# Time-function,one reactive

Discussion in 'Homework Help' started by Alice10, Jan 6, 2015.

Oct 3, 2014
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2. ### WBahn Moderator

Mar 31, 2012
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Which side of Rs is the more positive?

3. ### Alice10 Thread Starter New Member

Oct 3, 2014
2
0
I would say the "south" side on Rs would be positive because the current goes in to the positive side of a resistor?

But then the next question gets wrong, when I want to replace everything except the capacitor with a "two terminal equivalent".
(Doing KVL at t=0^+ , starting at bottom through U,Rs,Rc(no current so=0),Uab
Since then my KVL around is Uab=-U-(-IRs)

Uab=UT

4. ### WBahn Moderator

Mar 31, 2012
20,059
5,655
Okay, so now just track the math.

Let's call the bottom node 'g', the node at the top of the capacitor 'b' and the top node 'a'.

As you noted, the voltage on the "south side" of the resistor (node 'g') is higher than the voltage on the north side (node 'a') because the current is entering from the bottom. So

Vga = Vg - Va = I·Rs = the voltage across the resistor.

The voltage across the capacitor, Vc, is defined such that it is equal to

Vc = Vb - Vg

Because there is no current flowing in Rc, the voltage across it is 0V, hence

Va - Vb = 0; hence Va = Vb

Combining these together, we have:

Vc = Vb - Vg = Va - Vg = -(Vg - Va) = -(I·Rs) = -I·Rs

For the next part, please show your work -- I'm not sure I'm following your verbal description well enough.