Two capacitors C1 = 1 µF and C2 = 2 µF are connected to a battery using the swit

Question

Two capacitors C1 = 1 µF and C2 = 2 µF are connected to a battery using the switch S. When the switch is in position A the capacitor C1 is connected to a VB = 12 Volt battery. When the switch is moved to position B the capacitor C1 is disconnected from the battery but remains with the same charge when it was connected to the battery. When the switch is moved to position C the charged capacitor C1 is connected to the uncharged capacitor C2.
(a) When the switch is in position B, what is the charge on the capacitor C1? µC
(b) When the switch is in position C (after stabilizing), what is the charge on capacitor C1?
µC
(c) When the switch is in position C (after stabilizing), what is the voltage across capacitor C1?
V
(d) When the switch is in position C (after stabilizing), what is the charge on capacitor C2?
µC
(e) When the switch is in position C (after stabilizing), what is the voltage across capacitor C

Explanation / Answer

(a) The charge is given by:    q = CV ( 1 - e-t/RC ) = 0.00003 * 26 ( 1 - e-0.060 / 960*0.00003 ) = 683x 10-6 Coulomb (b) Across the capacitor:    V = q/C = 0.000683 / 0.000030 = 22.77volts So across the resistor:   26 - 22.7 = 3.3 volts (c) When the switch is thrown back, the battery is nolonger in the circuit. At this instant, the voltage across thecapacitor is still 22.77 volts... but now the voltage acrossthe resistor (which is directly connected to the cap) is ALSO 22.77 volts (d) Now the capacitor is discharging. We use:    q = CV e-t/RC = 0.00003 * 22.77 e-.060/960*0.00003    = 85.06 x10-6 coulomb
(d) Now the capacitor is discharging. We use:    q = CV e-t/RC = 0.00003 * 22.77 e-.060/960*0.00003    = 85.06 x10-6 coulomb

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