need help will rate best answer, A capacitor with capacitance C = 6.00 x 10^-3 F

Question

need help will rate best answer,

A capacitor with capacitance C = 6.00 x 10^-3 F is initially uncharged. It is in series with a source of Emf of 5.00 volts, a resistor R, and a switch as shown in the figure below. At t = 0, the switch is closed. The graph below shows the potential difference across the capacitor as function of t, the time elapsed since the switch closed. Calculate the value of R. Note that the curve passes through a grid intersection point. What can you tell about the maximum power dissipated in R? The maximum power dissipated in R occurs at ...

Explanation / Answer

B) 1.27*10^3 ohms

here is the explanation

Let T is the time constant of the ckt.

we know potential across capacitor at any time t,

V = Vmax*(1 - e^(-t/T)

from graph, at t = 7s, v = 3 volts

so,

3 = 5*(1 - e^(-t/T))

0.6 = 1 - e^(-t/T)

e^(-t/T) = 1 - 0.6

t = -T*ln(0.4)

T = -t/ln(0.4)

= -7/ln(0.4)

= 7.64 s

but we know, T = R*C

==> R = T/C

= 7.64/(6*10^-3)

= 1.27*10^3 ohms

maximum power dissipated at t = 0

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