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.