The inverse of given Laplace transform is

Apply KVL to mesh 2

-2 + V_y + 1 = 0

? V_y = 1

Apply KVL to mesh 1

? -8 + V_x + 2 = 0

? V_x = 6 V

Apply KVL to mesh 3

-1 + V_z -2 = 0

? V_z = 3 V.

i(t) = V_r + (i_i - i_j)e^-1/p = 0.5 - 0.125e^-1000t .

There is negative number present in first column. Thus network is unstable.

.

If this converges, it implies that s = - 3 is in the ROC.

A casual and stable system must always have its ROC to the right of all its poles. However, s = - 3 is not to the right of the pole at s = - 2.

Statement 2 is false, because it is equivalent to stating that H(0) = 0. This contradicts the fact that H(s) does not have a zero at the origin.

Statement 3 is false. If h(t) is of finite duration, then if its Laplace transform has any points in its ROC, ROC must be the entire s-plane.

However, this is not consistent with H(s) having a pole at s = - 2.

Statement 4 is false. If it were true, then H(s) has a pole at s = - 2, it must also have a pole at s = 2.

This is inconsistent with the fact that all the poles of a causal and stable system must be in the left half of the s-plane.

r_c = 2.475 = 1.258 kV

Therefore holes are majority carriers.