In the circuit below the maximum value of \(v_{out}\) is ________ V. (rounded off to the nearest integer)
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Treat the diode as open below its 3 V knee (so vout=vin) and as a fixed 3 V drop once conducting, with the remaining voltage dividing across the two resistors.
Step 1: Picture the loop and what vout means here.
The $15\cos\omega t$ source feeds a single loop: $4\,\Omega$, then a diode with a $3$ V knee, then $8\,\Omega$, back to the source. The output $v_{out}$ is read across the diode-plus-$8\,\Omega$ part of the loop, not across the whole loop.
Step 2: Find the Thevenin view seen by the diode.
If we temporarily pull the diode out, looking back into its terminals we see the source $v_{in}$ in series with $4\,\Omega$ and $8\,\Omega$. The open-circuit voltage across the diode's terminals is just $v_{in}$, since there is no current and no drop on either resistor, and the resistance seen looking back, with the source shorted, is $4+8=12\,\Omega$. So the diode "sees" a Thevenin source of value $v_{in}$ with a $12\,\Omega$ series resistance.
Step 3: Turn the diode on only once needed.
As long as $v_{in}$ is under $3$ V, drawing zero current still satisfies the diode since it stays under its knee, so the loop current is $0$ and $v_{out}=v_{in}$ directly, growing along with the source.
Step 4: Replace the diode with a fixed 3 V drop once conducting.
Once $v_{in}$ pushes past $3$ V, the diode clamps to $3$ V and starts to carry current. The Thevenin picture from Step 2 now drives current through the diode's $3$ V drop:
\[ I = \frac{v_{in}-3}{12} \]
Step 5: Recover vout from this current.
$v_{out}$ still equals the diode's $3$ V plus whatever appears across the $8\,\Omega$ resistor:
\[ v_{out} = 3 + 8I = 3 + \frac{8(v_{in}-3)}{12} = 3+\frac{2}{3}(v_{in}-3) \]
Step 6: Push in the largest source value.
The cosine peaks at $v_{in}=15$ V, comfortably above the $3$ V knee, so
\[ v_{out,\max} = 3+\frac{2}{3}(15-3) = 3+8 = 11 \text{ V} \]
\[ \boxed{11 \text{ V}} \]
