Question

Question 9 In an RC series circuit, ε = 12.0 V, R = 1.07 MΩ, and C = 2.66 µF. (a) Calculate the time constant. (b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to 14.3 µC? (a) Number Enter your answer for part (a) in accordance to the question statement Units Choose the answer for part (a) from the menu in accordance to the question statement (b) Number Enter your answer for part (b) in accordance to the question statement Units Choose the answer for part (b) from the menu in accordance to the question statement (c) Number Enter your answer for part (c) in accordance to the question statement Units Choose the answer for part (c) from the menu in accordance to the question statement

Answer 1

a.) τ = 2.85 s b.) Q = 3.19 * 10^-5 C c.) t = 1.691 s

Step-by-step explanation:

So we are told that it is a RC circuit. We are told
`Q = C V [1 - e^(-t/RC)]` = 12.0 V, R = 1.07 MΩ and C = 2.66 µF.

a.) The time constant for RC circuit, τ = RC. Substituting our known values we get:

τ = RC where R = (1.07 * 10 ^ 6)Ω and C = (2.66 * 10 ^ -6) F

τ = (1.07 * 10 ^ 6)Ω * (2.66 * 10 ^ -6) F = 2.8462 s ≈ 2.85 s

τ = 2.85 s

b.) The relationship between capacitance, potential, charge is given:

`Q = CV[1-e^(-t/RC) ]`

The capacitor is fully charge when t approaches infinity, therefore:

`Q =  \lim_(t \to \infty) a_n CV[1-e^(-t/RC) ]`

When t approaches infinity, the term e becomes very small (e^-∞ = 0), therefore we can simplify the equation and plug in our values

`Q = (2.66*10^(-6)) F * (12.0)V *[1 - 0] = 3.192 * 10^(-5)`

Q = 3.19 * 10^-5 C

c.) Using the same equation as before, we can substitute Q in and solve for Q:

`(14.3 * 10 ^ 6) C = (2.66*10^(-6))F *(12.0)V*[1-e^(-t/(2.85s))]\\0.552 = e^(-t/(2.85s))\\t = -1 * 2.85 * ln(0.552) \\t = 1.69120678 s`

t = 1.691 s

Hope this helps! I'm not sure what the units you want, so convert to the desired units.