Question

1. A parallel plat capacitor with C= 20 ?F is being charged by a battery of ?=12 V. If total resistance of the circuit is 5.2 k?, then: a) How long does it take for the capacitor to charge to 90% of fully charged? b) Draw the current profile and determine the current at this time. c) How much energy is stored in the capacitor by this time? 3 d) To what value should we change the resistance so that the time in part (a) is reduced by 50%? e) If the capacitor plate separation distance d = 1.0 mm, calculate the E and the B field values and directions between the plates of the capacitor at this time? Do they depend on the location inside plates?

Answer #1

1.

C = Capacitance = 20 uF

E = battery Voltage = 12 Volts

R = resistance = 5200 ohm

Time constant for the circuit is given as

T = RC = 5200 x 20 x 10^{-6} = 0.104 sec

Given that : Q = 0.90 Q_{o}

Using the equation

Q = Q_{o} (1 - e^{-t/T})

0.90 Q_{o} = Q_{o} (1 - e^{-t/T}?)

0.90 = (1 - e^{-t/0.104})

t = 0.24 sec

b)

V_{c} = Voltage across the capacitor = 0.90 E = 0.90 x
12 = 10.8 Volts

V_{r} = Voltage across the resistor = 12 - 10.8 = 1.2
Volts

Using ohm's law , current is given as

i = V_{r} /R = 1.2 /5200 = 2.3 x 10^{-4} A

c)

U = energy stored = (0.5) C V_{c}^{2} = (0.5)
(20 x 10^{-6}) (10.8)^{2} = 0.0012 J

d)

t = 0.50 x 0.24 = 0.12 sec

Using the equation

Q = Q_{o} e^{-t/T}

0.90 Q_{o} = Q_{o} (1 - e^{-0.12/T})

T = 0.9 sec

RC = 0.052

R (20 x 10^{-6}) = 0.052

R = 2600 ohm

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