Detailed Solution

To find the maximum value of the output voltage \( v_{\text{out}} \), we need to analyze the circuit during the positive peak of the input signal.

1. Identify the Input Signal

The input voltage is given as:

$$ v_{\text{in}}(t) = 15 \cos \omega t $$

The maximum (peak) value of the input voltage is \( V_{p} = 15 \) V.

2. Analyze the Circuit for the Positive Half-Cycle

When \( v_{\text{in}} \) is positive, the top terminal of the AC source is at a higher potential than the bottom terminal. In this condition:

• The Zener diode is in reverse bias.
• The Zener diode will enter the breakdown region if the voltage across it exceeds its Zener voltage (\( V_Z = 3 \) V).
• Since the peak input is 15 V, which is significantly higher than the Zener voltage, the diode acts as a constant voltage source of \( 3 \) V in series with the \( 8 \, \Omega \) resistor.

3. Calculation of Output Voltage (\( v_{\text{out}} \))

During the positive peak (\( v_{\text{in}} = 15 \) V), the circuit consists of the \( 4 \, \Omega \) resistor in series with the Zener branch (the Zener diode and the \( 8 \, \Omega \) resistor). We can find the output voltage using the node voltage method or current calculation.

Method 1: Using Kirchhoff's Current Law (KCL) at the output node

Let the voltage at the output node be \( v_{\text{out}} \). Applying KCL:

$$ \frac{15 - v_{\text{out}}}{4} = \frac{v_{\text{out}} - 3}{8} $$

Multiply both sides by 8 to clear the denominators:

$$ 2(15 - v_{\text{out}}) = v_{\text{out}} - 3 $$ $$ 30 - 2v_{\text{out}} = v_{\text{out}} - 3 $$ $$ 33 = 3v_{\text{out}} $$ $$ v_{\text{out}} = 11 \text{ V} $$

Method 2: Using the Series Current

The total current \( I \) in the circuit during the peak is:

$$ I = \frac{V_{\text{in}} - V_Z}{R_1 + R_2} = \frac{15 - 3}{4 + 8} = \frac{12}{12} = 1 \text{ A} $$

The output voltage is the voltage across the Zener branch:

$$ v_{\text{out}} = V_Z + (I \times 8 \, \Omega) = 3 + (1 \times 8) = 11 \text{ V} $$

4. Conclusion

During the negative half-cycle, the Zener diode will be forward-biased (acting like a standard diode with a drop of approximately \( 0.7 \) V), resulting in a negative output voltage. Therefore, the absolute maximum value occurs during the positive peak.

The maximum value of \( v_{\text{out}} \) is 11 V.