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To solve this problem using the superposition principle, we will consider the effects of each source (the voltage source \( V_s \) and the current source \( I_s \)) separately. Then, we will sum the contributions of both sources to find the total current \( I \) through \( R_2 \).

### Given Data:
- \( V_s = 2 \, \text{V} \)
- \( I_s = 1 \, \text{A} \)
- \( R_1 = 2 \, \Omega \)
- \( R_2 = 3 \, \Omega \)

### Step 1: Contribution from \( V_s \) (Voltage Source Only)
1. **Turn off** the current source \( I_s \), meaning it becomes an open circuit.
2. The circuit becomes a simple series circuit with \( V_s \), \( R_1 \), and \( R_2 \) in series.
   - The total resistance in the series is:
     \[
     R_{\text{total}} = R_1 + R_2 = 2 \, \Omega + 3 \, \Omega = 5 \, \Omega
     \]
   - The total current \( I_{\text{V}} \) flowing in the series circuit (same current through both resistors) is:
     \[
     I_{\text{V}} = \frac{V_s}{R_{\text{total}}} = \frac{2 \, \text{V}}{5 \, \Omega} = 0.4 \, \text{A}
     \]
   - Since the current is the same in a series circuit, the current through \( R_2 \) due to \( V_s \) is \( I_{\text{V}} = 0.4 \, \text{A} \).

### Step 2: Contribution from \( I_s \) (Current Source Only)
1. **Turn off** the voltage source \( V_s \), meaning it becomes a short circuit.
2. The current source \( I_s = 1 \, \text{A} \) will split between \( R_1 \) and \( R_2 \) because they are now in parallel.
   - The equivalent resistance for \( R_1 \) and \( R_2 \) in parallel is:
     \[
     R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{2 \times 3}{2 + 3} = \frac{6}{5} = 1.2 \, \Omega
     \]
   - The current \( I_{\text{R2}} \) through \( R_2 \) is found using the current division rule:
     \[
     I_{\text{R2}} = I_s \times \frac{R_1}{R_1 + R_2} = 1 \, \text{A} \times \frac{2 \, \Omega}{2 \, \Omega + 3 \, \Omega} = 1 \times \frac{2}{5} = 0.4 \, \text{A}
     \]

### Step 3: Total Current Through \( R_2 \)
Now, applying the superposition principle, we sum the contributions of both sources to find the total current through \( R_2 \).

- The current through \( R_2 \) due to \( V_s \): \( I_{\text{V}} = 0.4 \, \text{A} \) (downward through \( R_2 \)).
- The current through \( R_2 \) due to \( I_s \): \( I_{\text{R2}} = 0.4 \, \text{A} \) (downward through \( R_2 \)).

Thus, the total current \( I \) through \( R_2 \) is:
\[
I_{\text{total}} = I_{\text{V}} + I_{\text{R2}} = 0.4 \, \text{A} + 0.4 \, \text{A} = 0.8 \, \text{A}
\]

### Final Answer:
The total current \( I \) downward through \( R_2 \) is **0.8 A**.

Would you like more details or have any questions? Here are 5 follow-up questions:

1. How does the current division rule apply when resistors are in parallel?
2. What happens to a current source when using superposition with a voltage source?
3. Why do we treat the voltage source as a short circuit when turned off?
4. How would the result change if \( R_1 \) and \( R_2 \) had different values?
5. Can superposition be applied to circuits with non-linear components?

**Tip**: Superposition is only valid for linear systems, so it doesn't apply to circuits with non-linear components such as diodes or transistors.

Using superposition, calculate the current I labelled in the circuit below (downward through R2), giving your answer in A, to one decimal place. The labelled quantities are as follows: Vs = 2 V, Is = 1 A, R1 = 2 Ω, R2 = 3 Ω.

Learn how to calculate the current I through R2 using the superposition theorem. The circuit involves a voltage source (Vs = 2V), a current source (Is = 1A), and resistors R1 = 2Ω, R2 = 3Ω. Detailed solution steps are provided, covering Ohm's law, current division, and the superposition principle.

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