Looking at C2, just before Q2 turns on, the left terminal of C2 is at the base-emitter voltage of Q1 (VBE_Q1) and the right terminal is at "V"CC (""V"CC" is used here instead of "+"V"" to ease notation). The voltage across C2 is "V"CC minus "V"BE_Q1 . 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"CC - "V"BE_Q1) or "V"BE_Q1 - "V"CC. From this instant in time, the left terminal of C2 must be charged back up to VBE_Q1. How long this takes is half our multivibrator switching time (the other half comes from C1). In the charging capacitor equation above, substituting: