Circuit theory: capacitor energy storage and discharging/charging times?

[Munnu]

Homework Statement

What is the initial energy store in the capacitor?

How long does it take for the capacitor to discharge to 50% of the initial stored energy?

In the following circuit, how long will it take the initial energy stored in the capacitor to increase to 50% of the final value?

Relevant Equations

u=0.5CV^2
V_c=V_0*e^(t/RC)
V_c=V_source*(1-e^(-t/RC))

This is not my homework. I took it upon myself to answer a textbook question for mental stimulation. I wanted to know if someone can verify if these were the correct values that needed to be solved for, process, and final answer, and if not, what needed to be considered.

For the initial voltage stored in the capacitor, I got 20V. I used thevenin’s theorem to get R_th=60 ohm, V_th=20V; V_c is V_th. Then I used capacitor discharge equation V_c=V_0*e^(t/RC) where t=0. V_c=20V*e^0=20V.

1. What is the initial energy store in the capacitor?
   1. Is it 0.1 watt-second (joule), are we looking for the joules unit of measurement? I calculated using u=0.5CV^2, where C=0.5mF and V_c=20V (calculated , thevenin)
2. How long does it take for the capacitor to discharge to 50% of the initial stored energy?
   1. Do I divide u=0.1 joule in half? What I did was u=0.5CV^2 where u=0.1joule/2. I solved for voltage using that formula. The voltage at the halfway stored energy is 14.14V. After solving for the voltage V_c=14.14V and a prior calculation of our V_0=20V (via thevenin’s), I used V_c=V_0*e^(-t/RC); RC=0.03s. I got t=0.010s.
3. In the following circuit, how long will it take the initial energy stored in the capacitor to increase to 50% of the final value?
   1. u=0.5CV^2 where u=0.1/2 joule, where the voltage at the halfway stored energy is 14.14V I used V_c=V_source*(1-e^(-t/RC)); RC=0.03s, V_source=20V. I got t=0.0368s.
   2. Note: V_c in this scenario is where the voltage would be when energy stored u=0.1/2 (50% energy stored) , V_source is the initial voltage stored in the capacitor.

 

[DaveE]

Good job, almost right. You have the correct concepts and calculations. Look at the time constants again. Where does the charging and discharging current flow?

 

[Munnu]

Good job, almost right. You have the correct concepts and calculations. Look at the time constants again. Where does the charging and discharging current flow?

Correct me if necessary:

At 50% Charging: There will be current through 3 branches—120 ohm resistor branch, 80 ohm resistor branch, and the 12ohm and 0.5mF capacitor branch. Current flows towards the capacitor’s positive terminal.

Charging:
RC=((120||80ohm)+12ohm)*0.5mF=60ohm *0.5mF=0.03sec
resistor calculation from thevenin’s theorem.

Upon discharging, current flows away from the capacitor’s positive terminal (towards the 12 ohm resistor and 80 ohm resistor). It won’t flow through the 120 ohm resistor since there’s an open circuit on the left end of that resistor due to the switch flipping open.

Discharging
RC=(12+80ohm)*0.5mF=0.046sec

The value of RC doesn’t change for the charging scenario for question 3, I won’t recalculate it. I’ll solve question 2.

question #2 retry:
V_c=V_0*e^(-t/RC)
14.14=20*e^(-t/0.046)
t=0.0159s (@50%
initial stored energy discharge)

 

[DaveE]

Yes, exactly right.

One additional comment. You don't need part 1 (the initial voltage) to solve part 2. You just need to know that the energy is a function of the voltage squared and that 50% of the initial energy will always be 0.7071x the initial voltage. You can see this in the formula when you end up using the ratio of the two voltages when you do the ln function. In this case that ratio will always be 0.7071 regardless of the initial conditions.

 

[Tom.G]

I think I detect a problem. For clarification:

Is that capacitor value, "0.5mF", in Farads, milliFarads, or microFarads?
How does that affect the answer?

Cheers,
Tom

 

[DaveE]

Is that capacitor value, "0.5mF", in Farads, milliFarads, or microFarads?

That must be millifarads, which REAL EEs NEVER USE. It confuses people. In ancient times, you may see that as microfarads, and mmF for picofarads, but then people got smarter. Now days you'll see things like 4700uF, 0.47F, 2200pF, 0.012uF, etc. Sometimes people (like me) use nF, because it's not confusing, but it's also not necessary. But no one in industry likes mF, when they do use it, it's millifarads in my experience, even though it used to be uF.

I can almost instantly tell if homework problems were written by physicists or EEs because of a bunch of practical issues like this. Things like 10H inductors, 1 Coulomb, or 250Ω just hardly ever exist in the real world.

How does that affect the answer?

It changes it by some factor of 1000. A 1000x larger capacitor takes 1000x longer to charge/discharge through resistors.

In the real world, you can almost always tell the factors of 1000 by the circuit it's in. No one builds an audio amp with a bandwidth of 100MHz or 100Hz, for example. In fact it isn't uncommon to see capacitors in schematics labeled "0.012" because they know you know it's uF, although that is bad practice IMO. They also expect that if you're confused you'll look it up on the BOM, which always comes with a real schematic these days.

