Question

1) If the capacitance in the circuit is doubled, how does this affect the half-life? Physically why does this make sense?

2) If the resistance in the circuit is doubled, how does this affect the half-life? Physically why does this make sense?

3) If the voltage that charges the capacitor is doubled, how is the half-life affected?

Answer 1

We know that Vc(t) = Vc(0)*e^-(t/RC) = Vc(0)/2 at the moment of half life
=> 0.5 = e^-(t/RC) Where Vc(0) is the capacitor's initial voltage.
ln(0.5) = -t/RC = -0.693 => t = 0.693*R*C where t is the moment when the capacitor voltage is half it's initial value.
Thus the half life is only determined by R and C. If R or C are larger, the half life is longer. It takes more time for the capacitor to discharge. That makes sense. A larger cap means it stores greater charge, a larger R means the discharge current encounters.
greater resistance. Let the half life time = Th
1 Doubling the cap size doubles the half life since Th = 0.693*R*C
2 Doubling the R size doubles the half life since Th = 0.693*R*C
3 No effect because the voltage does not appear in the equation for Th
4 No effect because the voltage does not appear in the equation for Th