In an RC circuit as depicted in the figure above, what happens to the time requi

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 5.90 s
µC

(d) the potential difference across the resistor after 5.90 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

First question -

Solution -

Since, the expression for time constant, T = R* C

And, greater the time constant, greater time it will take to charge

So, the answers are :
A.Increasing the resistance increases the time.
C.Increasing the capacitance increases the time

Second question -

(a) Time constant, T = R*C = 7.90x10^5 x4.70x10^-6 = 3.71 s

(b) The maximum charge occurs after a very long time, when the capacitor voltage approaches 9V.

Q = C*V = 4.70x10^-6 * 14 = 65.8x10^-6 C = 65.8 µC

(c) use the expression -

Vc(t) = V*e^(-t/RC)
time constant = RC = 3.71

V = 14

at t = 5.90 s -

Vc(t) = 14*e^(-5.9/3.71) = 2.85 V

charge on the capacitor = CV = 4.70*2.85 = 13.396 F = 1.34 x 10^-5 C

(d) Potential difference across the resistor at t = 5.90 s, Vc(R) = 14 - 2.85 = 11.15 V

(e) Current in the resistor, I = 11.15/(7.90x10^5) = 1.41 x 10^-5 A

Third question -

(a) The same process as mentioned above -

Vc(t) = 14*e^(-1.90/3.71) = 8.39 V

charge on the capacitor = CV = 4.70*8.39 = 39.43 F = 3.94 x 10^-5 C

(b) Potential difference across the resistor at t = 1.90 s, Vc(R) = 14 - 8.39 = 5.61 V

(e) Current in the resistor, I = 5.61/(7.90x10^5) = 7.10 x 10^-6 A

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