LEARN MORE REMARKS In solving this problem, we paid scrupulous attention to sign

Question

LEARN MORE REMARKS In solving this problem, we paid scrupulous attention to signs. These signs must always be chosen when applying Kirchhoff's loop rule and must remain consistent throughout the problem Alternately, magnitudes can be used and the signs chosen by physical intuition. For example, the magnitude of the potential difference across the resistor must equal the magnitude of the potential difference across the battery minus the magnitude of the potential difference across the capacitor. QUESTION In an RC circuit as depicted in the figure above, what happens to the time required for the capacitor to be charged to half its maximum value if either the resistance or capacitance is increased with the same applied voltage? (Select all that apply.) Increasing the resistance increases the time Increasing the resistance decreases the time. Increasing the capacitance increases the time Increasing the capacitance decreases the time PRACTICE IT Use the worked example above to help you solve this problem. An uncharged capacitor and a resistor are connected in series to a battery, as shown in the figure above. If = 13.0 V, C = 5.30 F, and R = 7.40 x 10 Q, find the following. (a) the time constant of the circuit (b) the maximum charge on the capacitor HC (c) the charge on the capacitor after 6.50 s (d) the potential difference across the resistor after 6.50 s (e) the current in the resistor at that time EXERCISE HINTS: GETTING STARTED I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. (a) Find the charge on the capacitor after 1.90 s have elapsed (b) Find the magnitude of the potential difference across the capacitor after 1.90 s LVC= (c) Find the magnitude of the potential difference across the resistor at that same time LVR=

Explanation / Answer

Question: Time constant of an RC circuit = RC

so increasing the capacitance increases the time. Increasing the resistance increases the time

Practise IT

a)Time constant = RC

=5.3*10^-6*7.4*10^5

=3.922 s

b)maximum charge on the capacior = V*C

=13*5.3*10^-6

=68.9 uC

c)The equation for cvharge on the capacitor is given by Q = (VC)*(1-e^-t/RC))

=68.9 * (1-e^(-6.5/3.922))

=55.762 uC

d)Potential drop across the capacitor at 6.5s = Q/C

=55.762/5.3

=10.52 V

so potential drop across the resistor = 13-10.52

=2.48 V

e)Current = V/R

=2.48/(7.4*10^5)

=3.35 uA

=3.35*10^-6 A

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