Question
The variation with potential difference V of the charge Q on one of the plates of a capacitor is shown in Fig. 5.1.
The capacitor is connected to an 8.0V power supply and two resistors R and S as shown in Fig. 5.2.
The resistance of R is \(25\) kΩ and the resistance of S is \(220\) kΩ.
The switch can be in either position X or position Y.
(a) The switch is in position X so that the capacitor is fully charged.
Calculate the energy E stored in the capacitor.
(b) The switch is now moved to position Y.
(i) Show that the time constant of the discharge circuit is \(3.3\)s.
(ii) The fully charged capacitor in (a) stores energy E.
Determine the time t taken for the stored energy to decrease from E to E/9
(c) A second identical capacitor is connected in parallel with the first capacitor.
State and explain the change, if any, to the time constant of the discharge circuit.
Answer/Explanation
Ans:
(a) (energy stored =) area under line or \(1/2\) QV
= \(1/2 \times 8.0 \times 1.2 \times 10^{-4}\)
= \(4.8 \times 10^{–4}\) J
(b) (i) (τ=) RC
(τ=) \(220 \times 10^{3} \times (1.2 \times 10^{-4}/8.0)\) = \(3.3\) s
(ii) \(E\) ∝ \(V^{2}\)
(so time to) \(V_{0}/3\)
\(V\) = \(V_{0}e^{-t/RC}\)
\(\frac{V_{0}}{3}\) = \(V_{0}e^{-t/3.3}\)
\(\frac{1}{3}\) = \(e^{-t/3.3}\)
\(t\) = \(3.6\) s
(c) (total) capacitance is doubled