna geçince, 50V kaynağı ve 10Ω direnç devreden çıkıyor. - Devre: 1H indüktör, 5Ω direnç, \( \frac{1}{9} \) F kapasitör bir seri RLC devresi oluşturuyor. - **Endüktör akımı süreklidir:** \( i_L(0^+) = i_L(0^-) = 5A \) --- ### **Adım 3: Diferansiyel Denklemin Kurulması** Seri RLC devresi için KVL yazalım: \[ L \frac{di}{dt} + Ri + \frac{1}{C} \int i dt = 0 \] Verilenler: - \( L = 1 \) H - \( R = 5 \) Ω - \( C = \frac{1}{9} \) F Denklem: \[ 1 \frac{di}{dt} + 5i + 9 \int i dt = 0 \] Türevi alalım: \[ \frac{d^2 i}{dt^2} + 5 \frac{di}{dt} + 9i = 0 \] --- ### **Adım 4: Karakteristik Denklemin Çözümü** Karakteristik denklem: \[ s^2 + 5s + 9 = 0 \] Kökler: \[ s = \frac{-5 \pm \sqrt{25 - 36}}{2} = \frac{-5 \pm j\sqrt{11}}{2} \] Kökler karmaşık: \[ s_{1,2} = -\frac{5}{2} \pm j\frac{\sqrt{11}}{2} \] --- ### **Adım 5: Genel Çözüm** Çözüm: \[ i(t) = e^{-\frac{5}{2}t} \left[ A \cos\left(\frac{\sqrt{11}}{2}t\right) + B \sin\left(\frac{\sqrt{11}}{2}t\right) \right] \] --- ### **Adım 6: Başlangıç Şartlarını Bulma** - \( i(0^+) = 5 \) A (başlangıç akımı) - \( \frac{di}{dt} \bigg|_{t=0^+} \) için devreyi analiz edelim: KVL: \( L \frac{di}{dt} + Ri + v_C = 0 \) İlk anda kondansatör üzerindeki gerilim sıfırdır (\( v_C(0) = 0 \)), çünkü daha önce açık devreydi. \[ 1 \cdot \frac{di}{dt}\bigg|_{t=0} + 5 \cdot 5 + 0 = 0 \implies \frac{di}{dt}\bigg|_{t=0} = -25 \] --- ### **Adım 7: Sabitleri Bulma** \[ i(0) = 5 = A \] Türevi: \[ \frac{di}{dt} = e^{-\frac{5}{2}t} \left[ -\frac{5}{2}A \cos(\omega t) - A \omega \sin(\omega t) - \frac{5}{2} B \sin(\omega t) + B \omega \cos(\omega t) \right] \] \( \omega = \frac{\sqrt{11}}{2} \) \( t = 0 \) için: \[ \frac{di}{dt}\bigg|_{t=0} = -\frac{5}{2}A + B\omega = -25 \] \[ -\frac{5}{2} \cdot 5 + B \cdot \frac{\sqrt{11}}{2} = -25 \] \[ -\frac{25}{2} + B \cdot \frac{\sqrt{11}}{2} = -25 \] \[ B \cdot \frac{\sqrt{11}}{2} = -25 + \frac{25}{2} = -\frac{25}{2} \] \[ B = -\frac{25}{2} \cdot \frac{2}{\sqrt{11}} = -\frac{25}{\sqrt{11}} \] --- ### **Adım 8: Sonuç** \[ \boxed{ i(t) = e^{-\frac{5}{2}t} \left[ 5 \cos\left( \frac{\sqrt{11}}{2} t \right) - \frac{25}{\sqrt{11}} \sin\left( \frac{\sqrt{11}}{2} t \right) \right], \quad t > 0 } \] Her adımı detaylıca gösterdim, başka bir nokta açıklama isterseniz memnuniyetle yardımcı olurum!

