{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chp31_5 - PHY2061 R D Field Discharging a Capacitor After...

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

View Full Document Right Arrow Icon
PHY2061 R. D. Field Department of Physics chp31_5.doc University of Florida Discharging a Capacitor After the switch is closed the current is leaving the capacitor so that I = -dQ/dt , where Q is the charge on the capacitor and summing all the potential changes in going around the loop gives Q C IR = 0 , where I(t) and Q(t) are a function of time. If the switch is closed at t=0 then Q(0)=Q 0 and Q C R dQ dt + = 0 , which can be written in the form dQ dt Q = − 1 τ , where I have defined τ =RC . Dividing by Q and multiplying by dt and integrating gives
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}