A

Step 1: Analyze the circuit using Kirchhoff's Laws.
In the circuit, we have a node at point C with currents entering and leaving the node. Applying Kirchhoff's Current Law (KCL) at node C:

• The current entering the node from the left is $I_s$ (which is 1 A).
• The current leaving the node to the right is $I_6$ (4 A).
• The current through the 10 Ω resistor is denoted as $I_1$.

Applying KCL, we have:
$$I_s = I_6 + I_1$$
Substituting the values:
$$1 = 4 + I_1$$
From this equation, we find:
$$I_1 = 1 - 4 = -3 ext{ A}$$
Hence, the current through the 10 Ω resistor is 3 A (in the opposite direction to that assumed).

Step 2: Calculate the voltage V1 across the 10 Ω resistor.
Using Ohm's law:
$$V_1 = I_1 imes R = 3 imes 10 = 30 ext{ V}$$

Step 3: Determine the value of source voltage Vs.
We can apply KVL (Kirchhoff's Voltage Law) around the loop including $Vs$, the 5 Ω resistor, and the 30 V source:
$$Vs - I_s imes 5 - 30 = 0$$
Substituting the known values:
$$Vs - 1 imes 5 - 30 = 0$$
Then:
$$Vs = 5 + 30 = 35 ext{ V}$$

Final Results:
The current in the 10 Ω resistor is 3 A (option A).
Therefore, the correct answer is A.