Calculating the Charge on Capacitor

[AnonBae]

Homework Statement

The circuit has been completed for several minutes. Suppose that ε = 55 V .

Calculate the magnitude of charge on the 100 μF capacitor plate.

Homework Equations

Q = C * V

The Attempt at a Solution

Since the capacitors have been fully charged, current has stopped flowing at the 300 Ω and 100 Ω resistors. But for the other resistors, current still flows by the junction and loop rules. So, Q = 100 μF * 55 V = .0055 C.
But that is wrong, and I do not understand. If the capacitor is already fully charged, does that not mean it equals the voltage of the battery? Why am I not getting the right answer?

 

[gneill]

There will still be current flowing through any paths that do not have capacitors... Where will the current be flowing in your circuit?

 

[AnonBae]

The current will be flowing through the branches that do not have the capacitors, or the 525, 100, and 1500 resistors, which I did say in the original post. How will this help though in determining the charge of the capacitor if it's already fully charged? Like Q = C * V, but that does not give the right answer, and I still do not know what is the problem then.

 

[gneill]

What is the voltage across the branch containing the 100 μF capacitor? If you removed the capacitor, what voltage would you measure at the open terminals where it was connected?

 

[AnonBae]

The voltage across the branch containing the 100 μF capacitor should be 55 V, stored in the capacitor.
I am not quite understanding what you mean by measuring the voltage at the open terminals. At the open terminals, the voltage would drop depending on which resistor comes before it?

 

[gneill]

The voltage across the branch containing the 100 μF capacitor should be 55 V, stored in the capacitor.

No, the capacitor voltage never reaches 55 V. There are potential drops in the circuit before (and after) where the capacitor connects to the circuit.

I am not quite understanding what you mean by measuring the voltage at the open terminals. At the open terminals, the voltage would drop depending on which resistor comes before it?

Let's try this. You've identified a path where the current is still flowing. Mark in the potential drops across each of the resistors around that path. What do they add up to? What's the drop across the resistor that parallels the capacitor branch?

Edit: Here's a diagram to show what I mean:

What are v1, v2, and v3?

 

[AnonBae]

Oh. Okay
For the loop specified, Each potential drop will sum to the emf or the 55 V battery.
So, since 55 V = I*525 + I*1500 + I*100, since current in the same everywhere in a series loop. Thus, current would be after simplification around 0.026 from sig. figs. So V1 would be 13.65, V2 = 39, and V3 = 2.6.

Then would that mean since voltage is the same for parallel branches, the voltage after the voltage drops for V1 and V2 would be the voltage on the branch containing the 100 micro-farad capacitor?

I am still confused on how the voltage drops would help me find the voltage the capacitor contains.

 

[AnonBae]

I figured it out! The voltage of the 1500 ohm resistor is the same amount that charged up the 100 microfarad capacitor. Thank you so much for the help

 

[gneill]

I figured it out! The voltage of the 1500 ohm resistor is the same amount that charged up the 100 microfarad capacitor. Thank you so much for the help

Well done!