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.

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: $V_B = V_{CC} \times \frac{R_2}{R_1 + R_2}$ $V_B = 15 \, V \times \frac{4.7 \, k\Omega}{22 \, k\Omega + 4.7 \, k\Omega}$ $V_B = 15 \, V \times \frac{4.7}{26.7} \approx 2.64 \, V$

(b) $V_E$ - Emitter Voltage

[ V_E = V_B - V_{BE} \approx 2.64 , V - 0.