PHYS-102 Homework-04 Solutions Electric Potential and Capacitance P20.41(a) 1Cs=115.0+13.00Cs=2.50 μFCp=2.50+6.00=8.50 FCeq=18.50 F+120.0 F⎛⎝⎜⎞⎠⎟−1=5.96 F(b) Q=CΔV=5.96 F()15.0 V=89.5 Con 20.0 F ΔV=QC=89.5 C20.0 F=4.47 V15.0−4.47=10.53 VQ=CΔV=6.00 F10.53 V=63.2 Con 6.00 F89.5−63.2=26.3 Con 15.0 F and 3.00 FIG. P20.41P20.49U=12CΔV2The circuit diagram is shown at the right. (a) Cp=C1+C2=25.0 F+5.00 F=30.0 U=1230.0×10−61002=0.150 J(b) Cs=1C1+1C2⎛⎝⎜⎞⎠⎟−1=125.0 F+15.00 F⎛⎝⎜⎞⎠⎟−1=4.17 U=12CΔV2ΔV=2UC=20.1504.17×10−6
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 09/03/2008 for the course PHYS 102 taught by Professor N/a during the Spring '08 term at Drexel.