Proving time constant formulas of a capacitor using Kirchoff's laws

[estudent1371]

Homework Statement

Using R1, R2, ℰ and C along with Kirchhoff’s laws, prove the formulas in equations (7) and (8).

Relevant Equations

time constant charging
(7) t_c=(R_1 R_2)/(R_1+R_2 ) C
time constant discharging
(8) t_d=R_2 C

the attached files are a diagram of the circuit. also I believe I figured out how to proof equation (7) but I'm lost on how to proof (8)

 

[berkeman]

Welcome to the PF.

Homework Statement:: Using R1, R2, ℰ and C along with Kirchhoff’s laws, prove the formulas in equations (7) and (8).
Relevant Equations:: time constant charging
(7) t_c=(R_1 R_2)/(R_1+R_2 ) C
time constant discharging
(8) t_d=R_2 C

I'm lost on how to proof (8)

For equation 8, in order for the capacitor to discharge, what has to happen to the voltage source? (Hint: think "open circuit" for the source when it's off in this problem)

 

[TSny]

Your derivation of (7) looks good to me. The second line of the second page could be fixed up a bit. It is missing a parenthesis on the right-hand side and the limits of integration are not indicated.

 

[estudent1371]

thank you for the help so I have made some head way but I feel like i may have made a mistake in my math?

 

[TSny]

Looks like you were a little sneaky with inserting a negative sign on the left side of the third equation below the crossed-out box. How can you justify that?

I would go back to the equations at the top of the page and think about signs. Are you still considering I₂ to be positive if it flows downward through R₂? Are you still considering Q to be the charge on the upper plate of C?

 

[estudent1371]

Looks like you were a little sneaky with inserting a negative sign on the left side of the third equation below the crossed-out box. How can you justify that?

I would go back to the equations at the top of the page and think about signs. Are you still considering I₂ to be positive if it flows downward through R₂? Are you still considering Q to be the charge on the upper plate of C?

I didn't show the step but my integral on the left side is going from charged Q to completely discharged. but I used the integral property to switch them and put a negative in ∫ ⁰_Q=-∫^Q₀. However, I do see your point in that the current of the mini loop would flow in the opposite direction.

 

[TSny]

Your equation $\frac{dQ}{Q} = \frac{dt}{R_2C}$ should actually be $\frac{dQ}{Q} = -\frac{dt}{R_2C}$. Without the negative sign, the equation would yield a solution in which $Q$ grows exponentially with time rather than decays.

Integrating $\frac{dQ}{Q} = -\frac{dt}{R_2C}$ gives $$\int_{Q_0}^{Q(t)} \frac{dQ}{Q} = -\int_0^t\frac{dt}{R_2C}$$

At time $t = 0$, the charge on the capacitor is $Q_0$. At the later time $t$, the charge is $Q(t)$.

Note that your integral $-\int_{0}^{Q} \frac{dQ}{Q}$ leads to trouble with the lower limit of $0$ since $\ln{0}$ is undefined. This is related to the fact that it takes an infinite amount of time for the charge to decay to zero.