 

[DaveE]

I didn't really follow your reasoning here. Although the truth is I'm too lazy to work it out. Here's my take:

Initially: V₀ = 20V, C = 500uF then u₀ = CV₀²/2 = 0.1J

Later: u₁ = u₀/2 = CV₀²/4
At what voltage (V₁) on C do you get u₁?
u₁ = CV₁²/2 = CV₀²/4 ⇒ V₁ = √(V₀²/2) = V₀/√2
so V₁/V₀ = 1/√2 = √(u₁/u₀) since energy is proportional to voltage squared in capacitors.

In your method remember √(2u/C) is a voltage, I'm not sure why you would divide that by 2.

 

[Munnu]

Yes, exactly right.

One additional comment. You don't need part 1 (the initial voltage) to solve part 2. You just need to know that the energy is a function of the voltage squared and that 50% of the initial energy will always be 0.7071x the initial voltage. You can see this in the formula when you end up using the ratio of the two voltages when you do the ln function. In this case that ratio will always be 0.7071 regardless of the initial conditions.

I might need a walkthrough, I will try to show what I think I understand from this specific reply.

You don't need part 1 (the initial voltage) to solve part 2

You mean I didn’t need the V_c_init=20V solved for in order to find “how long does it take for the capacitor to discharge to 50% of the initial stored energy”?

You just need to know that the energy is a function of the voltage squared

What would this look like in function notation? What I think is said is using u=0.5CV^2, I can see this as u(V)=0.5CV^2. Can it also be denoted as u = u(V^2) (in this case would that require a rewrite of the equation to u=0.5CV?)

50% of the initial energy will always be 0.7071x the initial voltage.

0.5u_0=0.7071V_c_init, yeah?

You can see this in the formula when you end up using the ratio of the two voltages when you do the ln function.

From 14.142=20*e^(-t/0.046) to 14.142/20=e^(-t/0.046), 14.142/20=0.7071

 

[DaveE]

I might need a walkthrough, I will try to show what I think I understand from this specific reply.You mean I didn’t need the V_c_init=20V solved for in order to find “how long does it take for the capacitor to discharge to 50% of the initial stored energy”?What would this look like in function notation? What I think is said is using u=0.5CV^2, I can see this as u(V)=0.5CV^2. Can it also be denoted as u = u(V^2) (in this case would that require a rewrite of the equation to u=0.5CV?)0.5u_0=0.7071V_c_init, yeah?From 14.142=20*e^(-t/0.046) to 14.142/20=e^(-t/0.046), 14.142/20=0.7071

You know Vc(t) = Vc(0)⋅e^(-t/τ) also, you know τ = RC from examination (solving) the circuit.

So how long does it take to discharge 50% of it's initial energy?

You have also found that 50% of the initial energy occurs at 70.7% of the initial voltage. So the duration of that discharge is the time from 100% to 70.7% of the initial voltage. Let's define that as Vc(t₁) = Vc(0)/(√2).

Then you have the equation:
Vc(t₁) = Vc(0)⋅e^(-t₁/τ) = Vc(0)/(√2)

Divide both sides by Vc(0) to get e^(-t₁/τ) = 1/(√2) ⇒ t₁ = -τ⋅ln(1/(√2))
Which can be solved with no knowledge of the value of Vc(0).

Also, note that 70.7% isn't magical it is the square root of the ratio on the energies. So the more general form of this answer is t₁ = -τ⋅ln(√(E₁/E₀)).

If they'd asked for 1/5 of the initial energy, the answer would be -τ⋅ln(1/(√5)).

If they'd asked for the time from 1/2 to 1/4 of the initial energy, the answer would be -τ⋅ln((√2)/(√4)) = -τ⋅ln(1/(√2)), which is, again, a 50% discharge from when you started measuring the duration.

The time from 100% to 25%? Twice the time from 100% to 50%, or also half of the time from 50% to 6.25%. It's all about the ratios.

 

[Munnu]

Also, note that 70.7% isn't magical it is the square root of the ratio on the energies. So the more general form of this answer is t1 = -τ⋅ln(√(E1/E0)).

This was useful. Is this some sort of technique-rule in an electronics textbook?

The time from 100% to 25%? Twice the time from 100% to 50%, or also half of the time from 50% to 6.25%. It's all about the ratios.

Are we talking about E1/E0 values?
When I plug in values to solve for each t using t1 = -τ⋅ln(√(E1/E0)), I get:
t_100_to_50=0.0159424
t_100_to_25=0.0318848
t_50_to_6.25=0.0478272

I can see that t_100_to_50’s result is half of t_100_to_25’s result. I can’t see the “also half of the time from 50% to 6.25%“ part.

 

[DaveE]

This was useful. Is this some sort of technique-rule in an electronics textbook?Are we talking about E1/E0 values?
When I plug in values to solve for each t using t1 = -τ⋅ln(√(E1/E0)), I get:
t_100_to_50=0.0159424
t_100_to_25=0.0318848
t_50_to_6.25=0.0478272

I can see that t_100_to_50’s result is half of t_100_to_25’s result. I can’t see the “also half of the time from 50% to 6.25%“ part.

Oops! Yes you and the formula are correct. I was careless.