Answer in Electric Circuits for Rashinda #159587
October 16th, 2022
(i) Let’s first find the time constant:
“tau=RC=1.3cdot10^3 Omegacdot2.0cdot10^{-9} F=2.6cdot10^{-6} s=2.6 mu s.”
The discharge current in the circuit as a function of time can be found as follows:
“I=dfrac{Q_0}{RC}e^{-dfrac{t}{tau}},”“I=dfrac{5.1cdot10^{-6} C}{2.6cdot10^{-6} s}cdot e^{-dfrac{9.0 mu s}{2.6 mu s}}=0.06 A.”
(ii) The charge on the capacitor discharging through a resistance as a function of time can be found as follows:
“Q=Q_0e^{-dfrac{t}{tau}},”“Q=5.1cdot10^{-6} Ccdot e^{-dfrac{8.0 mu s}{2.6 mu s}}=0.235 mu C.”
The current in the circuit is the maximum at “t=0”:
“I=dfrac{Q_0}{RC}e^{-dfrac{0}{tau}}=dfrac{Q_0}{RC},”“I=dfrac{5.1cdot10^{-6} C}{2.6cdot10^{-6} s}=1.96 A.”
