not a homework problem ,seen this somewhere and wanted to understand itWhat's the context of the question? Otherwise, "It works by applying 9 V between the top and bottom nodes," would be a valid response.
Is this a homework problem?
So I would recommend taking a look at the link that crutschow posted, if you haven't already, and see if that satisfies you. If not, then by all means come back and try to post a question or three that get at what part of the operation is vexing you. That should provide much better targeted help to address your confusion.not a homework problem ,seen this somewhere and wanted to understand it
Assuming Q-left comes on first the collector moves toward +9 , discharging C-left, holding Q-right off. Eventually the capacitor drains and the resistor R-right pulls the base of Q-right to ground, turning it on. This discharges C-right, turning Q-left off. Eventually C-left discharges and resistor R-left turns Q-left back on which turns Q-right off ... all the magic is in the timing of the capacitors and the base resistors.
Emulate? Only ten parts. Build it.To me, this circuit is a perfect candidate for simulation in LTspice that the OP could use as a playground, so that he could see for himself what would happen if he were to change or tweak the values of the components, and learn something along the process.
That tooBuild it.
Easy for you to say. You have a scope.Emulate? Only ten parts. Build it.