Answer
a) $t=0.686s$
b) Increase
c) See solution.
d) See solution.
Work Step by Step
a) $V_C=E(1-e^{-t/RC})$
$90.0V=(105V)(1-e^{-t/(2.35\times10^6\Omega)(0.150\times10^{-6}F)})$
$t=0.686s$
b) If R is increased, the time will increase
c) The flashing of the light is very brief because the light will turn off the moment the capacitor voltage becomes smaller than the voltage required to flash the lamp. The capacitor discharges very quickly because the neon lamp has very low resistance.
d) After the lamp flashes for the first time, the capacitor will start charging again.