Problem in Circuit

Since the time constant of the circuit is on the order of microseconds, the capacitors behave like open circuits. The rest is easy

Writing down the mesh equations for the four meshes and solving, we see that the currents in the said branches do not depend on time .

(The 2nd. mesh current being $i_2$ and the 4th. mesh current being $i_4$ , the two mesh equations are

$i_2-i_4=\dfrac{3}{2}=1.5, 2i_4-i_2=3$ . Solving we get $i_2=6, i_4=4.5\implies I_{BC}=i_4=4.5, I_{GD}=i_2-i_4=6-4.5=1.5$ ).

We simply get the values of the currents as

$I_{BC}=4.5, I_{GD}=1.5$ amps, and the ratio as

$\dfrac{4.5}{1.5}=\boxed 3$ .