Kirchhoff loop rule and Ohm's Law

In summary: C.In summary, the homework statement is to understand the behavior of the current and voltage in a simple R-C circuit. A capacitor with capacitance C is initially charged with charge q0. At time t=0 a resistor with resistance R is connected across the capacitor. PART A is to use the Kirchhoff loop rule and Ohm's law to express the voltage across the capacitor V(t) in terms of the current I(t) flowing through the circuit. PART B is to use the relation V(t) = I(t)R to find the voltage and current in the circuit as functions of time. To do so, the mathematician uses the fact that current can be expressed in terms of the voltage.
  • #1
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Homework Statement



To understand the behavior of the current and voltage in a simple R-C circuit

A capacitor with capacitance C is initially charged with charge q0. At time t=0 a resistor with resistance R is connected across the capacitor

PART A
Use the Kirchhoff loop rule and Ohm's law to express the voltage across the capacitor V(t) in terms of the current I(t) flowing through the circuit

PART B

We would like to use the relation V(t) = I(t)R to find the voltage and current in the circuit as functions of time. To do so, we use the fact that current can be expressed in terms of the voltage. This will produce a differential equation relating the voltage V(t) to its derivative. Rewrite the right-hand side of this relation, replacing I(t) with an expression involving the time derivative of the voltage.

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Homework Equations





The Attempt at a Solution



For PART A I got the answer to be V(t) = I(t)R, but I'm having problems with the PART B. I'm not sure what the "time derivative of the voltage" is. Any suggestions?
 
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  • #2
Remember that C=Q/V, and dQ/dt=I(t)

I need to stop here if the problem was to actually FIND the differential equation, because any more and I'd pretty much finish it >_> Take a stab keeping in mind what I typed
 
  • #3
The only thing that comes to mind is:

[tex]V(t) = \frac{dQ}{dt}R[/tex]

but I don't think that's the right answer.
 
  • #4
Ok more prodding
Q=C*V, take the derivative of both sides, dQ/dt=C*dV/dt (because C is constant)

oh look an expression for dQ/dt that involves the time derivative of V...
 
  • #5
Power wouldn't happen to be the same as dV(t)/dt?
 
  • #6
No

All part B wants is V(t)=some stuff involving dV/dt

right now you have V(t)=some stuff involving I(t)

You have seen how dQ/dt = somet stuff involving dV/dt

dQ/dt = ? *pokes you with cattle prod*
 
  • #7
dQ/dt=C*dV/dt

therefore dV/dt = (dQ/dt)/C

I(t) = dQ/dt = V(t)/R

dV/dt = (V(t)/R)/C

C(dV/dt) = V(t)/R

V(t) = RC(dV/dt)

If this is it, I'm going to kick myself...
 
  • #8
That's it

You could've also saved some steps and algebra and just gone from I=C*dV/dt, and plugged that into V=IR and gotten V=RC*dV/dt

Someday soon you might do a similar problem with LC circuits, and even take it a step further to RLC circuits, just don't forget the basic procedure you used here, just different things will be derivatives and different things will be constants
 
  • #9
For some reason the correct answer is V(t) = -RC(dV/dt)
 
  • #10
Yah that negative slips in there somewhere because the capacitor is discharging

So charge is flowing out of the capacitor so...-dQ/dt=C*dV/dt?
 
  • #11
So another question unlocked itself after you finish this one:

Now solve the differential equation V(t) = -RC(dV/dt) for the initial conditions given in the problem introduction to find the voltage as a function of time for any time t.

Express your answer in terms of q, R, C, and t.

so I know:

V(t) = -RC(dV/dt) and dv/dt = (dQ/dt)/C = I(t)/C

V(t) = -RC(I(t)/C)
V(t) = -R(I(t)) and I know Q = I(t)*t => I(t) = q/t
V(t) = -R(q/t), but I can't seem to get the C in there.

Any suggestions?
 
  • #12
I'm not sure where you were going starting with "so I know", it's separation of variables

So you have V=-RC*dV/dt, separate them to get dV/V=-RC*dt

Remember RC is a constant(it's got a special name that I forget too)integrate both sides

You'll have a natural log on the left, and just -RCt + C on the right, use your initial conditions to find C, then exponentiate both sides(raise both sides as powers of e)remembering that e^(ln(anything))=anything
 
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  • #13
Bit rusty on my integration (havent don't it in about a year or so)

so the integral of v/dv would be ln|v+C|.

and the "+C" is just a constant.
 
