capacitordualinductor
simulate this circuit – Schematic created using CircuitLab
The book I'm reading is noting that I should recognize the initial voltage across the \$C_1\$ capacitor is \$10 V\$ but I can't see why this is. I see that by setting up the mesh equations for \$i_2\$ I would get:
\$-4\frac{di_1}{dt}\ + 4 \frac{di_2}{d_t} + \frac{1}{8} \int^t_{t_0}i_2dt' + 5i_2 + v_{C_1}(t_0) = 0\$
Could someone explain to my how \$10V\$ is derived?
It appears that with a square wave in, the output (OUT1, below) is:
$$V_o = \frac{C_2}{C_1 +C_2} (V_n)$$
The spikes From OUT2 do have a decay time, and if you want to find out what it is, the LTspice circuit list follows:
Version 4
SHEET 1 880 680
WIRE 144 -192 64 -192
WIRE 272 -192 208 -192
WIRE 352 -192 272 -192
WIRE 272 -160 272 -192
WIRE 272 -64 272 -96
WIRE 64 80 64 -192
WIRE 144 80 64 80
WIRE 272 80 208 80
WIRE 384 80 272 80
WIRE 464 80 384 80
WIRE 384 96 384 80
WIRE 64 112 64 80
WIRE 272 112 272 80
WIRE 64 208 64 192
WIRE 272 208 272 176
WIRE 272 208 64 208
WIRE 384 208 384 176
WIRE 384 208 272 208
WIRE 64 256 64 208
FLAG 64 256 0
FLAG 272 -64 0
FLAG 352 -192 OUT1
FLAG 464 80 OUT2
SYMBOL cap 256 112 R0
SYMATTR InstName C1
SYMATTR Value 1µ
SYMBOL cap 208 64 R90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName C2
SYMATTR Value 1µ
SYMBOL voltage 64 96 R0
WINDOW 0 36 94 Left 2
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value PULSE(0 10 .1 1u 1u .1 .2)
SYMBOL res 368 80 R0
SYMATTR InstName R1
SYMATTR Value 10
SYMBOL cap 256 -160 R0
SYMATTR InstName C3
SYMATTR Value 1µ
SYMBOL cap 208 -208 R90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName C4
SYMATTR Value 1µ
TEXT 30 280 Left 2 !.tran 1
Analysis with impedances assumes a sinusoidal steady-state operation. Since the voltages in your circuit will not change direction unless you switch the switches (thus changing the equivalent circuit), this analysis is not appropriate.
You'll need to analyze the transient behavior of your circuit, not the sinusoidal steady state behavior. You can do this with mesh analysis, nodal analysis, or an ad hoc combination of the two.
Best Answer
There isn't a "why". To start analyzing the circuit at any point in time, anything with "stored" state has to be known. In this case these are the current thru the inductor, the voltage across the capacitor, and the phase of the sine voltage source.
10 V is something they simply picked. It's a starting state they want you to analyze the circuit from.