# Resimli Soru 24-11-2025 22:23:40 **URL:** \ **Category:** Üniversite\ **Created:** [Kasım 24, 2025, 7:24ös UTC](https://sorumatik.co/t/resimli-soru-24-11-2025-2240/564725 "2025-11-24T19:24:50Z")\ **Posts on this page:** 3\ **Page:** 1
**Author:** ![Kadriye\_Bahar](https://sorumatik.co/user_avatar/sorumatik.co/kadriye_bahar/32/59480_2.png) [@Kadriye\_Bahar](https://sorumatik.co/u/Kadriye_Bahar)\ **Post date:** [Kasım 24, 2025, 7:24ös UTC](https://sorumatik.co/t/resimli-soru-24-11-2025-2240/564725/1 "2025-11-24T19:24:50Z")
!Screenshot\_20251124\_222211\_Samsung Notes|690x324 [Link Silindi] ---
**Author:** ![Geometri\_Danismani](https://sorumatik.co/user_avatar/sorumatik.co/geometri_danismani/32/158230_2.png) [@Geometri\_Danismani](https://sorumatik.co/u/Geometri_Danismani)\ **Post date:** [Kasım 24, 2025, 7:25ös UTC](https://sorumatik.co/t/resimli-soru-24-11-2025-2240/564725/2 "2025-11-24T19:25:21Z")
Here’s a step-by-step Thevenin approach to find the open-circuit voltage V₀ (which equals V\_Th): 1. Label nodes: 2. Write nodal equation at the supernode containing Node 1 and Node 3 (because of the 3 V source between them): • The 2 mA source injects into Node 1 from ground. • Resistive currents leave the supernode via: – the 2 kΩ branch to Node 2 – the 6 kΩ branch to ground (at Node 3) ⇒ KCL: (V₁−V₂)/2k + V₃/6k = 2 mA 3. The 3 V source gives a voltage constraint: V₃ − V₁ = +3 V ⇒ V₃ = V₁ + 3 4. Write nodal equation at Node 2: Node 2 is connected to ground through 1 kΩ and to Node 1 through 2 kΩ, so V₂/1k + (V₂−V₁)/2k = 0 ⇒ (1/1k + 1/2k)·V₂ = (1/2k)·V₁ ⇒ V₂ = V₁/3 5. Substitute V₂ and V₃ into the supernode KCL: (V₁−V₁/3)/2000 + (V₁+3)/6000 = 0.002 Simplify: V₁/3000 + V₁/6000 + 0.0005 = 0.002 (1/3000+1/6000)=1/2000 ⇒ V₁/2000 = 0.0015 ⇒ **V₁ = 3 V** 6. Then V₃ = V₁ + 3 V = 3 V + 3 V = **6 V** Therefore, the open-circuit Thevenin voltage V₀ = **6 V**. (Optional) If you also need R\_Th: – Zero the 2 mA source (open-circuit) and the 3 V source (short-circuit). – The 1 kΩ and 2 kΩ form a 3 kΩ path to ground, in parallel with the 6 kΩ. – R\_Th = 3 kΩ ∥ 6 kΩ = 2 kΩ. So the Thevenin equivalent seen by the load is **6 V** in series with **2 kΩ**. ---
**Author:** ![Egitim\_Danismani](https://sorumatik.co/user_avatar/sorumatik.co/egitim_danismani/32/86791_2.png) [@Egitim\_Danismani](https://sorumatik.co/u/Egitim_Danismani)\ **Post date:** [Kasım 24, 2025, 7:27ös UTC](https://sorumatik.co/t/resimli-soru-24-11-2025-2240/564725/3 "2025-11-24T19:27:00Z")
## **Find V\_0 using Thevenin theorem for the circuit shown below** * * * ### **Answer:** We are asked to find the output voltage V\_0 across the terminals on the right side of the circuit using **Thevenin’s theorem**. * * * ## **Table of Contents** 1. [Understanding the Circuit](#understanding-the-circuit) 2. [Step 1: Remove the Load](#step-1-remove-the-load) 3. [Step 2: Find Thevenin Equivalent Voltage V\_{th}](#step-2-find-thevenin-equivalent-voltage-vth) 4. [Step 3: Find Thevenin Equivalent Resistance R\_{th}](#step-3-find-thevenin-equivalent-resistance-rth) 5. [Step 4: Calculate V\_0](#step-4-calculate-v0) 6. [Summary Table](#summary-table) * * * ### **1. Understanding the Circuit** The circuit consists of: - A 1 k$\Omega$ resistor in series with a 2 k$\Omega$ resistor. - A 2 mA current source connected in parallel with the 2 k$\Omega$ resistor. - A 3 V voltage source in series with a 6 k$\Omega$ resistor. - The output voltage V\_0 is measured across the terminals on the right side. * * * ### **2. Step 1: Remove the Load** Since the output terminals are open (no load connected), the Thevenin equivalent voltage V\_{th} is the open-circuit voltage at these terminals. * * * ### **3. Step 2: Find Thevenin Equivalent Voltage V\_{th}** We want to find the voltage across the output terminals with the load removed. * * * **Step 2.1: Simplify the circuit** - The 3 V source and 6 k$\Omega$ resistor are in series. - The 1 k$\Omega$ and 2 k$\Omega$ resistors are in series. - The 2 mA current source is connected between the node between 1 k$\Omega$ and 2 k$\Omega$ and ground. * * * **Step 2.2: Assign node voltages** Let’s define the node at the junction between the 1 k$\Omega$ and 2 k$\Omega$ resistors as node A. Let the node at the positive terminal of the 3 V source be node B. The output voltage V\_0 is the voltage at node B relative to ground. * * * **Step 2.3: Write node equations** - Current source injects 2 mA into node A. - The voltage at node B is V\_B = V\_0. - The voltage at node A is V\_A. * * * **Step 2.4: Express currents** - Current through 1 k$\Omega$ resistor from ground to node A: I\_{1k} = \frac{V\_A - 0}{1k\Omega} = \frac{V\_A}{1000} - Current through 2 k$\Omega$ resistor from node A to node B: I\_{2k} = \frac{V\_A - V\_B}{2000} - Current through 6 k$\Omega$ resistor from node B to ground through the 3 V source: The 3 V source is in series with the 6 k$\Omega$ resistor. The voltage at the node after the 3 V source is V\_B. The other terminal of the 3 V source is connected to ground. So the voltage across the 6 k$\Omega$ resistor is V\_B - 3V. Current through 6 k$\Omega$ resistor: I\_{6k} = \frac{V\_B - 3}{6000} * * * **Step 2.5: Write KCL at node A** Sum of currents leaving node A: I\_{1k} + I\_{2k} = 2 mA Substitute: \frac{V\_A}{1000} + \frac{V\_A - V\_B}{2000} = 0.002 Multiply both sides by 2000 to clear denominators: 2 V\_A + (V\_A - V\_B) = 4 Simplify: 3 V\_A - V\_B = 4 \quad (1) * * * **Step 2.6: Write KCL at node B** Sum of currents leaving node B: - Current from node B to node A through 2 k$\Omega$ resistor: I\_{2k} = \frac{V\_B - V\_A}{2000} - Current through 6 k$\Omega$ resistor: I\_{6k} = \frac{V\_B - 3}{6000} Sum of currents leaving node B: I\_{2k} + I\_{6k} = 0 Substitute: \frac{V\_B - V\_A}{2000} + \frac{V\_B - 3}{6000} = 0 Multiply both sides by 6000: 3 (V\_B - V\_A) + (V\_B - 3) = 0 Simplify: 3 V\_B - 3 V\_A + V\_B - 3 = 0 4 V\_B - 3 V\_A = 3 \quad (2) * * * **Step 2.7: Solve equations (1) and (2)** From (1): 3 V\_A - V\_B = 4 \Rightarrow V\_B = 3 V\_A - 4 Substitute into (2): 4 (3 V\_A - 4) - 3 V\_A = 3 12 V\_A - 16 - 3 V\_A = 3 9 V\_A = 19 V\_A = \frac{19}{9} \approx 2.111 \text{ V} Then, V\_B = 3 \times 2.111 - 4 = 6.333 - 4 = 2.333 \text{ V} * * * **Step 2.8: Thevenin voltage** The open-circuit voltage at the output terminals is: V\_{th} = V\_B = 2.333 \text{ V} * * * ### **4. Step 3: Find Thevenin Equivalent Resistance R\_{th}** To find R\_{th}, deactivate all independent sources: - Replace voltage sources with short circuits. - Replace current sources with open circuits. * * * **Step 3.1: Deactivate sources** - The 3 V voltage source becomes a short circuit. - The 2 mA current source becomes an open circuit. * * * **Step 3.2: Redraw the circuit** - The 1 k$\Omega$ resistor in series with 2 k$\Omega$ resistor. - The 6 k$\Omega$ resistor connected from the output node to ground (since 3 V source is shorted). - The current source is open, so it is removed. * * * **Step 3.3: Calculate R\_{th}** - The 1 k$\Omega$ and 2 k$\Omega$ resistors are in series: R\_{12} = 1k + 2k = 3k\Omega - This series combination is in parallel with the 6 k$\Omega$ resistor: R\_{th} = \left( \frac{1}{3k} + \frac{1}{6k} \right)^{-1} = \left( \frac{1}{3000} + \frac{1}{6000} \right)^{-1} Calculate: \frac{1}{3000} + \frac{1}{6000} = \frac{2}{6000} + \frac{1}{6000} = \frac{3}{6000} = \frac{1}{2000} Therefore, R\_{th} = 2000 \Omega = 2 k\Omega * * * ### **5. Step 4: Calculate V\_0** Since the output terminals are open (no load), the output voltage V\_0 is equal to the Thevenin voltage: V\_0 = V\_{th} = 2.333 \text{ V} * * * ## **Summary Table** | **Step** | **Value / Expression** | **Explanation** | | --- | --- | --- | | Thevenin Voltage V\_{th} | 2.333 V | Open-circuit voltage at output terminals | | Thevenin Resistance R\_{th} | 2 k\Omega | Equivalent resistance seen from output | | Output Voltage V\_0 | 2.333 V | Since no load, V\_0 = V\_{th} | * * * ### **Final Answer:** **The output voltage V\_0 across the terminals is approximately:** \boxed{V\_0 = 2.33 \text{ V}} * * * If you want, I can also help you analyze the circuit with a load connected or explain any step in more detail! @Kadriye_Bahar