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  • #14
You'd usually just stick the constant on the other side

I messed up rearranging the equation fyi, it's dV/V=-dt/RC, but 1/RC is still a constant so it's not problem. The absolute value on the natural log is to avoid a domain error, but you usually just ignore it since you're going to get rid of the ln anyways, so you end up with ln(V)=-t/RC+C, and it's not "just" a constant, you have initial conditions you have to utilize to determine it, basically at t=0 V=such and such, so you can solve for C
 
  • #15
At time t = 0, V would have to be 0 since voltage is a measure of the pressure under which electricity flows, and at time t = 0, there isn't any flow of electricity. There is also no current at time t = 0, so I = 0 as well.

so C = q

but that's also wrong. *sigh*
 
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  • #16
There's still a V at t=0

if there wasn't then electricity would never flow
 
  • #17
so if there's voltage and charge, then C = VQ

I apologize if it's wrong, but I'm next to useless when it comes to Circuits
 
  • #18
Right, at t=0 the V=C/Qo

that's the initial voltage across the capacitor and hence the circuit
 
  • #19
I submitted e^((-t/RC)+V(t)Q), but it says the answer does not depend on the variable v

I think it's e^((-t/RC)+tQ) instead
 
  • #20
You didn't find c right(lower case c=the constant from integration, upper case is capacitance)

You plug in t=0, then plug in C/Qo for V, and solve for c
 
  • #21
Ah... that makes more sense

V = e^((-t/RC)+x) ------- x = C to avoid confusion on my behalf
C/q = e^x ------- t = 0, V = C/q
ln(c/q) = lne^x
ln (c/q) = x lne
x = ln(c/q)
 
  • #22
That looks right, then in the process of simplifying, remember that e^(x+y)=(e^x)*(e^y) so you can get the e^(ln(x)) by itself

Sadly I've got this problem worked out completely in an electronics textbook roughly 400 miles away
 
  • #23
so I inputted the following into the assigment

e^(-(t/RC) + ln(C/Q)) and it says "Your answer does not have dimensions of voltage"
 
  • #24
haha, Q=CV, not C=VQ. I'm tooootally blaming that one on you!

That C/Q is going to end up in front(that was my simplifying hint, be careful, I'm not sure that negative is in the right place)except it's not going to be C/Q, it's going to be Q/C, charge/capacitance is voltage.

in the exponent, resistance times capacitance is time, so the units cancel so the units are defined by that Q/C
 
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  • #25
I would love to say I'm surprised at this mistake, but I just can't

so it's going to be e^(-(t/RC) + ln(Q/C))

once simplified:

e^(-(t/RC)) * e^(ln(Q/C))
 
  • #26
In order for the current in a conductor to exist continuously, the conductor must be part of a loop, that is, a closed path through which the charged particles can move without creating a "build-up." Such build-up, if it occurs, creates its own electric field that cancels out the external electric field, ultimately causing the current to stop.

However, having a loop, or a closed circuit, is not enough to maintain the current; there must also be a source of energy. Its necessity is fairly obvious: As charged particles move along the circuit, they lose potential energy. In fact, electrostatic forces always push the particles in the direction that leads to a decrease in potential energy. At some point, each charged particle would reach the location in the circuit where it has the lowest possible potential energy. How can such a particle move toward a point where it would have a higher potential energy?

Such a move requires that nonelectrostatic forces act upon the charged particle, pushing it toward higher potential energy despite the presence of electrostatic forces. In circuits, such forces exist inside a device commonly known as a battery. In a circuit, the battery serves as the energy source that keeps the charged particles in continuous motion by increasing their potential energy through the action of some kind of nonelectrostatic force.

The amount of work that the battery does on each coulomb of charge that it "pushes through" is called (inappropriately) the electromotive force (pronounced "ee-em-ef" and abbreviated emf or denoted by EMF). Batteries are often referred to as sources of emf (rather than sources of energy, even though they are, fundamentally, sources of energy). The emf of a battery can be calculated using the definition mentioned above: {\cal{E}}={W}/{q}. The units of emf are joules per coulomb, that is, volts.

