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1.) What is the effective capacitance of this circuit? Answer in units of μF.
2.) What is the total energy stored by the 27 μF capacitor on the right-hand side of the circuit? Answer in units of mJ.
If someone could help me out with these two part I'd really appreciate it!
CwCc
Favorite Answer
Capacitors in parallel add. 26+27 = 53 μF on each side of the circuit. Capacitors in series combine like resistors in parallel. Two equal capacitors in series will have an equivalent capacitance of half. 53/2 = 26.5 μF is the capacitance of the circuit.
Only half the voltage drop will occur over the parallel combination on the right. The energy stored in a capacitor is given by
1/2 CV^2 = 1/2 (27 μF)(32 V)^2 = 13.8 mJ
?
C1 = 26 μF , C2 = 27 μF , C3 = 26 μF , C4 = 27 μF
C1 and C2 are connected in parallel and C3 and C4
are connected in parallel.
Left hand combination and the Right hand combination
are in series.
Left hand side : Cx = C1 + C2 = (26 + 27) μF = 53 μF
Similarly, Right hand side :
Cy = C3 + C4 = (26 + 27) μF = 53 μF
1/ Ceff = 1/Cx + 1/Cy = 2/53
=> Ceff = 53/2 μF = 26.5 μF
Vx = Vy = 64/2 = 32 V
Energy stored in the C4 = 27 μF capacitor
= E = (1/2)C4 (Vy)² = (1/2)(27)(32)² μJ
= 13824 μJ = 13.824 mJ