Resimli Soru 24-11-2025 22:23:40

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1

!Screenshot_20251124_222211_Samsung Notes|690x324 [Link Silindi]

2

Here’s a step-by-step Thevenin approach to find the open-circuit voltage V₀ (which equals V_Th):

  1. Label nodes:

    • Let the bottom rail be 0 V (ground).
    • Call the central top node “Node 1” (voltage V₁).
    • Call the left top node “Node 2” (voltage V₂).
    • The output node (top right) is “Node 3” ⇒ V₀ = V₃.

  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.

3

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
  2. Step 1: Remove the Load
  3. Step 2: Find Thevenin Equivalent Voltage V_{th}
  4. Step 3: Find Thevenin Equivalent Resistance R_{th}
  5. Step 4: Calculate V_0
  6. 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