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yikes_physics
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capcitance again FTL :(
All capacitors of the open circuit in Figure 3 are discharged at time t = 0, when the switch
S is turned on and capacitors starts to be charged. What is the time constant \tau of the charging process? What is the charge on capacitor C2 at an arbitrary time t > 0? If the resistances R1 and R2 are increased by a factor of 2, while capacitances C1 and C2 are decreased by a factor of 2, would the charging process occur faster or slower?
Resistors in sereis = R1 + R2 +...
capacitors in series = C1 + C2 + ...
capacitors in parallel = 1/C1 + 1/C2 + ...
Voltage across a capacitor = V(o)e^-(delta t/\tau)
capacitor discharging = I(o)e^-(t/\tau) = Q(o)e^-(t/\tau)
ok, bak with another one of these variable infested things, lol. First, i find R(eq) by adding the resistors in series and find Ceq by first combining the capacitors in series and then adding them in parallel. Now i have R(eq) and C(eq). Multiplying these together nets me \tau, correct?
if that's right, i then take \tau and place it in the equation V(o)e^-(delta t/\tau) where my voltage is half of C(eq). This is because both pieces of parallel capacitors get equal voltage.
As for the final part of the question, i would assume that the charge occurs faster due to the lower capacitance, but I am not sure by how much. since the resistors add in series, they simple become 2R(eq) while C(eq) goes up much more, or is this wrong to assume? any direction would be awesome.
ps. you guys are THE WIN...end of story
Homework Statement
All capacitors of the open circuit in Figure 3 are discharged at time t = 0, when the switch
S is turned on and capacitors starts to be charged. What is the time constant \tau of the charging process? What is the charge on capacitor C2 at an arbitrary time t > 0? If the resistances R1 and R2 are increased by a factor of 2, while capacitances C1 and C2 are decreased by a factor of 2, would the charging process occur faster or slower?
Homework Equations
Resistors in sereis = R1 + R2 +...
capacitors in series = C1 + C2 + ...
capacitors in parallel = 1/C1 + 1/C2 + ...
Voltage across a capacitor = V(o)e^-(delta t/\tau)
capacitor discharging = I(o)e^-(t/\tau) = Q(o)e^-(t/\tau)
The Attempt at a Solution
ok, bak with another one of these variable infested things, lol. First, i find R(eq) by adding the resistors in series and find Ceq by first combining the capacitors in series and then adding them in parallel. Now i have R(eq) and C(eq). Multiplying these together nets me \tau, correct?
if that's right, i then take \tau and place it in the equation V(o)e^-(delta t/\tau) where my voltage is half of C(eq). This is because both pieces of parallel capacitors get equal voltage.
As for the final part of the question, i would assume that the charge occurs faster due to the lower capacitance, but I am not sure by how much. since the resistors add in series, they simple become 2R(eq) while C(eq) goes up much more, or is this wrong to assume? any direction would be awesome.
ps. you guys are THE WIN...end of story