This is a two part question!!! Show all work! A capacitor with capacitance c 4.0

Question

This is a two part question!!! Show all work!

A capacitor with capacitance c 4.00 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 0 1 2 3 4 5 6 7 8 9 Time t (in seconds) Calculate the value of R. Note that the curve passes through a grid intersection point. (in

Explanation / Answer

1) H) 1.91*10^3 ohms

given
Vmax = 5volts

C = 4*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 = 7 s, VC = 3 volts

we know, while charging

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

3 = 5*(1 - e^(-7/T))

0.6 = 1 - e^(-7/T)

e^(-7/T) = 1 - 0.6

-7/T = ln(0.4)

T = -7/ln(0.4)

= 7.639 s

R*C = 7.639

R = 7.639/C

= 7.639/(4*10^-3)

= 1.91*10^3 ohms

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

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