Kirchoff's rules and capacitors

[BOAS]

Hello,

I thought I had questions concerning Kirchoff's rules down to a tee, but now capacitors have entered the scene and I'm stumped.

Homework Statement

The circuit in the drawing (attached) shows two resistors, a capacitor, and a battery. When the capacitor is fully charged, what is the magnitude q of the charge on one of its plates?

Homework Equations

The Attempt at a Solution

I apologise for the terrible resolution of the image, but I think it's still readable.

From the junction rule;

I_A = I_B + I_C

For loop A;

12V = 2I_A + 4I_C

My textbook says a loop does not need to contain a battery, so my expression for loop B is;

4I_c = Q/C

I am confused about whether the capacitor represents a voltage drop, because to me, it seems like a fully charged capacitor should create a potential rise.

I am unsure about whether I need to make an expression for the loop created by the 'outer' sides of the circuit.

Please can you help me in this regard?

Thanks.

 

[willem2]

When the capacitor is fully charged, no more current will go through it, so you can ignore the loop B.

Even if the capacitor isn't fully charged, you can treat the resistor and the capacitor as one circuit element, with I = V/R + C (dV/dt), and you can still ignore loop B.

 

[NascentOxygen]

When the capacitor is fully charged, you can imagine removing it from the circuit and replacing it with a voltmeter, an ideal voltmeter. Whatever reading the voltmeter gives is the same voltage as on the fully charged capacitor that is/was there.

* remember, an ideal voltmeter has infinite resistance and it draws no current

 

[CWatters]

Are you familiar with potential divider circuits? Work out the voltage at the junction of R1 and R2 without the capacitor present. What happens if the voltage on the capacitor reaches this voltage?

Once you figure that out you can work out the charge on the capacitor.

 

[HalfLostAndSearching]

Dear student,

Thank you for your question. Kirchoff's rules are fundamental principles in circuit analysis that allow us to solve complex circuits and understand the behavior of electrical components. In this case, we are dealing with a circuit that contains both resistors and a capacitor.

Firstly, let's clarify the role of the capacitor in this circuit. A capacitor is an electrical component that can store and release electrical energy in the form of electric charge. In this circuit, the capacitor is initially uncharged and acts as an open circuit. As the capacitor charges, it acts as a closed circuit, allowing current to flow through it. Therefore, the capacitor does not represent a voltage drop, but rather a voltage rise as it charges.

Now, let's apply Kirchoff's rules to solve for the charge on the capacitor's plates. We can use the junction rule to find the current flowing through the capacitor, as you have correctly done in your attempt at a solution. However, we also need to consider the loop created by the outer sides of the circuit. This loop includes the battery, both resistors, and the capacitor. Applying the loop rule to this loop, we get:

12V = 2IA + 4IC + QC

where QC is the voltage across the capacitor, which is equal to the charge on its plates divided by its capacitance (Q/C).

Combining this with the junction rule, we get:

12V = 2IA + 4IC + 4Ic

Solving for IC, we get:

IC = 3IA + 2Ic

Substituting this into our original junction rule equation, we get:

IA = IB + 3IA + 2Ic

Simplifying this, we get:

IB = -2IA - 2Ic

Now, we can substitute this into our loop rule equation to eliminate IB, and we get:

12V = 2IA + 4IC + QC

12V = 2IA + 4IC + 4Ic

Substituting in our expression for IB, we get:

12V = 2IA + 4IC + 4Ic

12V = 2IA + 12Ic

Solving for IA, we get:

IA = 2Ic

Now, we can substitute this back into our original junction rule equation to solve for IC:

2Ic = IB + 3(