EMF of the battery is 50.0V A) At what time t after the switch is closed is the

Question

EMF of the battery is 50.0V

A) At what time t after the switch is closed is the charge on the capacitor equal to 40.0 × 106 C?

B) When the current has magnitude 3.00 A, at what rate is energy being stored in the capacitor?

C) When the current has magnitude 3.00 A, at what rate is energy being supplied by the battery?

Exercise 26.42 Constants You connect a battery, resistor, and capacitor as in (Figure 1), where R = 14.0 and C-5.00 × 10-6 F. The switch S is closed at t = 0, when the current in the circuit has magnitude 3.00 A, the charge on the capacitor is 40.0 × 10-6 C. Figure 1 of 1 Switch open i=0 q=0

Explanation / Answer

i = 3 A

Q = charge on capacitor = 40 x 10-6 C

Vr = potential difference across the resistor = iR = 3R

Vc = potential difference across the capacitor = Q/C = ( 40 x 10-6)/(5 x 10-6) = 8 Volts

E = battery emf = 50 volts

E = Vr + Vc

50 = 3R + 8

R = 14 ohm

a)

T = time constant = RC = 14 (5 x 10-6) = 70 x 10-6 sec

t = ?

Q(t) = charge on capacitor after time "t" = 40 x 10-6 C

Qmax = maximum charge on capacitor = CE = (5 x 10-6) (50) = 250 x 10-6 C

Using the equation

Q(t) = Qmax (1 - e-t/T)

40 x 10-6 = 250 x 10-6 (1 - e-t/(70 x 10-6))

t = 12.2 x 10-6 sec

B)

i = 3 A

Vc = 8

rate of energy being stored is given as

P = i Vc = 3 x 8 = 24 Watt

c)

i= 3

E = 50 Volts

energy supplied by battery is given as

P = iE = 3 x 50 = 150 watt

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