In Figure P18.50, suppose that the switch has been closed for a length of time sufficiently long enough for the capacitor to become fully charged. (R = 12.0 k, R2 = 19.0 k, R3 = 3.00 k, V = 9.10 V)
Figure P18.50
http://www.webassign.net/sercp/p18-50alt.gif
(a) Find the steady-state current in each resistor. IR = µA
IR2 = µA
IR3 = µA
(b) Find the charge on the capacitor.
µC
2007-03-01 12:51:34 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
(a) IR == V/(R+R2) = 9.10/(12E3 + 19 E3)
IR=294 µA
IR2 = IR1
IR3 = 0 (capacitor doe3s not conduct the voltage across R1=V(capasitor))
(b) Find the charge on the capacitor.
Q=Vc C=
Vc=V [R2/(R1+R2)]
Or
Vc=IR2 R2
Q= (IR2 R2 ) C=
Q=(19 E3 x 294 E-6) 10E-6
Q= 5.586E-6 Coulombs
2007-03-05 08:07:41 · answer #1 · answered by Edward 7 · 1⤊ 0⤋