Certainly! Here's a step-by-step solution to the problem in English: --- ### **Problem Overview:** - At \( t = 0 \), the switch is suddenly moved from position "a" to position "b". - Before \( t=0 \), the circuit is in steady state. - The goal: Find the inductor current \( i(t) \) for \( t > 0 \). --- ### **Step 1: Steady-State (for \( t < 0 \))** - When the switch is at position "a" (for \( t<0 \)), the circuit consists of: - 50 V voltage source - 10 Ω resistor - 1 H inductor - **In steady state**, the inductor acts as a short circuit (since current is constant). - **Calculate initial current \( i(0^-) \):** \[ i(0^-) = \frac{V}{R} = \frac{50\,V}{10\,Ω} = 5\,A \] - The initial current through the inductor at \( t=0^- \) is **5 A**. --- ### **Step 2: At \( t=0^+ \) (immediately after switching)** - The circuit switches to a different configuration (position "b"). - The voltage source and resistor are disconnected. - The inductor current is **continuous** at the instant of switching: \[ i(0^+) = i(0^-) = 5\,A \] - The circuit now is a series RLC circuit with: - 1 H inductor - 5 Ω resistor - Capacitance \( C = \frac{1}{9} \,F \) --- ### **Step 3: Write the differential equation** - The series RLC circuit obeys: \[ L \frac{d^2 i}{dt^2} + R \frac{di}{dt} + \frac{1}{C} i = 0 \] - Substituting values: \[ 1 \cdot \frac{d^2 i}{dt^2} + 5 \frac{di}{dt} + 9 i = 0 \] --- ### **Step 4: Solve the characteristic equation** - Characteristic equation: \[ s^2 + 5 s + 9 = 0 \] - Roots: \[ s = \frac{-5 \pm \sqrt{25 - 36}}{2} = \frac{-5 \pm j \sqrt{11}}{2} \] - Roots are complex conjugates: \[ s_{1,2} = -\frac{5}{2} \pm j \frac{\sqrt{11}}{2} \] --- ### **Step 5: Write the general solution** - The solution for \( i(t) \): \[ i(t) = e^{-\frac{5}{2} t} \left[ A \cos\left( \frac{\sqrt{11}}{2} t \right) + B \sin\left( \frac{\sqrt{11}}{2} t \right) \right] \] --- ### **Step 6: Apply initial conditions** - At \( t=0 \): \[ i(0) = A = 5\,A \] - To find \( B \), differentiate \( i(t) \): \[ \frac{di}{dt} = e^{-\frac{5}{2} t} \left[ -\frac{5}{2} A \cos(\omega t) - A \omega \sin(\omega t) + -\frac{5}{2} B \sin(\omega t) + B \omega \cos(\omega t) \right] \] where \( \omega = \frac{\sqrt{11}}{2} \). - Evaluate at \( t=0 \): \[ \frac{di}{dt}\bigg|_{t=0} = -\frac{5}{2} A + B \omega \] - From circuit analysis (initial inductor behavior): \[ L \frac{di}{dt}\bigg|_{t=0^+} + R i(0^+) = 0 \Rightarrow 1 \cdot \frac{di}{dt} + 5 \times 5 = 0 \] \[ \Rightarrow \frac{di}{dt} = -25\,A/s \] - Now substitute known values: \[ -\frac{5}{2} \times 5 + B \times \frac{\sqrt{11}}{2} = -25 \] \[ - \frac{25}{2} + B \times \frac{\sqrt{11}}{2} = -25 \] \[ B \times \frac{\sqrt{11}}{2} = -25 + \frac{25}{2} = -\frac{25}{2} \] \[ B = -\frac{25/2}{\sqrt{11}/2} = -\frac{25}{\sqrt{11}} \] --- ### **Final answer:** \[ \boxed{ i(t) = e^{-\frac{5}{2} t} \left[ 5 \cos\left( \frac{\sqrt{11}}{2} t \right) - \frac{25}{\sqrt{11}} \sin\left( \frac{\sqrt{11}}{2} t \right) \right], \quad t > 0 } \] This describes the inductor current after the switch is moved at \( t=0 \). --- If you need further clarification on any step, feel free to ask!

