https://file.aiquickdraw.com/mathbot/file/17265102025262wfr96mt.jpg

Let's break down and analyze the problem from the image:

### Part 1: Identifying the Amplitude \( V_m \) and the Peak-to-Peak Voltage \( V_{pp} \)

The given source voltage is:
\[
v_s(t) = 2 \cos(2\pi 1000t + 15^\circ) \, \text{V}
\]

This is a standard sinusoidal function. In such a case:
- **Amplitude** \( V_m \) is the coefficient of the cosine function, which is **2 V**.
- **Peak-to-Peak Voltage** \( V_{pp} \) is twice the amplitude. Therefore:
\[
V_m = 2 \, \text{V}, \quad V_{pp} = 2 \times V_m = 4 \, \text{V}
\]

### Part 2: Calculating \( v_{R1}(t) \), \( v_{R2}(t) \), and \( i(t) \)

This is an **AC voltage divider** circuit. The total impedance of the resistors \( R_1 \) and \( R_2 \) in series is given by:
\[
Z_{\text{total}} = R_1 + R_2 = 5.6 \, \text{k}\Omega + 12 \, \text{k}\Omega = 17.6 \, \text{k}\Omega
\]

The current \( i(t) \) in the circuit can be found using Ohm's law:
\[
i(t) = \frac{v_s(t)}{Z_{\text{total}}}
\]

Substituting the values:
\[
i(t) = \frac{2 \cos(2\pi 1000t + 15^\circ)}{17.6 \times 10^3} = \frac{2 \cos(2\pi 1000t + 15^\circ)}{17600} \, \text{A}
\]
\[
i(t) = 1.136 \times 10^{-4} \cos(2\pi 1000t + 15^\circ) \, \text{A}
\]

Now, using this current, we can calculate the voltage drops across each resistor.

#### Voltage across \( R_1 \):
\[
v_{R1}(t) = i(t) \times R_1 = (1.136 \times 10^{-4} \cos(2\pi 1000t + 15^\circ)) \times 5600
\]
\[
v_{R1}(t) = 0.636 \cos(2\pi 1000t + 15^\circ) \, \text{V}
\]

#### Voltage across \( R_2 \):
\[
v_{R2}(t) = i(t) \times R_2 = (1.136 \times 10^{-4} \cos(2\pi 1000t + 15^\circ)) \times 12000
\]
\[
v_{R2}(t) = 1.36 \cos(2\pi 1000t + 15^\circ) \, \text{V}
\]

### Summary:
- **Amplitude** \( V_m \): \( 2 \, \text{V} \)
- **Peak-to-Peak Voltage** \( V_{pp} \): \( 4 \, \text{V} \)
- **Current** \( i(t) \): \( 1.136 \times 10^{-4} \cos(2\pi 1000t + 15^\circ) \, \text{A} \)
- **Voltage across \( R_1 \)**: \( 0.636 \cos(2\pi 1000t + 15^\circ) \, \text{V} \)
- **Voltage across \( R_2 \)**: \( 1.36 \cos(2\pi 1000t + 15^\circ) \, \text{V} \)

Would you like a more detailed explanation of any part of the process?

Here are five questions related to this analysis:
1. What would change if \( R_2 \) were smaller than \( R_1 \)?
2. How does the phase angle \( 15^\circ \) affect the waveform of \( v_s(t) \)?
3. What happens to \( v_{R1}(t) \) and \( v_{R2}(t) \) if the input frequency changes?
4. How is the power dissipated across \( R_1 \) and \( R_2 \) calculated?
5. Can this voltage divider circuit be used to filter AC signals?

**Tip:** The voltage divider equation is a key tool in AC circuit analysis, where the impedance of the components may be frequency-dependent.

For the source voltage in Figure 1.1.3, what is the amplitude Vm? What is the peak-to-peak voltage Vpp? Calculate the voltages vR1(t), vR2(t), and the current i(t) in the circuit.

Solve an AC voltage divider circuit with source voltage vs(t) = 2cos(2π1000t + 15°) V. Learn how to calculate the amplitude Vm, peak-to-peak voltage Vpp, and find the voltages across resistors R1 and R2 as well as the current i(t) in the circuit using Ohm's Law and the voltage divider rule.

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