A capacitor with capacitance C = 7.00 times 10^-3 F Is Initially uncharged. It I

Question

A capacitor with capacitance C = 7.00 times 10^-3 F Is Initially uncharged. It Is In series with a source of Emf of 4.00 volts, a resistor R, and a switch as shown In the figure below. At t=0, the switch is dosed. The graph below shows the potential difference across the capacitor as function oft, the time elapsed since the switch closed. Calculate the value of R. Note that the curve passes through a grid intersection point, (in ohm) 5.83 times 10^2 5.94 times 10^2 6.06 times 10^2 6.18 times 10^2 6.31 times 10^2 6.43 times 10^2 6.56 times 10^2 6.69 times 10^2 What can you tell about the maximum power dissipated in R? The maximum power dissipated in R occurs at ... t = 0 t = RC t = infinity

Explanation / Answer

1) D) 6.18*10^2 ohms

given
Vmax = 4 volts

C = 7*10^-3 F

let R is the resistance of resistor.

let T is the time constant of the ckt. so, T = R*C

at t = 6 s, VC = 3 volts

we know, while charging

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

3 = 4*(1 - e^(-6/T))

0.75 = 1 - e^(-6/T)

e^(-6/T) = 1 - 0.75

-6/T = ln(0.25)

T = -6/ln(0.25)

= 4.328 s

R*C = 4.328

R = 4.328/C

= 4.328/(7*10^-3)

= 6.18*10^2 ohms

2) A) t = 0 .
because, at t = 0, the capacitor acts as short ckt.

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