 # Electricity and Magnetism Questions

## Transcribed Text

8.1 Charging a capacitor Figure 1: The positive and negative terminals of a capacitor. the Ir: lectures, we found that the voltage across a capacitor is given by Vacross cap = 8 where in charge in coulombs on one plate of the capacitor and C is the capacitance of the capacitor is farads. From this relationship, we cau see that as we put more charge onto the plates of capacitor, the voltage across the capacitor will increase (and vice versa if we remove charge from a the plates of a capacitor). Consider the following circuit: -q / Switch Vo les Figure 2: Charging a capacitor. Catneli, J. D. and K. W Johnson (2012). Physics. Hobaken, NJ, Wiley. capacitor has no charge on it initial'y Because thers is no excess charge will on flow the Say that the the voltage V across the capacitor is vero (*). When the switch is closed charge plates, 271 and Increase with on time(* the capacitor Because plates. This will mean that the voltage across the capacitor will collect in the citeuit is limited this of the resistor in the circuit, the amount of charge that can flow it repels capacitor increases slowly(*). As builds up on the plates, the voltage across the means that the charge on the capacitor builds up slowly and heuce across be deposited incoming capacitor on the charge capacitor will with plates bigger and time bigger force charge This means that less and capacitor less charge will the as goes by. Therefore, the rate of increase of voltage also why our capacitor computers that explains the slow propagation of nerve signals in the human body and the plates of a decrease as time goes by. It is this slow increase of voltage across over this in lecture are limited in how fast they can compute if we have time we will go It charged can be the shown voltage (see across your text the capacitor for the derivation) varies as: that when an initially uncharged capacitor is (1) obms) the resistor, C is the capacitance (in farads) of the capacitor, and t is the time the whete V of is the voltage across the capacitor, Vo is the battery voltage, R is the resistance (in v.S. capacitor has been charged for in the circuit. If we plot a graph of voltage across the capacitor time the capacitor has been charged for, we get the plot in Figure 3. Time (t) Figure 3: Charging a capacitor. Exercise 2 Suppose Vo = 9 volts, R = 270 ks. C = 820 uF in Figure 2, calculate the voltage accoss the capacitor, 37 seconds after the switch is closed Notice that if we wait a time t = RC, the voltage across the capacitor is: V = Vo(1 RE) Vo(1 = Vo(1 0.368) - 0.63V6 i.e. the voltage across the capacitor has increased to 63% of Vo after t RC seconds This time is called the time constant of the circuit The time constant of a circuit is useful as it gives you a quick idea of how fast voltages can change in a circuit. Exercise 3: Suppose Vo = 9 volts, R = C = 820 F in Figure 2, calculate the time constant of the circuit. Exercise 5: Suppose R = 270k9, C = 820 uF in Figure 4. calculate the voltage across the capacitor, 37 seconds after the resistor is connected across the charged capacitor. The capacitor is initially charged to 9 volts. Notice that if we wait a time = RC after the resistor has been placed across the capacitor, then the voltage across the capacitor is: V6(0.368) R i.e. the voltage across the capacitor has decreased by 63% of Vo after t = RC seconds. This time is called the time constant T of the circuit. The time constant of a circuit is useful as it gives you a quick idea of how fast voltages can change in a circuit. Exercise 6: Suppose R = 270 kn C = 820 uF in Figure 4, calculate the time constant of the circuit. The capacitor is initially charged to 9 volts.

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