Capacitor and Dependent Voltage Source

In summary, the student is trying to solve for the voltage across the capacitor, but is struggling to do so. They find the correct equation after substituting in the values for charge and current.
  • #1
Drakkith
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Homework Statement


Select the correct expression for ##i_o(t)## for ##t≥0^+##.
Figure_P07.30.jpg

Homework Equations


##V(t) = V(∞)+[v(0)-v(∞)]e^{\frac{-(t-t_0}{\tau}}##
##i(t) = C\frac{dV(t)}{dt}##
##\tau = RC##

The Attempt at a Solution



I'm a bit stuck on this problem. Specifically, I'm not sure how to deal with the dependent source.

At ##t=0##, the voltage of the capacitor should be 15 volts since the capacitor is directly in series with the voltage source with no branches leading off anywhere.

After ##t=0## the current through the capacitor would be ##C\frac{dV(t)}{dt}##, but I'm sure the dependent source changes things somehow. I'm just not sure how.
 

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  • #2
Drakkith said:
At t=0t=0, the voltage of the capacitor should be 15 volts since the capacitor is directly in series with the voltage source with no branches leading off anywhere.
Correct.
Drakkith said:
After t=0t=0 the current through the capacitor would be CdV(t)dtC\frac{dV(t)}{dt}, but I'm sure the dependent source changes things somehow. I'm just not sure how.
Try writing the KVL loop equation for that right-hand loop after the switch changes, and then factor in the initial condition that you have.
 
  • #3
berkeman said:
Try writing the KVL loop equation for that right-hand loop after the switch changes, and then factor in the initial condition that you have.

For my KVL loop I get ##5i_0 + V -15i_0 = 0##
## V-10i_0 = 0##
##V=10i_0##

After this I'm not sure what to do. I thought my capacitor voltage equation would be ##V(t) = 15e^{-3333t}## but using that doesn't lead me to a correct current expression.
 
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  • #4
Drakkith said:
For my KVL loop I get ##5i_0 + V -15i_0 = 0##
## V-10i_0 = 0##
##V=10_i0##

After this I'm not sure what to do. I thought my capacitor voltage equation would be ##V(t) = 15e^{-3333t}## but using that doesn't lead me to a correct current expression.
To get the voltage across the capacitor, you need to use the differential equation version. There may be a better way to approach it, but I'd start out my try adding up the voltage drops across the components in that loop and setting the sum equal to zero. what is the voltage drop across the capacitor as a function of the current?
 
  • #5
berkeman said:
To get the voltage across the capacitor, you need to use the differential equation version. There may be a better way to approach it, but I'd start out my try adding up the voltage drops across the components in that loop and setting the sum equal to zero.

Didn't I just do that?

berkeman said:
what is the voltage drop across the capacitor as a function of the current?

In general, or for this problem?
 
  • #6
Drakkith said:
In general, or for this problem?
In general, as a function of time.

Or, set up a DE in terms of charge on the capacitor.
 
  • #7
cnh1995 said:
In general, as a function of time.
It should be ##V(t) = \frac{1}{c}\int i(t)dt##
 
  • #8
Drakkith said:
It should be ##V(t) = \frac{1}{c}\int i(t)dt##
Yes, so what is i(t)?
 
  • #9
Drakkith said:
For my KVL loop I get ##5i_0 + V -15i_0 = 0##
## V-10i_0 = 0##
##V=10_i0##

After this I'm not sure what to do. I thought my capacitor voltage equation would be ##V(t) = 15e^{-3333t}## but using that doesn't lead me to a correct current expression.
You have arrived at the correct expression for V.
What is i0(t) in terms of charge on the capacitor? What is voltage V in terms charge on the capacitor?
 
  • #10
Drakkith said:
I'm a bit stuck on this problem. Specifically, I'm not sure how to deal with the dependent source.
I'm confused, too. You show a voltage arrow across the capacitor and the arrow appears to be labelled ##i_0##. Is that what you see in your textbook?
 
  • #11
cnh1995 said:
Yes, so what is i(t)?
It should be just what I put in my original post: ##i(t) = c\frac{dv}{dt}##
cnh1995 said:
You have arrived at the correct expression for V.
What is i0(t) in terms of charge on the capacitor? What is voltage V in terms charge on the capacitor?

##i(t) = \frac{dQ(t)}{dt}##
##V = \frac{Q}{C}##

NascentOxygen said:
I'm confused, too. You show a voltage arrow across the capacitor and the arrow appears to be labelled ##i_0##. Is that what you see in your textbook?
Yes, this picture is directly from the book. ##i_0## is the current through the capacitor.
 
  • #12
Your equations for i(t) and V(t) in terms of charge are correct.
Now substitute these values in the pink equation in #9 and solve the DE (for Q(t)).
 
  • #13
@Drakkith, I just realized that I wasn't careful with the polarities.
To avoid any sign errors, I would reverse the the direction of Io (since the capacitor is discharging), call it I1 (So I1= -Io) and reverse the dependent source polarity (it will now be 5I1).

Now you can see the dependent source is not really a source but is actually a 5 ohm resistor in disguise.
 
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  • #14
cnh1995 said:
Your equations for i(t) and V(t) in terms of charge are correct.
Now substitute these values in the pink equation in #9 and solve the DE (for Q(t)).

Solving for the DE I get: ##i_0(t) = 50,000Ae^{50,000Q(t)}##
##Q(0) = CV(0) = 2*10^{-6}(15) = 3*10^{-5}##
So ##i_0(0) = 50,000Ae^{1.5}##

To find A, I need ##i_0(0)##.
Doing a KVL loop: ##15-5i_0+15i_0=0##
##i_0 = -1.5A##

Plugging that in:
##-1.5=50,000Ae^{1.5}##
That would make ##A = -6.69*10^{-6}##
and:
##i(t) = -0.335e^{50,000Q(t)}##

Well, that's closer to one of the possible answers. The exponential matches some of them, but the constant doesn't.

