CourseNotes.31 - RC time constant is = RC 8.4.2 Discharging...

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Figure 25: A circuit for charging and discharging a capacitor through a resistor. 8.4.1 Charging a Capacitor If we set the switch to position a in figure 25 and begin walking around the circuit, then the potential differences of the battery, resistor, and capacitor must sum to zero. V - iR + q C = 0 Everything in this equation is a constant except for the current and the charge. Recall that the current is related to the charge however via i = dq dt . Plugging this in gives R dq dt + q C = V This is a inhomogeneous, linear differential equation whose solution (though not difficult) is beyond the scope of this course. Suffice it to say that the solution to such an equation is: q ( t ) = CV ± 1 - e - t RC ² i = dq dt = ³ V R ´ e - t RC Notice that this is a time dependent process. This is the first way we have seen in which a circuit can have a sense of time. This situation pops up often enough that we assign the thing on the bottom of the fraction in the exponential its own name. The
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Unformatted text preview: RC time constant is = RC 8.4.2 Discharging a Capacitor If we move the switch in gure 25 to position b and repeat our analysis, we nd R dq dt + q C = 0 The solution to this homogeneous linear dierential equation is q = q e-t RC i =- q RC e-t RC So, we see that discharging a capacitor also has an exponential time dependence with the same characteristic time constant. 8.5 Problems Problem 27.33 In gure 26 the ideal batteries have emfs E 1 = 5 V and E 2 = 12 V , the resistances are each 2, and the potential is dened to be zero at the grounded point of the circuit. What are potentials (a) V 1 and (b) V 2 at the indicated points? The rst step in this problem is to simplify the two resistors in the upper right corner of the circuit. Since they are in parallel their equivalent resistance is P 1 = 1 1 R + 1 R = R 2 31...
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