The terminals of a battery are often labeled + and - for "higher potential" and "lower potential," respectively. The potential difference between the terminals is called the terminal voltage of the battery. If no current is running through a battery, the terminal voltage is equal to the emf of the battery: V_{\rm ab}={\cal{E}}.

However, if there is a current in the circuit, the terminal voltage is less than the emf because the battery has its own internal resistance (usually labeled r). When charge q passes through the battery, the battery does the amount of work EMF q on the charge; however, the charge also "loses" the amount of energy equal to Ir ( I is the current through the circuit); therefore, the increase in potential energy is {\cal{E}}q-qIr, and the terminal voltage is

V_{\rm ab}={\cal{E}}-Ir.

In order to answer the questions that follow, you should first review the meaning of the symbols describing various elements of the circuit, including the ammeter and the voltmeter; you should also know the way the ammeter and the voltmeter must be connected to the rest of the circuit in order to function properly.

Note that the internal resistance is usually indicated as a separate resistor drawn next to the "battery" symbol. It is important to keep in mind that this resistor with resistance r is actually inside the battery.

In all diagrams, EMF stands for emf, r for the internal resistance of the battery, and R for the resistance of the external circuit. As usual, we'll assume that the connecting wires have negligible resistance. We will also assume that both the ammeter and the voltmeter are ideal: That is, the ammeter has negligible resistance, and the voltmeter has a very large resistance.1 For the circuit shown in the diagram View Figure , which potential difference corresponds to the terminal voltage of the battery?

between points K and L
between points L and M
between points K and M

2 n which diagram(s) (labeled A - D) does the ammeter correctly measure the current through the battery?
Enter the letter(s) of the correct diagram(s) in alphabetical order. For example if both A and C are correct enter AC.
 
  • #27
can you guys please help me with dis question i dunt now how to put the pictures here :(
 
  • #28
RC is called time constant...

ravi, what is your understanding? are you here to look for answers ONLY??
 
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Related to Kirchhoff loop rule and Ohm's Law

1. What is Kirchhoff's loop rule and how does it relate to Ohm's Law?

Kirchhoff's loop rule, also known as Kirchhoff's voltage law, states that the sum of the voltage drops in a closed loop in a circuit must be equal to the sum of the voltage sources in that loop. This rule is based on the principle of conservation of energy. Ohm's Law, on the other hand, states that the current through a conductor is directly proportional to the voltage across it and inversely proportional to its resistance. These two laws are related in that they both help us understand and analyze the behavior of electrical circuits.

2. How do I apply Kirchhoff's loop rule to a circuit?

To apply Kirchhoff's loop rule to a circuit, you must first identify all the loops in the circuit. Then, you can start at any point in the loop and follow the path, taking into account the direction of current flow and the voltage drops across components. As you move around the loop, you will add up all the voltage drops and equate them to the sum of the voltage sources in that loop. This will give you a mathematical equation that you can use to solve for the unknown values in the circuit.

3. Can Kirchhoff's loop rule be violated?

No, Kirchhoff's loop rule is a fundamental law in circuit analysis and cannot be violated. It is based on the principle of energy conservation and has been proven to hold true in all electrical circuits. If the rule appears to be violated in a particular circuit, it is likely due to an error in measurement or analysis.

4. How does Ohm's Law help us understand the behavior of resistors in a circuit?

Ohm's Law is essential in understanding the behavior of resistors in a circuit because it tells us that the current through a resistor is directly proportional to the voltage across it. This means that as the voltage increases, so does the current, and vice versa. It also tells us that the resistance of a resistor is inversely proportional to the current, meaning that as the current increases, the resistance decreases. This helps us predict how different resistors will affect the flow of current in a circuit.

5. How are Kirchhoff's loop rule and Ohm's Law used in practical applications?

Kirchhoff's loop rule and Ohm's Law are used extensively in the design and analysis of electrical circuits, from simple household circuits to complex electronic devices. They allow engineers and scientists to understand and predict the behavior of circuits and design them to meet specific requirements. They are also used in troubleshooting and diagnosing problems in circuits, as they help identify faulty components or connections.

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