Question:

Please convert the below content to rendered equations without any errors.Alright, let’s walk through it cleanly and **step by step**, exactly like you’d write it in your notebook ? --- ## 1. Understand the given equivalent circuit From the figure: ### Input side * The input signal is a **Thévenin source**: * ( V_s = \pm 0.3 , \text{V} ) * ( R_s = 50 , \Omega ) * This drives a **270 Ω resistor**. * The voltage across the 270 Ω resistor is labeled **(V_x)**. ### Dependent source * There is a **voltage-dependent current source**: [ I_d = 2.2,\text{mS} \cdot V_x = 0.0022,V_x \quad (\text{A}) ] ### Output side * Load resistor: (10,\text{k}\Omega) * Oscilloscope input resistance: (1,\text{M}\Omega) These two are in **parallel**. --- ## 2. Combine the output resistances [ R_{out} = 10,\text{k}\Omega \parallel 1,\text{M}\Omega ] [ R_{out} = \frac{(10,000)(1,000,000)}{10,000 + 1,000,000} \approx 9.9,\text{k}\Omega ] (Almost the same as 10 kΩ) --- ## 3. Find (V_x) (input voltage across 270 Ω) The input forms a voltage divider: [ V_x = V_s \cdot \frac{270}{270 + 50} ] [ V_x = V_s \cdot \frac{270}{320} = 0.84375,V_s ] --- ## 4. Current from the dependent source [ I_d = 0.0022 \cdot V_x ] Substitute (V_x): [ I_d = 0.0022 \cdot (0.84375,V_s) = 0.001856,V_s \quad (\text{A}) ] --- ## 5. Output voltage calculation The dependent current flows **downward**, producing a voltage across (R_{out}): [ V_{out} = I_d \cdot R_{out} ] [ V_{out} = (0.001856,V_s)(9900) ] [ V_{out} \approx 18.38,V_s ] ? This is the **voltage gain** of the circuit. --- ## 6. Numerical results ### Case 1: (V_s = +0.3,\text{V}) [ V_{out} = 18.38 \times 0.3 \approx \boxed{+5.51,\text{V}} ] --- ### Case 2: (V_s = -0.3,\text{V}) [ V_{out} = 18.38 \times (-0.3) \approx \boxed{-5.51,\text{V}} ] --- ## 7. Final answers (Table 1) | (V_{in}) of source | (V_{out}) | | ------------------ | ----------- | | (+0.3,\text{V}) | **+5.51 V** | | (-0.3,\text{V}) | **−5.51 V** | --- ### ✔ Key takeaway * The circuit behaves like a **linear amplifier** * Gain ≈ **18.4** * Output polarity follows input polarity If you want, I can also: * redraw the small-signal equivalent, * show KCL at the output node, or * explain how this models an **op-amp transconductance stage** ?