Electronic – Capacitor RC Charging

Your nodal analysis is incorrect. For steady-state resistive analysis, we can remove all of the capacitors from the circuit. This quickly shows us that the green node is at 4V with respect to the negative terminal of the 12V supply. This bold part is important.

Now, on the left hand side, we effectively have a 15V supply into three series 5k resistors - the 2k and the 3k combine in series and the capacitor has no effect. This puts the red node at 10V with respect to the negative terminal of the 15V supply. Notice that our reference point for the potential is different for the red and green node - so the voltages can not be used in the same equation just yet.

For that, we have to pick a common reference point (which if I were doing this analysis I would mark using a signal ground symbol to help me think). For convenience, let's make the negative terminal of the 12V supply the reference point - which means the green node voltage needs no transformation.

The red node does, however. We know that 5V is dropped across each of the 5k resistors from Kirchoff's laws, so across the 5k resistor between the red node and our newly chosen reference is 5V. So now we can see that with 5V wrt our reference at the red node and 4V wrt the same reference at the green node, we have 5V - 4V = 1V across the 3uF capacitor.

The trick can be to notice which Kirchhoffs-law to use. I suggest you to try with Kirchhoff's Loop Rule instead of Kirchhoffs current law.

There are two simple loops in which if you calculate each voltages, you will get two equations from which the transmission characteristic (and matrix) can be determined.

^{simulate this circuit – Schematic created using CircuitLab}

The transmission characteristic looks like:

$$ V_1 = A \times V_2 + B \times I_2 $$ $$ I_1 = C \times V_2 + D \times I_2 $$

The two equations for the two loops:

$$ V_1 = j\omega L \times I_1 + \frac{1}{j\omega C} \times (I_1 + I_2) $$ $$ V_2 = \frac{1}{j\omega C} \times (I_1 + I_2) $$

Your task is to order them to get A,B,C,D. Though, sometimes it can be a little difficult/messy but not in this case.

Furthermore, there is an other method to calculate the transmission matrix of a two-port. There two special case; when \$ V_2 = 0 \$ and when \$ I_2 = 0 \$ , and in these cases siplified equations can be used to determine A,B,C,D:

These two documents describe this method:

• First, only about transmission characteristic.
• Second, generally about two-ports, transmission characteristic starts on page 19.

So all in all, you should try to answer the question using one of the methods and the check your results with the other. This way you can practice both of them while making sure that your homework will be correct.