If the initial current in the inductor is \$i(0^+)= 0 A \$ why is it that when the switch occurs the initial voltage in the inductor is \$ V_L(0^+)=121.4 V\$
Electrical – How does the inductor have an initial condition in this circuit
capacitorcircuit analysisinductorswitching
Related Solutions
The way you drew it, I and \$\frac{dI}{dt}\$ are both positive in the direction of the arrow and the correct expression is
\$V_L = L\frac{dI}{dt}\$
The negative sign depends on the direction you define your voltage and current.
simulate this circuit – Schematic created using CircuitLab
Found the problem. Need to have "UIC" in the "tran" line.
.tran 50us 30ms 0s 500us UIC
works fine.
Guess I need to spend more time with the spice user manual.
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Best Answer
The circuit you have is this:
simulate this circuit – Schematic created using CircuitLab
I have removed the component values because they are not important for this question.
ASSUMING the switch as been in position #1 to full charge the capacitor, the voltage at the capacitor is 20V
The AC voltage source follows \$ 100 \sqrt(2) Cos (100 t) \$, ie at t=0 has an instantaneous value of: 141.421V
The moment you throw the switch to connect the AC source to the RLC network you will have (141.421 - 20) 121.421 volts across the R-L, but how much across the inductor? Remember that for DC an inductor is a short circuit and for infinite frequency the inductor is open-circuit. Likewise remember on of the fundamental equations of an inductor: \$V = L\frac{d I }{d t}\$ as a result the inductor appears as an open-circuit, initially and thus all 121.421V appears across the inductor and the inductor's current is 0