QUESTION In an RC circuit as depicted in the figure above, what happens to the t
ID: 1430379 • Letter: Q
Question
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
(a) the time constant of the circuit
s
(b) the maximum charge on the capacitor
µC
(c) the charge on the capacitor after 6.20 s
µC
(d) the potential difference across the resistor after 6.20 s
V
(e) the current in the resistor at that time
A
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.
Q = C
(b) Find the magnitude of the potential difference across the capacitor after 1.90 s.
VC = V
(c) Find the magnitude of the potential difference across the resistor at that same time.
VR = V
Explanation / Answer
As given in the question,
Voltage accross battery: Vo = 15 V
Capacitance: C = 5.1 µF = 5.1*10^-6 F
Resistance: R = 8.9*10^5
Time constant of the circuit: T = RC = 4.539 s
Maximum charge on capacitor: Qmax = C*Vo = 7.65*10^-5 C
EXERCISE HINTS: GETTING STARTED
(a) The charge on the capacitor after 1.9 s have elapsed (t = 1.9 s),
Q = Q(t) = Qmax[ 1 - e^-(t / T) ]
= (7.65*10^-5)*[ 1 - e^-(1.9 / 4.539) ] = 2.617*10^-5 C
(b) Te magnitude of the potential difference across the capacitor after 1.9 s,
VC = Q(t) / C = (2.617*10^-5) / (5.1*10^-6) = 5.13 V
(c) The magnitude of the potential difference across the resistor at that same time,
VR = Vo - VC = 15 - 5.13 = 9.87 V
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.