This circuit consists of a current source (8 A) in series with two resistors: $1 \, \Omega$ and $3 \, \Omega$. To calculate the current through the $3 \, \Omega$ resistor, we first need to understand that this is a series circuit. In a series circuit, the current is the same through all components.

Steps:

1. Since the current source provides $8 \, \text{A}$, and all elements are in series, the same $8 \, \text{A}$ flows through the $1 \, \Omega$ and the $3 \, \Omega$ resistors.
2. Therefore, the current through the $3 \, \Omega$ resistor is also $8 \, \text{A}$.

No further calculation is required, as the series arrangement ensures that all components have the same current.

Would you like more details or have any further questions?

Here are 5 related questions for practice:

1. What is the voltage drop across the $3 \, \Omega$ resistor?
2. What is the total resistance in the circuit?
3. What is the voltage drop across the $1 \, \Omega$ resistor?
4. How would the current change if the $3 \, \Omega$ resistor was replaced with a $6 \, \Omega$ resistor?
5. What happens to the total current if a parallel branch is added?

Tip: In series circuits, the current remains the same across all components, but the voltage divides across each resistor based on their resistances.