In order to determine the current I in the circuit, we need to apply Kirchhoff's Current Law (KCL), which states that the sum of currents entering a node must be equal to the sum of currents leaving the node.
Looking at the circuit diagram, we see that there are two nodes: the one at the top, where the current source and resistor meet, and the one at the bottom, where the two resistors meet.
Let's start with the top node. We know that the current entering the node is equal to the current leaving the node, since there are no other connections to this node. Therefore, we have:
I = I1
Next, let's move to the bottom node. Here, we know that the current leaving the node is equal to the current entering the node. Therefore, we have:
I1 = I2 + I3
Now, we can use Ohm's Law to relate the currents to the resistances and voltages:
I1 = (V1 - V2) / R1
I2 = V2 / R2
I3 = V2 / R3
Substituting these expressions into the equation for the bottom node, we get:
(V1 - V2) / R1 = V2 / R2 + V2 / R3
Multiplying both sides by R1R2R3 and simplifying, we get:
R2R3(V1 - V2) = R1R3V2 + R1R2V2
Expanding and simplifying further, we get:
(V1 - V2) / 8 = (2V2) / 9
Solving for V2, we get:
V2 = V1 / 3
Substituting this value back into our expressions for I1, I2, and I3, we get:
I1 = 2V1 / 27
I2 = V1 / 27
I3 = V1 / 9
Finally, substituting these values back into our equation for the top node, we get:
I = I1 = 2V1 / 27 = 2/27 * 24 = 16/9
Therefore, the current I is 16/9 or 1.78, which corresponds to option (B) 9/11.