Certainly! Here's the complete step-by-step solution in mathematical equations: --- ### **Initial Conditions (\( t < 0 \))** - Circuit: 50 V source, 10 Ω resistor, 1 H inductor in series. - **Steady state inductor current:** \[ i(0^-) = \frac{V}{R} = \frac{50\,V}{10\,Ω} = 5\,A \] --- ### **At \( t = 0 \), the switch moves from position "a" to "b"** - **Initial inductor current:** \[ i(0^+) = i(0^-) = 5\,A \] - The circuit after switching is a series RLC circuit with: \[ L = 1\,H,\quad R=5\,Ω,\quad C=\frac{1}{9}\,F \] --- ### **Differential Equation for \( t > 0 \)** Applying Kirchhoff's Voltage Law: \[ L \frac{d^2 i}{dt^2} + R \frac{di}{dt} + \frac{1}{C} i = 0 \] Substitute known values: \[ 1 \cdot \frac{d^2 i}{dt^2} + 5 \frac{di}{dt} + 9 i = 0 \] or: \[ \boxed{ \frac{d^2 i}{dt^2} + 5 \frac{di}{dt} + 9 i = 0 } \] --- ### **Characteristic Equation** \[ s^2 + 5s + 9 = 0 \] Discriminant: \[ \Delta = 25 - 36 = -11 \] Roots: \[ s = \frac{-5 \pm j \sqrt{11}}{2} \] which are complex conjugates: \[ \boxed{ s_{1,2} = -\frac{5}{2} \pm j \frac{\sqrt{11}}{2} } \] --- ### **General Solution** \[ \boxed{ i(t) = e^{-\frac{5}{2} t} \left[ A \cos\left( \frac{\sqrt{11}}{2} t \right) + B \sin\left( \frac{\sqrt{11}}{2} t \right) \right] } \] --- ### **Initial Conditions** - **At \( t=0 \):** \[ i(0) = A = 5 \] - **Derivative \( \frac{di}{dt} \) at \( t=0 \):** Calculate: \[ \frac{di}{dt} = e^{-\frac{5}{2} t} \left[ -\frac{5}{2} A \cos\left( \frac{\sqrt{11}}{2} t \right) - A \cdot \frac{\sqrt{11}}{2} \sin\left( \frac{\sqrt{11}}{2} t \right) + -\frac{5}{2} B \sin\left( \frac{\sqrt{11}}{2} t \right) + B \cdot \frac{\sqrt{11}}{2} \cos\left( \frac{\sqrt{11}}{2} t \right) \right] \] At \( t=0 \): \[ \frac{di}{dt}\bigg|_{t=0} = -\frac{5}{2} A + B \cdot \frac{\sqrt{11}}{2} \] From circuit considerations: \[ L \frac{di}{dt}\bigg|_{0^+} + R i(0^+) = 0 \Rightarrow 1 \times \frac{di}{dt}\bigg|_{0^+} + 5 \times 5 = 0 \] \[ \Rightarrow \frac{di}{dt}\bigg|_{0^+} = -25 \] Substitute \( A=5 \): \[ - \frac{5}{2} \times 5 + B \cdot \frac{\sqrt{11}}{2} = -25 \] \[ - \frac{25}{2} + B \cdot \frac{\sqrt{11}}{2} = -25 \] Multiply through by 2: \[ -25 + B \sqrt{11} = -50 \] \[ B \sqrt{11} = -50 + 25 = -25 \] \[ \boxed{ B = - \frac{25}{\sqrt{11}} } \] --- ### **Final Expression for \( i(t) \)** \[ \boxed{ i(t) = e^{-\frac{5}{2} t} \left[ 5 \cos\left( \frac{\sqrt{11}}{2} t \right) - \frac{25}{\sqrt{11}} \sin\left( \frac{\sqrt{11}}{2} t \right) \right], \quad t > 0 } \] This describes the inductor current after switching at \( t=0 \).