Question

in order to prevent an electric current surge in a circuit, the resistance R is stepped down by 4.0 ohm after each 0.1 s. if the voltage V is constant at 120 V, do the resulting currents I from an arthimetic sequence is V=IR

Answer 1

R = resistance at t = 0sec

$R' = R - 0.1 * \frac{n}{10} = R - 0.01 n$  
R' is    resistance at n = 1 sec
$I = \frac{V}{R'} = \frac{120}{R - 0.01 n}$

When R and R' are large, we can write:

$I = 120 \frac{[ 1 + \frac{0.01 n}{R} ]}{R} \\ \\ \frac{120}{R} + \frac{1.2}{R^{2}} * n$    is the current at nth second.

it is arith progr.  with first term and difference as shown in the above equation.