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The image contains a transistor amplifier circuit with labeled resistors, capacitors, and voltage sources. Your task asks for DC values related to the circuit: 

**DC Values to Determine:**
(a) \(V_B\) (Base voltage)  
(b) \(V_E\) (Emitter voltage)  
(c) \(I_E\) (Emitter current)  
(d) \(I_C\) (Collector current)  
(e) \(V_C\) (Collector voltage)

Let's proceed step by step to determine these values.

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### 1. **Assumptions and Key Relations**
- This is a **common-emitter amplifier** with resistive biasing.
- **DC Beta (\(\beta_{DC}\))** = 90
- \(V_{CC} = 15 \, V\)
- \(R_1 = 22 \, k\Omega\), \(R_2 = 4.7 \, k\Omega\), \(R_C = 2.2 \, k\Omega\), \(R_E = 1 \, k\Omega\)

Key relationships:
- **Base voltage (\(V_B\))**: Voltage divider across \(R_1\) and \(R_2\)
- **Emitter voltage**: \(V_E = V_B - V_{BE}\), where \(V_{BE} \approx 0.7 \, V\)
- **Emitter current**: \(I_E \approx \frac{V_E}{R_E}\)  
- **Collector current**: \(I_C \approx I_E\) (since \(I_C \approx I_E / (1 + \beta)\))
- **Collector voltage**: \(V_C = V_{CC} - I_C R_C\)

---

### 2. **Calculate DC Values Step-by-Step**

#### (a) \(V_B\) - Base Voltage
Use the **voltage divider rule**:
\[
V_B = V_{CC} \times \frac{R_2}{R_1 + R_2}
\]
\[
V_B = 15 \, V \times \frac{4.7 \, k\Omega}{22 \, k\Omega + 4.7 \, k\Omega}
\]
\[
V_B = 15 \, V \times \frac{4.7}{26.7} \approx 2.64 \, V
\]

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#### (b) \(V_E\) - Emitter Voltage
\[
V_E = V_B - V_{BE} \approx 2.64 \, V - 0.

Determine the following DC values for the amplifier: (a) VB, (b) VE, (c) IE, (d) IC, (e) VC. The given circuit includes resistors R1 = 22 kΩ, R2 = 4.7 kΩ, RC = 2.2 kΩ, and RE = 1.0 kΩ, with a power supply of VCC = 15 V. The DC Beta (βDC) is 90.

Learn how to determine the DC values for a transistor amplifier circuit with resistive biasing. The circuit parameters include resistors R1 = 22 kΩ, R2 = 4.7 kΩ, RC = 2.2 kΩ, and RE = 1.0 kΩ, with VCC = 15 V and βDC = 90. This tutorial covers key electronics concepts like transistor biasing and Ohm's Law.

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