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Yes. If voltage leads the current, the impedance is inductive (this would be the case if the load is a motor). If current leads the voltage, the impedance is capacitive (this would be the case for a CFL light bulb).
Yes, voltage matters when charging a capacitor. Capacitor charge rate is proportional to current and inversely proportional to capacitance. dv/dt = i/c So, voltage matters in terms of charge rate, if you are simply using a resistor to limit the current flow, because a larger voltage will attempt to charge faster, and sometimes there is a limit on the current through a capacitor. There is also a limit on voltage across a capacitor, so a larger voltage could potentially damage the capacitor.
The voltage is the same across parallel inductors for the same reason its the same across parallel anythings - resistors - capacitors - transformers - whatevers. Parallel means sharing common connections to another part of the circuit. That sharing involves conductors that have essentially zero impedance. (Zero for all practical considerations) In order for there to be a different voltage across those parallel components, the conductors connecting them together must have something other than zero impedance, and then we would not be talking about parallel anymore. (Ohm's Law: Voltage is Current times Resistance. If the resistance of the conductor is zero, then the voltage across the conductor must also be zero - it does not matter what the current is.)
bucking voltage is a voltage which is of opposite polarity to the voltage it acts .
DC Analysis: For this analysis, frequency is made zero and the voltage of the source is increased in small steps from 0V.And the output voltage is plotted. So, finally we get a Vout vs Vin curve. AC Analysis: In this analysis, we choose an AC source. We keep the Offset voltage = 0V, AC voltage or small signal voltage = 2V (You can take any voltage you wish and it doesnt matter). So, to plot the ac response or frequency response of the circuit, increase the frequency in steps and note the output voltage. from this analysis , we can find the gain of the circuit over frequency.