Example 21.9 Charging a Capacitor in an RC Circuit An uncharged capacitor and a resistor are connected in series to a battery as shown in the figure, where E = 12.0 V, C = 4.40 MF, and R = 7.1 x 103 0. The switch is thrown to position a. Find the time constant of the circuit, the maximum charge on the capacitor, the maximum current in the circuit, and the charge and current as functions of time. SOLVE IT + Conceptualize Study the figure and imagine throwing the a switch to position a as shown in figure (b). Upon doing so, the capacitor begins to charge. Categorize We evaluate our results using equations developed in this section, so we categorize this example as a substitution problem. 2 R + b C (a) A capacitor in series with a resistor, switch, and battery. (b) When the switch is thrown to position a, the capacitor begins to charge up. (c) When the switch is thrown to position b, the capacitor discharges. Evaluate the time constant of the circuit: T = RC = (7.1 x 103 n) (4.40 HF) = 3.12 S Evaluate the maximum charge on the 0 = CE = (4.40 AF) (12.0 V) = 52.8 capacitor: Evaluate the maximum current in the 12.0 V circuit: I =- 7.1 x 10* n I = 16.9 V HA Find the charge and current as functions of time (in seconds): q(1) = 52.8 1 - e (Use the following as necessary: z.) I(:) = 16.9 e 3.12 (Use the following as necessary: z.) MASTER IT HINTS: GETTING STARTED | I'M STUCK! Now, a second resistor R, of 3.0 x 103 0 is connected in parallel to the existing resistor in the circuit, and a second capacitor Cy = 4.20 #F is connected to the existing capacitor in parallel. (a) What will be the new time constant? T = (b) What will be the maximum current in the circuit (leaving the battery terminal)? Imax = Now, we connect the second resistor R, of 3.0 x 10 n and the second capacitor C, = 4.20 AF both in series to the other elements of the circuit. (c) What will be the new time constant? T = (d) What will be the maximum current in the circuit (leaving the battery terminal)? max =