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To solve this problem, we need to analyze the given circuit and determine the behavior of \( v_o \) (output voltage) and \( i_R \) (current through the resistor) for the provided input voltage \( v_i \), which is a sinusoidal waveform oscillating between \( +10 \, \text{V} \) and \( -10 \, \text{V} \).

The circuit consists of:
- A series combination of a \( 1 \, \text{k}\Omega \) resistor and a silicon (Si) diode.
- A parallel resistor of \( 1 \, \text{k}\Omega \).

The behavior of the circuit depends on the forward biasing of the silicon diode, which typically starts conducting when the voltage across it exceeds approximately \( 0.7 \, \text{V} \).

### Steps to solve the problem:

1. **Input voltage behavior**:
   The input voltage \( v_i \) is a sinusoidal waveform alternating between \( +10 \, \text{V} \) and \( -10 \, \text{V} \), as shown in the figure.

2. **Diode behavior**:
   - **When \( v_i \) is positive**: The silicon diode will be forward biased when \( v_i \) exceeds \( 0.7 \, \text{V} \). This means the diode will start conducting after \( v_i > 0.7 \, \text{V} \).
   - **When \( v_i \) is negative**: The diode will be reverse biased, and it will not conduct. In this case, the current will flow only through the parallel \( 1 \, \text{k}\Omega \) resistor.

3. **Current through \( i_R \)**:
   - **For \( v_i \geq 0.7 \, \text{V} \)**: The diode is forward biased, and current flows through both the diode and the parallel resistor. The voltage drop across the diode will be approximately \( 0.7 \, \text{V} \), so the rest of the voltage appears across the series resistor.
   - **For \( v_i < 0.7 \, \text{V} \)** (positive and negative regions): The diode is off (non-conducting), and the entire voltage is dropped across the series and parallel resistors.

4. **Output voltage \( v_o \)**:
   - When the diode conducts, the voltage across the \( 1 \, \text{k}\Omega \) parallel resistor will be approximately \( 0.7 \, \text{V} \), because the diode clamps the voltage. 
   - When the diode does not conduct (in the negative half cycle), the output voltage \( v_o \) will be determined by the voltage divider formed by the two \( 1 \, \text{k}\Omega \) resistors.

### Sketch of \( v_o \) and \( i_R \):

- **\( v_o \)** will be clamped at approximately \( 0.7 \, \text{V} \) during the positive half-cycle of \( v_i \) when \( v_i > 0.7 \, \text{V} \). In the negative half-cycle, the diode will be off, and the output will behave according to the resistor voltage divider.
  
- **\( i_R \)**: The current through the series \( 1 \, \text{k}\Omega \) resistor will be proportional to the input voltage minus the forward voltage drop of the diode during the forward bias condition. When the diode is reverse biased (negative half-cycle of \( v_i \)), the current will flow only through the parallel resistor.

Would you like a detailed mathematical explanation or the plotted sketch for \( v_o \) and \( i_R \)? Let me know if you have any questions! 

Here are some additional questions to explore this circuit behavior:
1. What is the role of the silicon diode in shaping the output waveform?
2. How does the forward voltage drop of the diode affect the output voltage?
3. Can we replace the silicon diode with an ideal diode? How would that affect the output?
4. What would happen if the input voltage were increased beyond \( 10 \, \text{V} \)?
5. How would the circuit behave if the resistor values were changed?

**Tip**: Diodes are often used for clipping circuits, where part of an input signal is "clipped" or cut off. This can shape the output waveform into a desired form.

For the given circuit with a sinusoidal input voltage oscillating between +10V and -10V, sketch the output voltage Vo and the current iR through the resistor.

Learn how to analyze a circuit with a sinusoidal input voltage and a silicon diode. This problem covers key concepts such as diode clipping behavior, Ohm's Law, and the voltage divider rule. Suitable for undergraduate electrical engineering students.

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