CHAPTER 2. ALTERNATING CURRENT CIRCUITS25case where the capacitor is initially charged and then the circuit is closed and the chargeis allowed to drain oﬀ the capacitor (eg. closing a switch). The resulting current will ﬂowthrough the resistor.Solving for the current we obtainI(t)= I0e−t/RC ,(2.10)where I(t =0)= I0 is the initial current given by Ohm’s lawI0 =V0R.(2.11)Using a time dependent version of Ohm’s law we can solve for the voltage across theresistorV (t)= RI(t)= RI0e−t/RC = V0e−t/RC = V0e−t/τ ,(2.12)where V (t =0)≡ V0 is the initial voltage across the capacitor and τ≡ RC is the commonlydeﬁned time constant of the decay. You should also be able to solve for the voltage acrossthe capacitor and charge on the capacitor.For the case of an applying voltage VB being suddenly placed into the circuit (insertinga battery) the capacitor is initially not charged and the voltage across the capacitor isV (t)= VB(1− e−t/τ ).(2.13)In the ﬁrst case, current and voltage exponentially decay away with time constant τ whenthe switch is closed. The charge ﬂows oﬀ the capacitor and through the resistor. The energyinitially stored in the capacitor is dissipated in the resistor.In the second case the capacitor charges to a voltage VB until no current ﬂows and hencethe voltage drop across the resistor is zero. Energy from the battery is stored in the capacitor.In both cases the characteristic RC time constant occurs. In general this is true of allresistor-capacitor combinations and will be important throughout the course.2.2.2RL CircuitThe response of the RL circuit, shown in ﬁgure 2.2, is similar to that of the RC circuit.There are however some signiﬁcant diﬀerences.RLFigure 2.2: RL circuit.