- #1
Eduardo Leon
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Homework Statement
Hi mates, I have problems solving the third part of this exercise, I've already done all the previous calculations.
Given the following circuit, where the switch S is open, the power supply = 50 volts and:
- The initial charge in the C capacitor: QC = 0 coulombs
- The initial charge in the 2C capacitor: Q2C = 20^-6 coulombs
- Switch⇒I find the current passing by the amperemeter (IA) and the voltage across the voltmeter VV After a long time t (t>5 time constants). (Done).
- Switch⇒II find the voltage across the voltmeter VV passed 10 milliseconds. (Done).
- Switch⇒III next get the final reading of the voltmeter VV at a given time t (t >0 ∧ t< 5 time constants) and the energy dissipated in the resistors till that instant.
Homework Equations
- Ohm's law: R=ΔV/i
- Capacitance: C=q/ΔV
- Kirchoff's laws: Σi=0 ∧ ΣΔV=0
- Energy gived by the battery until an instant t
- Energy dissipated in the resistor until an instant t
- Energy stored in the capacitor until an instant t
The Attempt at a Solution
As stated before, I've done the first and the second parts.
For the first:
Well, here is where i get lost, because I dont't know how to study and analize the circuit formed at this point, I've only seen discharging RC circuits whis a single current, one resistor and one capacitor due those elements could be joined into their equivalent ones, but here, I get 2 resistors, 2 capacitors and 2 meshes, like this one.
I know that the initial charges at this point are:
In order to get the initial charge in the C capacitor I used the voltage obtained in the voltmeter in the second part at the 10 milliseconds, because are connected parallel. So, multiplying that voltage times the capacitance, we get
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