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R-C circuits So far we considered stationary situations only meaning: dI dV 0, 0, ... dt dt Next step in generalization is to allow for time dependencies For consistency we follow the textbook notation where time dependent quantities are labeled by lowercase symbols such as i=I(t), v=V(t), etc Let’s charge a capacitor again -q +q Vab=Va-Vb Vb Va What happens during the charging process The capacitor starts to accumulate charge A voltage vab starts to build up which is given by vbc q C The transportation of charge to the capacitor means a current i is flowing The current gives rise to a voltage drop across the resistor R which reads: vab iR Using Kirchhoff’s loop rule we conclude E iR q 0 C With the definition of current as i dq q E dt RC R dq dt Inhomogeneous first order linear differential equation Since in the very first moment when the switch is closed there is not charge in the capacitor we have the initial condition q(t=0)=0 Lets consider common strategies to solve such a differential equation We use intuition to get an idea of the solution. We know that initially q=0 and after a long time when the charging process is finished i=0 and hence q(t ) CE q CE t t Let’s guess q(t ) CE 1 e and check if it works and, if so, let’s determine the time constant t q(t ) CE 1 e dq CE t e dt Substitution into dq q E dt RC R CE e t t CE 1 e RC CE RC RC Our guess works if we chose RC t q(t ) CE 1 e RC Guessing is a common strategy to solve differential equations In the case of a linear first order differential equation there is a systematic integration approach which always works dq q E dt RC R dq CE - q dt RC dq RC dt CE - q 0 q ( t 0) q with t dq t 0 CE - q RC q with q(t=0)=0 CE - q z, dz dq q CE CE dz t z RC CE - q t ln CE RC t CE - q e RC CE t RC q(t ) CE 1 e q Qf / e RC Qf CE t After time RC q(t ) CE 1 e 1 0.63 CE 63% of Qf reached How does the time dependence of the current look like t q(t ) CE 1 e RC i I0 dq E RCt i (t ) e dt R E R I0/e RC t Now let’s discharge a capacitor Using Kirchhoff’s loop rule in the absence of an emf we conclude iR q 0 C dq q 0 dt RC homogeneous first order linear differential equation In the very first moment when the switch is closed the current is at maximum and determined by the charge Q(t=0)=Q0 initially in the capacitor Q0 RCt q dq 1 Q q RC 0 dt 0 I (t 0) I 0 ln q t Q0 RC t RC q Q0e t Q0 RCt RC i e I 0e RC Current is opposite to the direction on charging We close the chapter with an energy consideration for a charging capacitor When multiplying E iR q 0 C i by the time dependent current i iq Ei i R 0 C 2 Rate at which energy is stored in the capacitor Power delivered by the emf Rate at which energy is dissipate by resistor Total energy supplied by the battery (emf) U emf = E idt E idt E 0 0 Qf dq E Q 0 Total energy stored in capacitor is q U C = idt C 0 Qf 0 1 Qf 2 f Half of the energy delivered by the battery is stored in the capacitor no matter what the value of R or C is ! q 1 1 dq = E Q f U emf C 2 C 2 2 Let’s check this surprising result by calculating the energy dissipated in R which must be the remaining half of the energy delivered by the emf U diss = i Rdt 2 with i (t ) dt with 0 U diss = E2 R e 2t RC R e t RC 2t RC x , dt dx RC 2 0 RC E 2 U diss = 2 R = E CE Q f 2C 2 C E x e 0 dx = 2 0 2 C E e dx = 2 x 1 1 E Q f U emf U C 2 2