Question 24.16: In Fig. 24.24, let R = 2 kΩ, C = 5µF, and Q = 50 µC. (a) Aft......
In Fig. 24.24, let R = 2 kΩ, C = 5µF, and Q = 50 µC. (a) After how many time constants, τ = R C, will the charge on the capacitor be half of its initial value when the switch is closed? (b) When the stored energy in the capacitor becomes half of its initial value?
(a) The time constant of the circuit is:
\tau=R\,C=(2\times10^{3}\,\Omega)(5\times10^{-6}\,{F})=10^{-2}\,s=10\,\mathrm{ms}After closing the switch at t = 0, the charge on the capacitor is given by Eq. 24.49,
q=Q e^{-t/R C}=Q e^{-t/\tau} (24.49)
q=Q e^{-t/\tau}. To find the time interval during which q drops to one-half its initial value, we substitute q = Q/2 into this equation and solve for the time t as follows:
\frac{Q}{2}=Q e^{-t/\tau}\;\;\;\Rightarrow\;\;\frac{1}{2}=e^{-t/\tau}Taking the logarithm of both sides, we find:
-\ln2=-{\frac{t}{\tau}} ⇒ t = (ln 2)τ = 0.69τ = 0.69 × (10 ms) = 6.9 ms
(b) From Eq. 23.24,
C_{\mathrm{eq}}={\frac{Q}{\Delta V}}=C_{1}+C_{2} (Parallel combination) (23.24)
the initial stored energy in the capacitor is U_{\circ}=Q^{2}/2C.
Using Eq. 24.49, the energy stored at time t is:
U=\frac{q^{2}}{2C}=\frac{Q^{2}}{2C}e^{-2t/\tau}=U_{o}e^{-2t/\tau}As in part (a), we set U = U_{\circ}/2 and solve for t as follows:
\frac{U_{o}}{2}=U_{o}e^{-2t/\tau}\ \ \Rightarrow\ \ \frac{1}{2}=e^{-2t/\tau}Again, taking the logarithm of both sides and solving for t, we find:
ln 2 = −2t/τ ⇒ t={\textstyle{\frac{1}{2}}}(\ln2)\tau={\textstyle{\frac{1}{2}}}\times0.69\tau={\textstyle{\frac{1}{2}}}\times0.69\times (10 ms) = 3.45 ms
Note that the results of both parts are independent on the value of Q.