The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric constant $K_1$ and $K_2$ with thickness $\frac{3}{8} d$ and $\frac{d}{2}$, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If $K_1=1.25 K_2$, the value of $K_1$ is:

The steady state current in the circuit shown below is :