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We are tasked with calculating the current labeled \( I \) using the superposition theorem for the given circuit. The labeled quantities are:

- \( V_s = 3 \, \text{V} \)
- \( I_s = 2 \, \text{A} \)
- \( R_1 = 2 \, \Omega \)
- \( R_2 = 3 \, \Omega \)

### Applying Superposition Theorem

The superposition theorem states that in a linear circuit with multiple independent sources, the total current or voltage for any element can be found by summing the individual contributions from each independent source, with all other sources replaced by their internal resistances (i.e., voltage sources are short-circuited and current sources are open-circuited).

#### Step 1: Contribution of the Voltage Source (\( V_s \))
- Set the current source \( I_s \) to 0 A (open circuit).
- The circuit becomes a simple series circuit with \( V_s \), \( R_1 \), and \( R_2 \).

The total resistance in the series is:
\[
R_{\text{total}} = R_1 + R_2 = 2 \, \Omega + 3 \, \Omega = 5 \, \Omega
\]

Using Ohm's Law, the current through the series is:
\[
I_{\text{Vs}} = \frac{V_s}{R_{\text{total}}} = \frac{3 \, \text{V}}{5 \, \Omega} = 0.6 \, \text{A}
\]

Since \( R_1 \) and \( R_2 \) are in series, this is the current through both resistors. The current through \( R_2 \) due to \( V_s \) is:
\[
I_{R_2}(\text{from} \, V_s) = 0.6 \, \text{A} \, (\text{downward through} \, R_2)
\]

#### Step 2: Contribution of the Current Source (\( I_s \))
- Set the voltage source \( V_s \) to 0 V (short circuit).
- The circuit now consists of \( R_1 \) and \( R_2 \) in parallel with \( I_s = 2 \, \text{A} \).

The equivalent resistance of \( R_1 \) and \( R_2 \) in parallel is:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2 \, \Omega} + \frac{1}{3 \, \Omega} = \frac{5}{6} \, \Omega
\]
Thus,
\[
R_{\text{eq}} = \frac{6}{5} \, \Omega = 1.2 \, \Omega
\]

The current divides between \( R_1 \) and \( R_2 \). Using the current division rule, the current through \( R_2 \) is:
\[
I_{R_2}(\text{from} \, I_s) = I_s \times \frac{R_1}{R_1 + R_2} = 2 \, \text{A} \times \frac{2 \, \Omega}{2 \, \Omega + 3 \, \Omega} = 2 \, \text{A} \times \frac{2}{5} = 0.8 \, \text{A}
\]

This current is also downward through \( R_2 \).

#### Step 3: Total Current through \( R_2 \)
Now, we sum the contributions from both sources:
\[
I_{R_2} = I_{R_2}(\text{from} \, V_s) + I_{R_2}(\text{from} \, I_s) = 0.6 \, \text{A} + 0.8 \, \text{A} = 1.4 \, \text{A}
\]

### Final Answer:
The current through \( R_2 \) (downward) is \( I = 1.4 \, \text{A} \).

---

Would you like more details on any of the steps? 

Here are 5 related questions you might find interesting:
1. How would the result change if \( R_1 \) or \( R_2 \) were different values?
2. How do you calculate current division in a parallel circuit?
3. What would happen if \( V_s \) or \( I_s \) were reversed in polarity?
4. How does superposition apply to circuits with dependent sources?
5. Can superposition be used for circuits with nonlinear components?

**Tip:** Always remember to turn off all but one independent source at a time when using superposition.

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

Learn how to calculate the current through resistor R2 using the superposition theorem for a circuit with Vs = 3V, Is = 2A, R1 = 2Ω, and R2 = 3Ω. This solution covers Ohm's Law, current division, and applying the superposition theorem.

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