How Do Parallel Resistors and Capacitors Affect RC Circuit Half-Life?

AI Thread Summary
In an RC circuit with two identical resistors and capacitors connected in parallel, the equivalent resistance becomes R/2 and the equivalent capacitance becomes 2C. The half-life formula for the circuit is derived from the relationship t1/2 = RC ln2. When substituting the new values into the equation, the time constant remains unchanged, leading to the conclusion that the original and new half-lives are the same. Thus, the half-life does not change despite the alterations in resistance and capacitance. Understanding these relationships is crucial for analyzing RC circuits effectively.
poweroffive
Messages
5
Reaction score
0
Hey Everyone, I have a question regarding physics.

Say you're given an RC circuit with two identical capictors and two identical resistors and that the two resistors are connected in parallel and the two capacitors are connected in parallel and both are used in the circuit.

How can you figure out what the ratio of the new t 1/2 to the original t 1/2 would be, as in the "half-life"?

Any help is appreciated. THank you in advance.



Sorry if I accidently posted this elsewhere, I am new to these wonderful forums and this mistake won't happen again.


PS: I attempted to use the equations with e and the -t 1/2 but kept getting odd ratios...
 
Physics news on Phys.org
Two identical resistors R connected in parallel have an equivalent resistance of R/2.

Two identical capacitors C connected in parallel have an equivalent capacitance of 2C.

Does that help? You need to show more specific work if you need more help.
 
Thanks for your reply, berkeman. It does, indeed, help.

So now that I have this, can i just plug it into this formula:

where the halflife (t 1/2) = (ln2)RC and that I derived from V(1/2)/Vi = e^((-t1/2)/RC)

so that the original equation is just that, t1/2 = RC ln2 and the new equation is the same since R/2 times 2C is equal to RC?

So, to summarize, wouldn't the original and the new half life be the same?
 
poweroffive said:
So, to summarize, wouldn't the original and the new half life be the same?

If I understand the question correctly, then yes, the time constant would stay the same.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top