Question

Suppose the same type of series circuit has R=7.00×10⁴Ω,C=4.00μF and an initial voltage across the capacitor of 4.5 V. (a) How long does it take the capacitor to lose half its initial charge? t= (b) How long does it take to lose all but the last 10 electrons on the negative plate? t= 16 How does the charge on a capacitor depend on time for a discharging circuit?

Answer 1

The charge on a capacitor in a discharging circuit depends on time according to an exponential function.

For a discharging circuit, the charge on a capacitor depends on time according to an exponential function. Initially, the charge on the capacitor is Q₀ (initial charge), and as time passes, the charge decreases exponentially. The equation that relates the charge on the capacitor to time for a discharging circuit is:

Q(t) = Q₀ * e(-t/RC)

where Q(t) is the charge on the capacitor at time t, Q₀ is the initial charge, R is the resistance in the circuit, C is the capacitance of the capacitor, and t is the time.