How to calculate R1 and R2 in this RC circuit

[derrickj585]

Homework Statement

Hey everyone,

I'm having difficulty calculating R1 and R2 in this RC Circuit, and I'd really appreciate if anyone could point out the flaws in my calculations.

The problem states: "The circuit contains an ideal battery, two resistors and a capacitor (C = 250 μF). The switch is closed at time t = 0, and the voltage across the capacitor is recorded as a function of time in the graph. Calculate the resistances R1 and R2".

We are provided with the circuit diagram and the Capacitor voltage vs time graph (both attached).

Relevant Equations

τ = RC
Q = VC
V = IR

Using the Capacitor voltage vs. time graph, I calculated the time constant τ when the switch is closed (capacitor charging) and open (capacitor discharging). I calculated τclosed = 2.5s and τopen = 3.75s.

Since the resistors are in parallel when the switch is closed, I assumed that 1/Req = 1/R1 + 1/R2, so τclosed = 2.5s = (1/R1 + 1/R2)^-1 * C; C = 250 uF
Since the resistors are in series when the switch is open/the capacitor is discharging, I assumed that Req = R1 + R2, and τopen = 3.75s = (R1 + R2) * C; C = 250 uF

This math didn't work at all (system of equations is impossible to solve), so I'm a bit stuck here. I would be extremely grateful if anyone could take the time to point out the mistakes in my approach and/or point me in the right direction. Thank you so much!

 

[berkeman]

Homework Statement:: Hey everyone,

I'm having difficulty calculating R1 and R2 in this RC Circuit, and I'd really appreciate if anyone could point out the flaws in my calculations.

The problem states: "The circuit contains an ideal battery, two resistors and a capacitor (C = 250 μF). The switch is closed at time t = 0, and the voltage across the capacitor is recorded as a function of time in the graph. Calculate the resistances R1 and R2".

We are provided with the circuit diagram and the Capacitor voltage vs time graph (both attached).
Relevant Equations:: τ = RC
Q = VC
V = IR

Using the Capacitor voltage vs. time graph, I calculated the time constant τ when the switch is closed (capacitor charging) and open (capacitor discharging). I calculated τclosed = 2.5s and τopen = 3.75s.

Since the resistors are in parallel when the switch is closed, I assumed that 1/Req = 1/R1 + 1/R2, so τclosed = 2.5s = (1/R1 + 1/R2)^-1 * C; C = 250 uF
Since the resistors are in series when the switch is open/the capacitor is discharging, I assumed that Req = R1 + R2, and τopen = 3.75s = (R1 + R2) * C; C = 250 uF

This math didn't work at all (system of equations is impossible to solve), so I'm a bit stuck here. I would be extremely grateful if anyone could take the time to point out the mistakes in my approach and/or point me in the right direction. Thank you so much!

Welcome to the PF.

Initially with the switch open, the R and C voltages will all be zero, since there is no source of energy for them.

Right after the switch is closed, the battery voltage will be placed across the resistor R2, and that will not change with time, since this is an ideal voltage source whose output voltage does not depend on its output current over time.

The voltages in the R1 and C branch will depend on time, as the capacitor charges up. Initially the capacitor is uncharged, so Vc=0 and all of the source voltage drops across the resistor R1. But as current flows through R1 to charge up the C, the voltage in that branch exponentially decays across R1 and exponentially increases across the capacitor and finally when there is no current through R1, all of the source voltage appears across the capacitor C.

Does that make sense? Have you had enough calculus yet where we can talk about this in terms of the differential equations for the voltage and current for a capacitor?

$i(t) = C\frac{dv(t)}{dt}$

 

[derrickj585]

Welcome to the PF.

Initially with the switch open, the R and C voltages will all be zero, since there is no source of energy for them.

Right after the switch is closed, the battery voltage will be placed across the resistor R2, and that will not change with time, since this is an ideal voltage source whose output voltage does not depend on its output current over time.

The voltages in the R1 and C branch will depend on time, as the capacitor charges up. Initially the capacitor is uncharged, so Vc=0 and all of the source voltage drops across the resistor R1. But as current flows through R1 to charge up the C, the voltage in that branch exponentially decays across R1 and exponentially increases across the capacitor and finally when there is no current through R1, all of the source voltage appears across the capacitor C.

Does that make sense? Have you had enough calculus yet where we can talk about this in terms of the differential equations for the voltage and current for a capacitor?

$i(t) = C\frac{dv(t)}{dt}$

Hi Berkeman,

Thank you so much for taking the time to help me out! Your explanation cleared a lot of it up for me.

So, since voltage is constant in R2 as soon as the switch is closed, does that mean that the time constant when the switch is closed is only dependent on R1 and C?

I don’t have much of a calculus background sadly, but I’ll try my best to understand :)

 

[phinds]

So, since voltage is constant in R2 as soon as the switch is closed, does that mean that the time constant when the switch is closed is only dependent on R1 and C?

Yes. R2 is basically nothing in the circuit other than an extra current drain on the source. It's irrelevant for any other circuit calculations.

Now, in the real world, where voltage sources have an internal resistance, then R2 does come into play.

EDIT: Oh, I should add that if, after a while you open the switch then R2 provides a current drain path for the cap.