C

To analyze the circuit, we can use Kirchhoff's Voltage Law (KVL) and Ohm's Law.

1. **Circuit Configuration**: There are three branches with resistances of 30Ω and 40Ω each. The voltage sources are 40V and 80V.
2. **Applying Kirchhoff's Law**: For the loop containing I1 and affected by the 40V source:
V = IR
KVL gives: $$ -80V + I_3(40 ext{Ω}) + I_2(40 ext{Ω}) = 0 $$
and for the loop containing I1, I2 with 40V,
$$ 40V - I_1(30 ext{Ω}) - I_2(40 ext{Ω}) = 0 $$
3. **Solving the equations**: Solving these equations gives: - I_1 = -0.8 A, - I_2 = 0.4 A
I3 (using the current at junction rule) results in 0.8 A.
Hence the correct option showing the current in the specified path is 0.4 A.
Therefore, the answer is C.