THEORY Capacitors store charge. As shown in Figure 1, equal and opposite charges
ID: 1433234 • Letter: T
Question
THEORY Capacitors store charge. As shown in Figure 1, equal and opposite charges, q, can be placed on the two plates of a capacitor. The charges produce a voltage difference, Vc given by where C is the capacitance. Capacitance has the units of Coulomb/Volt, which is called a Farad. The capacitance is the measure of how much charge can be stored Figure 1. The charge q distributed on two plates of a capacitor In the basic RC circuit shown Figure 2 there is initially no charge on the capacitor, and zero voltage across it. After the switch is closed, charges flow into the capacitor. The voltage Vc on the capacitor begins to increase, as shown in Figure 3. The shape of the curve arises from the fact that the voltage from the power supply is divided across the resistor and capacitor Taking Equation 2, plugging in Equation 1 for the capacitor voltage, using VR IR, and the fact that current is equal to charge flow Switch dt one obtains the differential equation dt C The solution to this differential equation is Figure 2. Charging a capacitor. Note that the where D is a constant. Dividing by C, and using Equation 1, the voltage across the capacitor can be rewritten as symbol for the capacitor, C, is two lines of equal length, unlike the symbol for the power supply, which is two unequal lines. pwr If the voltage is initially zero, then D =-CVpwr. Plugging this solution into Equation 6, the voltage across the charging capacitor is now pwr The product RC is the time constant of the circuit, and has units of seconds (2(Farad)- = RC, T Time ( Columb Volt )( Columb) Amp Volt Columb Figure 3. The voltage across a charging capacitor, V. rises towards the final ColumbIsFseconds value equal tothe power supply voltage, V The meaning of is shown in Figure 3 . By using equation 7, it can be seen that it takes a time for an initially uncharged capacitor to charge from 0 V to approximately 63% of the total .The time constant, T is also shownExplanation / Answer
a) R = 100 k.ohm, C= 30 uF
Time constant T =RC = 100*1000*30*10^-6 = 3 s
b) Vpwer = 5 V , Vc = 3.16 V
t = -RCln(1-Vc/Vpwer)
t = (-3)*ln(1-(3.16/5)) = 2.999s
c) C1 =30 uF, C2 = 50 uF
C= C1+C2 = 30+50 =80 uF
t = RC = 100*1000*80*10^-6 = 8s
(d) C =C1C2/(C1+C2) = (30*50)/(30+50)= 18.75 uF
t =RC = 100*1000*18.75*10^-6 = 1.875 s
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