Question

RC Circuit, with EMF. • The capacitance in the circuit shown is C = 0.30 μF, the total resistance is R = 20 kΩ, the battery emf is E = 12 V.   Calculate: (a) the time constant, (b) the maximum charge the capacitor could acquire, (c) the time it takes for the charge to reach 99% of this value, (d) the current I when the charge Q is half its maximum value, (e) the maximum current, (f) the charge Q when the current I is 0.20 of its maximum value. 

Answer 1

(a) $\tau=RC=20000\cdot 0.3\cdot10^{-6}=0.006(s)$ . Answer

(b) $q_0=C\cdot E=0.3\cdot10^{-6}\cdot12=3.6(\mu F)$ . Answer

(c) $0.99q_0=q_0(1-e^{-t/\tau})\to t=-\tau\ln0.01=0.028(s)$ . Answer

(d) $0.5q_0=q_0(1-e^{-t/\tau})\to t=-\tau\ln0.5$

$I=I_0e^{-t/\tau}=\frac{E}{R}e^{-t/\tau}=\frac{E}{R}e^{-(-\tau\ln0.5)/\tau}=$

$=\frac{E}{R}e^{\ln0.5}=\frac{12}{20000}e^{\ln0.5}=3\cdot10^{-4}(A)$ . Answer

(e) $I_0=\frac{E}{R}=\frac{12}{20000}=6\cdot10^{-4}(A)$ . Answer

(f) $0.2I_0=I_0e^{-t/\tau}\to t=-\tau\ln0.2$

$q=q_0(1-e^{-t/\tau})=q_0(1-e^{-(-\tau\ln0.2)/\tau})=$

$=q_0(1-e^{\ln0.2})=3.6\cdot 10^{-6}\cdot(1-e^{\ln0.2})=2.87(\mu F)$ . Answer