# Kirchoff's Voltage Law with Capacitor

1. Dec 6, 2015

### Haven Barnes

1. The problem statement, all variables and given/known data
Here is the problem in question. The only values given are those shown on the circuit labels. We are asked to form a voltage equation using Kirchoff's voltage law with specific loops. We were NOT asked to solve the equations for the currents.

2. Relevant equations
Sum of Voltage in a closed loop = 0.
V = IR
Voltage on an inductor = -L*(di/dt)
V = Q/C

Important equation for my question: I = dQ/dt

3. The attempt at a solution

-2I3 - I3/7 - 2I3 + 5*dI2/dt = 0

^This equation was marked completely wrong because he wanted Q/7 for the voltage on the capacitor in the diagram.

However, we are not given a value for Q, so I was wondering how having this unknown variable in the equation would help us solve for the currents at all?

When I asked, my professor told me that I was 100% wrong because if I tried to solve for the currents, it would not work. (Which confuses me as Q doesn't seem to be helpful either as we would be left with too many unknown variables).

Here's my attempt at arguing why it is right. I would appreciate any advice/direction on if I am way off or am in the right direction.

My idea is that since I = dQ/dt, this forms a direct relationship between the charge on that conductor and the current we would be trying to solve for. Using a differential equation and some other fiddling I feel that I could argue that this is true

Here is an example of the strategy I'm referring to, however this is used on an AC circuit.

2. Dec 6, 2015

### Staff: Mentor

Recall that the charge on a capacitor is equal to the integral of the current over time. So if $I_c$ is the current flowing into the capacitor, then
$$V_c = \frac{1}{C} \int I_c~dt$$

3. Dec 6, 2015

### Haven Barnes

My next thought is that we can assume from the circuit diagram (and no V(t) function given for the Battery), that this is a DC circuit, and that means that the current is constant.

Would it be ok to assume from this that the voltage on the capacitor is then : V_c = I(t) / C ?

4. Dec 6, 2015

### cnh1995

How can you say current is constant? DC circuit implies that the source voltage is constant.
Vc=Q/C always and not I/C.

5. Dec 6, 2015

### Staff: Mentor

At steady state (after a long time) the capacitor voltage will have reached its final value and no further current will flow into it. That will make that whole loop "dead", since no current will flow. Similarly, at steady state the current will stop changing in the inductor and it will end up with zero potential difference across it (it'll look like a short circuit). You can determine the steady state conditions by shorting out any inductors and removing any capacitors, then determine the potentials and currents in what's left of the circuit.

The interesting stuff for this sort of circuit is the period between switching on the power and steady state. This is known as the transient period, and the circuit behavior during that time is called the transient response. Solving the equations to determine the transient response involves solving simultaneous differential equations (your loop equations!).

6. Dec 6, 2015

### Haven Barnes

That is the normal value of Vc, but the main point of this thread is to see if in this case, there exists such a relationship through the I = dQ/dt equation where Vc could actually be I/C

Also, DC doesn't necessarily mean constant voltage only. It could refer to constant voltage and/or constant current inputs

7. Dec 6, 2015

### Haven Barnes

Thanks for your explanations. This has helped me understand RLC circuits and steady sate/transients exponentially greater :)

That said, in this specific case, if you were the grader would you think I had an argument to make for the presence of our current in the voltage value of the capacitor?

8. Dec 6, 2015

### Staff: Mentor

Unfortunately, no Even the units are not consistent, yielding volts/second instead of volts.

9. Dec 6, 2015

### Haven Barnes

Ah very true. Too bad. Thanks again for your help, it'll definitely be good for my final coming next week.