To solve my DE, I did:
##V-10i_0 = 0##
##\frac{Q}{C}-10\frac{dQ}{dt} = 0##
##-10\frac{dQ}{dt}+\frac{Q}{C} = 0##
##\frac{dQ}{dt}-\frac{Q}{10C}=0##
##\frac{dQ}{dt}-50,000Q=0##
My integrating factor is: ##exp(\int -50,000 dQ)=e^{-50,000Q}##
Multiplying the equation by the IF:
##e^{-50,000Q}\frac{dQ}{dt} -50,000e^{-50,000Q}Q = 0##
That's just:
##\frac{d({e^{-50,000Q}Q})}{dt} = 0##
Integrating and moving the constant over:
##e^{-50,000Q}Q = A##
##Q = \frac{A}{e^{-50,000Q}}##
##Q = Ae^{50,000Q}##

Since ##i_0 = \frac{dQ}{dt}##
##i_0 = 50,000Ae^{50,000Q}##
 
  • #15
Is the given answer I(t)=0.75e-25000t?
 
  • #16
cnh1995 said:
Is the given answer I(t)=0.75e-25000t?

Apparently it's ##-0.75e^{25,000t}##
 
  • #17
Drakkith said:
Apparently it's ##-0.75e^{25,000t}##
Yes, because I'd reversed Io in #13.

Your equation for i(t) in #14 is dimensionally incorrect (and you don't need that much math either).

Use the steps in #13. The fact that the dependent voltage source is actually just a 5 ohm resistor makes the solution a lot simpler.
 
  • #18
cnh1995 said:
Use the steps in #13.

Do you mean treat the dependent source as a resistor and solve the problem?

cnh1995 said:
The fact that the dependent voltage source is actually just a 5 ohm resistor makes the solution a lot simpler.

Hmm. Am I supposed to be able to recognize that this is a resistor in order to solve this?
 
  • #19
Drakkith said:
Do you mean treat the dependent source as a resistor and solve the problem?
Yes.
Drakkith said:
Hmm. Am I supposed to be able to recognize that this is a resistor in order to solve this?
Even if you don't recognize that, you'll realize it when you write the KVL equation. But since the capacitor is discharging, it would be better if you assumed a current I1 (anticlockwise) such that I1= -Io. Remember that you'll have to reverse the dependent source's polarity as well.

Now you'll have current I1 flowing throgh the dependent source from its positive terminal to negative terminal and this voltage "drop" is 5I1. Doesn't this suggest that the dependent source is acting as a 5 ohm resistor here?
 
  • #20
Alright, treating the dependent source as a resistor and using ##i_1 = -i_0 ##:
##V(t)=15e^{-25000t}##
##i_1(t)=-c\frac{dV}{dt}=-2*10^{-6}(-25000)(15)e^{-25000t}=0.75e^{-25000t}##

cnh1995 said:
Now you'll have current I1 flowing throgh the dependent source from its positive terminal to negative terminal and this voltage "drop" is 5I1. Doesn't this suggest that the dependent source is acting as a 5 ohm resistor here?

I suppose so. I'm not sure I would have recognized that had you not mentioned it. Correct me if I'm wrong, but it looks like this is only true as long as the polarity of the source is such that the current enters the positive terminal, correct? If the source voltage was proportional to ##-5i_0## then we wouldn't be able to treat this as a resistor.
 
  • #21
Drakkith said:
Correct me if I'm wrong, but it looks like this is only true as long as the polarity of the source is such that the current enters the positive terminal, correct? If the source voltage was proportional to −5i0−5i0-5i_0 then we wouldn't be able to treat this as a resistor.
That's right.
 
  • #22
Roger. Thanks all.
 
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1. What is a capacitor and how does it work?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It is made up of two conductive plates separated by an insulating material, known as a dielectric. When a voltage is applied across the plates, an electric field is created between them, causing one plate to accumulate positive charge and the other to accumulate negative charge. This process is known as charging, and the capacitor can store this energy until it is discharged.

2. What is the purpose of a dependent voltage source in a circuit?

A dependent voltage source is a type of voltage source that depends on another voltage or current in a circuit. Its purpose is to provide a voltage or current that is proportional to the voltage or current of another component in the circuit. This allows for more complex and precise control of the circuit, as the output of the dependent voltage source can change based on the behavior of other components.

3. What are the different types of capacitors?

There are several types of capacitors, including ceramic, electrolytic, film, and tantalum capacitors. Ceramic capacitors are small and inexpensive, but have a limited range of capacitance. Electrolytic capacitors have a higher capacitance and are often used in power supply circuits. Film capacitors have a wide range of capacitance and are commonly used in audio and radio frequency circuits. Tantalum capacitors are small and have a high capacitance, making them suitable for use in portable devices.

4. How do you calculate the capacitance of a capacitor?

The capacitance of a capacitor is determined by its physical characteristics, such as the area of the plates, the distance between the plates, and the type of dielectric used. The capacitance can be calculated using the equation C = εA/d, where C is the capacitance in farads, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates.

5. What is the relationship between voltage and capacitance in a capacitor?

The voltage across a capacitor is directly proportional to its capacitance. This means that if the capacitance increases, the voltage across the capacitor will also increase. Similarly, if the voltage decreases, the capacitance will also decrease. This relationship is described by the equation Q = CV, where Q is the charge stored in the capacitor, C is the capacitance, and V is the voltage across the capacitor.

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