Two parallel plate capacitors, C1 and C2 are connected in series to a 200-k Ohm

Question

Two parallel plate capacitors, C1 and C2 are connected in series to a 200-k Ohm resistor, a 60-V battery, and a switch S. Both capacitors are air gap capacitors and have plates with an area of 2.0 cm^2 and a separation of .10 mm. The capacitors are initially uncharged and switch S is open. After switch S is closed, how long will it take for the capacitors to charge to 50% of their maximum value? A) 8.9 *10^-7 s B) 1.2 *10^-6 s C) 1.8 * 10^-6 s D) 2.5 *10^-6 s E) 4.9 * 10^-6 s Must show work for rating

Explanation / Answer

C1 = eA/d = (8.85 x 10^-12)(2 x 10^-4)/1 x 10^-4 = 17.7 pf

C2 = 7C1 = 123.9 pf

Ceqv = 123.9*17.7/141.6 = 15.5 pf

q=Ceqv*60 = 0.92 x 10^-9 Coulombs

This is the charge on both capacitors

V C1 = 0.92 x 10^-9/1.77 x 10^-11 = 52 volts

V C2 = 60 -52 = 8 Volts

t = -RC*ln(1-(v(t)/v))

=4.9 * 10^-6 s

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