Page "Multivibrator" Paragraph 19
from
Wikipedia
Looking at C2, just before Q2 turns on the left terminal of C2 is at the base-emitter voltage of Q1 ( V < sub > BE_Q1 </ sub >) and the right terminal is at V < sub > CC </ sub > (" V < sub > CC </ sub >" is used here instead of "+ V " to ease notation ).
The voltage across C2 is V < sub > CC </ sub > minus V < sub > BE_Q1 </ sub >.
The moment after Q2 turns on, the right terminal of C2 is now at 0 V which drives the left terminal of C2 to 0 V minus ( V < sub > CC </ sub >-V < sub > BE_Q1 </ sub >) or V < sub > BE_Q1 </ sub >-V < sub > CC </ sub >.
From this instant in time, the left terminal of C2 must be charged back up to V < sub > BE_Q1 </ sub >.
How long this takes is half our multivibrator switching time ( the other half comes from C1 ).
In the charging capacitor equation above, substituting:
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1.956 seconds.