When CB is opened on 0 current, the Capacitor is charged to +V volts. When voltage source reaches its negative peak, the Voltage across CB is 2V & if then arc restrike, then what are the consequences of it. How the voltage across CB and Capacitor changes with time?
When you open the circuit to a capacitive load, the instantaneous voltage across the switch is simply the difference between the driving voltage and the capacitor voltage at the instant of opening. This is benign. Closing the switch is a hazard if it is done at an instant where the voltage difference is not zero because the capacitor will draw a large current spike to charge to the driving voltage.
Inductive loads are benign when you close the switch (other than core saturation hazards), but a problem when the current is interrupted by opening the switch. That causes the inductor voltage to zoom to extreme levels as the inductor attempts to maintain current constant.
Capacitive load = over-current hazard on closing switch, inductive load = over-voltage hazard on opening switch; the problems are duals of each other. For real components the peak capacitor current will be limited by short circuit impedance of the driving source, and peak inductor voltage will be limited by shunt conductance such as stray capacitance in the winding, or by insulation breakdown worst case.
The voltage across CB when capacitive current is interrupted grows like 1V, 3V, 5V Per Unit and so on indefinitely if the arc is not extinguished. But I have no clue how that happens. May i know any good books for switchgears wherein these concepts are illustrated well?
The voltage cannot grow indefinitely: it cannot exceed the source. The arc will self extinguish at some point. For AC, the voltage will cycle through zero difference twice per cycle. For DC the capacitor voltage should be the same as the source at disconnect, therefore no arc. Are you speaking of capacitive current in the sense of phase angle rather than a physical capacitor? Or are you actually thinking of the inductive case? With capacitors you deal with current peaks, with inductors you have voltage peaks.
Voltage produced in capecitive circuit is praposnal to C(dv/dt),where C the capecitance of circuit and is praposnal to dq/dv rate of change of charge with respect to voltage,hence voltage produce is praposnal to dq/dt ,the rate of variation of the charge with respect to time depends on capecitance of circuit and is considerable for high voltage transmission used today.
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