17_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

# 17_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

This preview shows page 1. Sign up to view the full content.

Electrical Energy and Capacitance 51 (b) From Figure 3: ( ) ( )( ) = 2.00 F 12.0 V 24.0 C ac ac ac QC V µ ∆= = = 24.0 C ab bc ac QQ Q From Figure 2: = = 3.00 F Thus, the charge on the capacitor is 3 =24 .0 C Q Continuing to use Figure 2, () 24.0 C =4 . 6.00 F ab ab ab Q V C 0 0 V ∆== and 3 24.0 C =8 . 0 0 V 3.00 F bc bc bc Q VV C = = ()() From Figure 1, 42 .00 V ab VVV ()( ) ( ) and 44 = QC 4 4.00 F 4.00 V 16.0 C V = ( ) 22 = 2 V 8.00 C = 16.32 (1) 12 1 9.00 pF 9.00 pF parallel CC C C =+= ⇒ = 2 C 11 1 2.00 pF series series C C C C =+ ⇒ = = + Thus, using equation (1), ( ) 9.00 pF 9.00 pF series C = −+ ( = which reduces to
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

Ask a homework question - tutors are online