The drawing shows two fully charged capacitors (C1 = 5 µF, q1 = 4 µC; C2 = 13 µF

Question

The drawing shows two fully charged capacitors (C1 = 5 µF, q1 = 4 µC; C2 = 13 µF, q2 = 4 µC). The switch is closed, and charge flows until equilibrium is reestablished (i.e., until both capacitors have the same voltage across their plates). Find the resulting voltage across either capacitor.
( The capacitor is a box with an open switch at the top .. on the right hand side it is labeled C2 with two positive and negative signs on the inside and outside of the box, on the left hand side there is one positive and one negative sign on each side of the box and it is labeled C1 )

Explanation / Answer

Given Capacitance of the two capacitors are C1 = 5F , C2 = 13F Charge on the two capacitors is q1 = 4C and q2 = 4C The relation between capacitance C, potential difference V and charge on the plates q is q = CV Charge is conserved during equilibrium, then      q1 + q2 = 8 C = q1f + q2f ----1 After equilibrium has reached , the capacitors have equal charge across them.And here they are connected in parallel                       Vf = q1f/C1 =q2f /C2                                 q1f = q2f (C1/C2)                                       = q2f (5 F /13F ) = 0.38 q2f substitute this value in eq1, we get                         0.38 q2f + q2f = 8C                                           q2f = 5.8 C The voltage across the capacitor is                                 Vf = q2f / C2 = 5.8 C / 13 F                                      = 0.45 V                                                   

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