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Example 18-11 A Charging Series RC Circuit A 10.0-M resistor is connected in series with a 5.00-mF capacitor. When a switch is thrown, these circuit elements are connected to a 24.0-V battery of negligible internal resistance. The capacitor is initially uncharged. (a) What is the current in the circuit immediately after the switch is moved so that charging begins? (b) What is the charge on the capacitor once it is fully charged? (c) Find the capacitor charge, current, power provided by the battery, power taken in by the resistor, and power taken in by the capacitor at t = 50.0 s. (d) When the capacitor is fully charged, find the total energy that has been delivered by the battery and the total energy that has been delivered to the capacitor. Set Up We are given R = 10.0 M = 10.0 * 106 , C = 5.00 mF = 5.00 * 1026 F, and e = 24.0 V. Equations 18-27 tell us the capacitor charge and current at any time, including at t = 0 (when the switch is first closed) and t S (long after the switch is closed, so the capacitor is fully charged). We’ll use Equations 18-23 and 18-24 to find the power out of the battery and into the resistor and capacitor. (Equation 17-14 will help us in this.) In order to charge the capacitor to its maximum charge qmax, the total charge that must pass through the battery is qmax; we’ll use this and Equation 17-6 to find the total energy delivered by the battery. Equation 17-17 tells us the total energy that is stored in the charged capacitor. Capacitor charge and current in a charging series RC circuit: q1t2 = Ce11 - e -t>RC 2 e i1t2 = e -t>RC R = 24.0 V R = 10.0 MΩ switch closed at t = 0 C = 5.00 µF (18-27) Power for a circuit element: (18-23) P = iV Power for a resistor: V2 (18-24) R Charge, voltage, and capacitance for a capacitor: P = i 2R = (17-14) q = CV Electric potential difference related to electric potential energy difference: V = Uelectric q0 (17-6) Electric potential energy stored in a capacitor: Uelectric = Solve (a) Find the current at t = 0. q2 1 1 qV = CV 2 = 2 2 2C (17-17) From the second of Equations 18-27, the current when the switch is first closed at t = 0 is i102 = e -102>RC e e = e0 R R Since any number raised to the power 0 equals 1, we have e0 = 1 and e 24.0 V = R 10.0 * 106 = 2.40 * 1026 A = 2.40 mA i102 = i max = (b) Find the capacitor charge long after the switch is closed (t S ). The first of Equations 18-27 tells us the capacitor charge q(t). As t S , the exponent -t>RC S - . Any number raised to the power - is zero, so e -t>RC S 0 q1t2 = Ce11 - e -t>RC 2 S q max = Ce11 - 02 = Ce = (5.00 * 1026 F)(24.0 V) = 1.20 * 1024 C = 0.120 mC (c) The time constant for this circuit is RC = 50.0 s, so we are actually being asked about the behavior of the circuit one time constant after the switch is closed. Use this to find charge q, current i, and the power out of or into each circuit element. The time constant for this circuit is RC = (10.0 * 106 )(5.00 * 1026 F) = 50.0 s so at t = 50.0 s, t>RC = 150.0 s2 > 150.0 s2 = 1.00 From Equations 18-27, q = Ce(1 2 e21.00) = (5.00 * 1026 F)(24.0 V)(1 2 0.368) = 7.58 * 1025 C = 0.632qmax e 24.0 V i = e -1.00 = 10.3682 R 10.0 * 106 = 8.83 * 1027 A = 0.368imax The voltage across the battery is e = 24.0 V, so from Equation 18-23 the power out of the battery is Pbattery = ie = (8.83 * 1027 A)(24.0 V) = 2.12 * 1025 W = 21.2 mW From the first of Equations 18-24, the power into the resistor is PR = i2R = (8.83 * 1027 A)2(10.0 * 106 ) = 7.80 * 1026 W = 7.80 mW Equation 17-14, q = CV, tells us that the voltage across the capacitor is V = q>C. Combining this with Equation 18-23 gives the power into the capacitor: q 7.58 * 10-5 C b = 18.83 * 10-7A2 a b C 5.00 * 10-6 F = 1.34 * 1025 W = 13.4 mW PC = i a Note that the net power into the resistor and capacitor combined equals the power out of the battery: PR + PC = 7.80 mW + 13.4 mW = 21.2 mW = Pbattery (d) Use the maximum charge stored by the capacitor, which is the total charge moved across the battery, and Equation 7-6 to calculate the change in electric potential energy imparted by the battery. Use Equations 17-7 to calculate the electric potential energy stored in the capacitor. Long after the switch is closed, the total amount of charge that has passed through the battery and to the positive capacitor plate is qmax = 1.20 * 1024 C. From Equation 17-6, the potential energy change that was imparted by moving this charge across the 24.0-V emf of the battery is Ubattery = qmax e = (1.20 * 1024 C)(24.0 V) = 2.88 * 1023 J = 2.88 mJ The last of Equations 17-17 tells us the amount of energy that went into the capacitor to store charge qmax there: 11.20 * 10-4 C2 2 q 2max = 2C 215.00 * 10-6 F2 = 1.44 * 1023 J = 1.44 mJ UC = Reflect So exactly one-half of the energy taken from the battery goes into the capacitor: UC = 11>22Ubattery. Our results for charge q and current i in part (c) agree with Figure 18-18: After one time constant the capacitor charge has reached 3 1 - 11>e2] = 0.632 = 63.2% of its fully charged value qmax, and the current has decreased to 1>e = 0.368 = 36.8% of its initial value imax. Can you show that after 5 time constants (t = 5RC = 250 s) the charge will have reached 99.3% of qmax and the current will have decreased to just 0.674% of imax? The power calculations in part (c) show that all of the power extracted from the battery is accounted for: Part of the energy extracted from the battery goes into the resistor, and the rest goes into adding to the electric potential energy stored in the capacitor. We’ve shown this for a specific instant, but it’s true at all times during the charging process. These results from part (c) also help us understand our calculations in part (d): Only one-half of the energy extracted from the battery goes into the capacitor, so the other half must have gone into the resistor. This is a general result for charging any series RC circuit.