vsage
Oct4-04, 05:44 PM
C = 7.20 micro-Farads, R1 = 10.5ohm, R2 = 22.5ohm, V0 = 16.0V. What is the current (to within 0.01A) through R1 immediately after the switch S is closed? I created two equations to represent the voltage drops in the system but Capacitors charge up over time which is confusing me:
16V - i1*R1 - i2*r2 = 0
16V - i1R1 - q/C = 0
Therefore i2*R2 = q/C subtracting the two. Unfortunately that's two unknowns. I also sort of (but not fully understand) that for a resitor and capacitor in series, q = C*Vapplied*(1-e^(-t*R*C)) so at t = 0, q = 0 and thus no voltage drop across the capacitor and therefore no current. This means that all of the current is flowing through R2 right? Therefore i2 = i1 so..
16V - I1(R1+R2) = 0
16V - I1(10.5+22.5) =0
I = 0.48A?
Am I right to apply this to this drawing? Edit 1.33 was wrong. So was 0.48. OK I'm really lost now.
16V - i1*R1 - i2*r2 = 0
16V - i1R1 - q/C = 0
Therefore i2*R2 = q/C subtracting the two. Unfortunately that's two unknowns. I also sort of (but not fully understand) that for a resitor and capacitor in series, q = C*Vapplied*(1-e^(-t*R*C)) so at t = 0, q = 0 and thus no voltage drop across the capacitor and therefore no current. This means that all of the current is flowing through R2 right? Therefore i2 = i1 so..
16V - I1(R1+R2) = 0
16V - I1(10.5+22.5) =0
I = 0.48A?
Am I right to apply this to this drawing? Edit 1.33 was wrong. So was 0.48. OK I'm really lost now.