Hey I was just thinking about the Rt expression. The general formula for Rt is R0*(1+a* ΔT) but you used Rt = only aΔT+R0. So did you just change the formula just like that because shouldn't it be R0+ R0*a*ΔT? Can you please confirm this.
Yes we can combine the capacitors in parallel and get a total capacitance. So is the total capacitance to be used with the total resistance in the relevant equation that I uploaded for this problem in order to get the charging time. Please confirm. Thanks
No its fine. I just don't understand how can I use only c2 when both the capacitors are in parallel. Just because they share same voltage does it mean I can ignore c1?
Oh yeah man sorry you're right
I suck at simplifying...if I ignore r3 I will have capacitors in parallel. But what about resistors will they be in parallel too?
And also how does having same voltage help...
Yes sorry you are right. Thanks alot.
can you also check out this forum https://www.physicsforums.com/threads/capacitor-charging-and-discharging.780167/#post-4903875
I can't get the understanding to this as well. Thanks
so dv/dT = 60 000/(T+40000)^2
and now do I put the two different T values in? As in should I get two dv/dT values? Because I have 2 different temperatures to use
Okay at the denominator I had v^2 which is (0.005T+200)^2 which gives me a quadratic equation 2.5*10^-5 T^2 + 2T +40000 so
dv/dT = 15/(2.5*10^-5 T^2 + 2T +40000)
Is that correct?
I have no idea sir. This is exactly the information I have. I think its supposed to be some general case. I have weak understanding of this whole circuit concept.
Is the total resistance of the circuit R3+R2 then parallel with R1 or is it R1+R2+R3?
I multiplyed with 5 to make it one fraction and then I took u and v. If I differentiate 5 and then take u and v won't it be some sort of double differentiation?
If I take u as 0.05T + 1000 then I get
dv/dT = 5.05-2.5*10^-4T/(0.005T+200)
Is this Dv/dT correct or the previous one?