Finding the time for charging Voltage

Just be careful with units, for example the top line should be written as ##.59 \cdot 7 \text{ volts} = 7 \text{ volts} \left(1- e^{-t / \tau} \right)##.Yes, that looks good. Just be careful with units, for example the top line should be written as ##.59 \cdot 7 \text{ volts} = 7 \text{ volts} \left(1- e^{-t / \tau} \right)##.In summary, the characteristic decay constant for a capacitor C discharging through a resistor R is the product of R times C. The voltage at time t = 2RC can be determined
  • #1
Mohamed Abdul

Homework Statement


1. The characteristic decay constant for a capacitor C discharging through a resistor R is the product of R times C. If the initial voltage at time t = 0 is Vo = 13Volts and what is the voltage at time t = 2RC?

2. A capacitor C = 9microFarads is charged from 0Volts to 7Volts through a resistor R = 39kOhms. How much time will it take to charge to 59% of full voltage? Express your answer in milliseconds.

Homework Equations


Charging => V = Vo(1-e^-t/Tau)
Discharge => V = Vo(e^-t/Tau)

Tau = RC

The Attempt at a Solution



For the first question I plugged in 2RC for time into the discharge equation, and canceled the RC with Tau to get 13*(e^-2) = 1.76

For the second question I'm thinking of plugging the same variables, but am unsure of what my initial and final voltages are. I thought they could be 0 and 7, but if that was the case V would equal 0 because Vo would equal 0, which I don't think is the case.
 
Physics news on Phys.org
  • #2
Mohamed Abdul said:

Homework Equations


Charging => V = Vo(1-e^-t/Tau)
Discharge => V = Vo(e^-t/Tau)

Tau = RC

The Attempt at a Solution



For the first question I plugged in 2RC for time into the discharge equation, and canceled the RC with Tau to get 13*(e^-2) = 1.76
OK. Don't forget the units.

For the second question I'm thinking of plugging the same variables, but am unsure of what my initial and final voltages are. I thought they could be 0 and 7, but if that was the case V would equal 0 because Vo would equal 0, which I don't think is the case.
For the second question use the discharging charging equation listed in your Relevant Equations section. But note that ##V_0## in that equation is not the initial potential. Consult your textbook or notes.
 
Last edited:
  • #3
TSny said:
OK. Don't forget the units.

For the second question use the discharging equation listed in your Relevant Equations section. But note that ##V_0## in that equation is not the initial potential. Consult your textbook or notes.
I'm confused, what else would V0 be besides the initial voltage? My notes say the same as well.
 
  • #4
Mohamed Abdul said:
I'm confused, what else would V0 be besides the initial voltage? My notes say the same as well.
Sorry, I meant to say use the formula for charging the capacitor for the second problem. In this equation ##V_0## is not the initial voltage. This cannot be zero since the equation would not make sense. As you noted, the initial voltage is zero. If your notes are saying ##V_0## is the initial voltage in the charging equation, then your notes are mistaken. Do a quick web search for "charging a capacitor".

Or, you can think about what the equation reduces to in the limit of t going to infinity.
 
  • #5
TSny said:
Sorry, I meant to say use the formula for charging the capacitor for the second problem. In this equation ##V_0## is not the initial voltage. This cannot be zero since the equation would not make sense. As you noted, the initial voltage is zero. If your notes are saying ##V_0## is the initial voltage in the charging equation, then your notes are mistaken. Do a quick web search for "charging a capacitor".

Or, you can think about what the equation reduces to in the limit of t going to infinity.
In that case would V0 be my max voltage, or the 59% of the full voltage?
 
  • #6
Mohamed Abdul said:
In that case would V0 be my max voltage, or the 59% of the full voltage?
Your charging equation is written as ##V = V_0 \left( 1- e^{-t / \tau }\right)##.

What is the limit of the right side for ##t## → ##\infty##? This should tell you the meaning of ##V_0##.
 
  • #7
TSny said:
Your charging equation is written as ##V = V_0 \left( 1- e^{-t / \tau }\right)##.

What is the limit of the right side for ##t## → ##\infty##? This should tell you the meaning of ##V_0##.
So that would mean when t approaches infinity it would be near zero, so V would equal Vo, right?
 
  • #8
Mohamed Abdul said:
So that would mean when t approaches infinity it would be near zero, so V would equal Vo, right?
Yes, ##e^{-t/\tau}## approaches zero as t → ∞. So, yes, V approaches V0 as t → ∞.
 
  • #9
TSny said:
Yes, ##e^{-t/\tau}## approaches zero as t → ∞. So, yes, V approaches V0 as t → ∞.
Doing that I set
.59(7) = 7(1-e^(-t/.351))
.59 = 1- e^-t/.351
.41 = e^-t/.351
ln(.41) = -t/.351
t=312.95 milliseconds

Is that the right line of thought for that, in terms of where I put my max voltage and where I put the 59% voltage?
 
  • #10
Mohamed Abdul said:
Doing that I set
.59(7) = 7(1-e^(-t/.351))
.59 = 1- e^-t/.351
.41 = e^-t/.351
ln(.41) = -t/.351
t=312.95 milliseconds

Is that the right line of thought for that, in terms of where I put my max voltage and where I put the 59% voltage?
Yes, that looks good.
 

Related to Finding the time for charging Voltage

1. What is the importance of finding the time for charging voltage?

The time it takes to charge voltage is important because it directly affects the efficiency and performance of a device or system. If the charging time is too long, it can lead to wasted energy and slower operation. On the other hand, if the charging time is too short, it can cause damage to the device and potentially create safety hazards.

2. How do you calculate the time for charging voltage?

The time for charging voltage can be calculated using the equation t = Q/I, where t is the time in seconds, Q is the charge in coulombs, and I is the current in amperes. This equation can be used to determine the time it will take for a specific voltage to be reached based on the charging rate.

3. What factors can affect the time for charging voltage?

Several factors can affect the time for charging voltage, including the charging rate, the capacity of the battery or capacitor, and the internal resistance of the device. Other external factors such as temperature and voltage fluctuations can also impact the charging time.

4. How can you optimize the time for charging voltage?

To optimize the time for charging voltage, you can adjust the charging rate and use a charger or power supply with the appropriate voltage and current output. It is also important to use high-quality components and maintain proper temperature and voltage levels to ensure efficient charging.

5. What are some common methods for charging voltage?

Some common methods for charging voltage include constant voltage charging, constant current charging, and pulse charging. Each method has its advantages and is suitable for different types of devices and batteries. It is important to choose the right charging method to ensure the best performance and longevity of the device.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
964
  • Introductory Physics Homework Help
Replies
9
Views
601
  • Introductory Physics Homework Help
Replies
2
Views
985
  • Introductory Physics Homework Help
Replies
3
Views
277
  • Introductory Physics Homework Help
Replies
4
Views
2K
Replies
1
Views
292
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
306
  • Introductory Physics Homework Help
Replies
6
Views
3K
Back
Top