Question of the day How does the general solution to a first-order differential equation describe the behavior of charge in an RC circuit?

First order differential equation These equations have the form 𝑑𝑦 𝑑𝑥 = F(x,y). In a circuit, the current I = 𝑑𝑄 𝑑𝑡 . In an RC circuit (not connected to an EMF), Vresistor = IR and Vcapacitor = Q/C. Using KVL, then, Vresistor + Vcapacitor = IR + Q/C = 0, which becomes: 𝑄 𝐶 + R 𝑑𝑄 𝑑𝑡 = 0 or 𝑑𝑄 𝑑𝑡 = (–𝑅𝐶) Q The solution for this group of differential equations is the exponential function, since 𝑑 𝑑𝑡 et = et

R-C circuits: Charging a capacitor: Slide 1 of 4 Shown is a simple R-C circuit for charging a capacitor. We idealize the battery to have a constant emf and zero internal resistance, and we ignore the resistance of all the connecting conductors. We begin with the capacitor initially uncharged.

R-C circuits: Charging a capacitor: Slide 2 of 4 At some initial time t = 0 we close the switch, completing the circuit and permitting current around the loop to begin charging the capacitor. As t increases, the charge on the capacitor increases, while the current decreases.

R-C circuits: Charging a capacitor: Slide 3 of 4 The charge on the capacitor in a charging R-C circuit increases exponentially, with a time constant τ = RC.

R-C circuits: Charging a capacitor: Slide 4 of 4 The current through the resistor in a charging R-C circuit decreases exponentially, with a time constant τ = RC.

R-C circuits: Discharging a capacitor: Slide 1 of 4 Shown is a simple R-C circuit for discharging a capacitor. Before the switch is closed, the capacitor charge is Q0, and the current is zero.

R-C circuits: Discharging a capacitor: Slide 2 of 4 At some initial time t = 0 we close the switch, allowing the capacitor to discharge through the resistor. As t increases, the magnitude of the current decreases, while the charge on the capacitor also decreases.

R-C circuits: Discharging a capacitor: Slide 3 of 4 The charge on the capacitor in a discharging R-C circuit decreases exponentially, with a time constant τ = RC.

R-C circuits: Discharging a capacitor: Slide 4 of 4

R-C circuits: Discharging a capacitor: Slide 5 of 4 The magnitude of the current through the resistor in a discharging R-C circuit decreases exponentially, with a time constant τ = RC.

Summary of Equations for RC circuits Note: the R in these equations is the total R the C sees. In general: Charging: ∆𝑽= ∆𝑽 𝒎𝒂𝒙 (𝟏− 𝒆 −𝒕/𝑹𝑪 ) 𝑸= 𝑸 𝒎𝒂𝒙 (𝟏− 𝒆 −𝒕/𝑹𝑪 ) Where Qmax= CDVmax and Imax = DVmax/R 𝑰= 𝑰 𝒎𝒂𝒙 𝒆 −𝒕/𝑹𝑪 Discharging: ∆𝑽= ∆𝑽 𝒐 𝒆 −𝒕/𝑹𝑪 DV0, Q0 and I0 are the values at t = 0 s (when the discharging starts) 𝑸= 𝑸 𝒐 𝒆 −𝒕/𝑹𝑪 𝑰= 𝑰 𝒐 𝒆 −𝒕/𝑹𝑪

Capacitors: Early and Late* Initially, when a switch closes there is a potential difference of 0 across an uncharged capacitor. After a long time, the capacitor reaches its maximum charge and there is no current flow through the capacitor. Therefore, at t=0 the capacitor behaves like a short circuit (R=0), and at t=∞ the capacitor behaves like at open circuit (R=∞). 100 V 100 V 12.5 A 100 V 2.5 A 10 A 0 V Circuit at t=0 at t=∞ Calculate initial currents. IB = 100 V/8 W = 12.5 A Calculate final potentials.

Example: Exponential Decay in a RC Circuit The switch has been in position a for a long time. It is changed to position b at t=0. What are the charge on the capacitor and the current through the resistor at t = 5.0 ms?

Neon Tube Oscillator Another old-fashioned circuit, but a good illustration of capacitor charging. The power supply is 80 to 100V. The neon tube is an insulator until the voltage across it reaches a critical value, below 80V, at which point the neon ionizes and becomes a good conductor, discharging the capacitor rapidly. The cycle repeats: a flashing light! In the first moments after the battery is connected, there is a big voltage drop IR1 across R1. As the capacitor charges and the current drops, the voltage across the neon tube builds up. After the capacitor discharges through the neon, the recharge again gives voltage drop across R1.