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While I was studying the use of Transistor as an Oscillator from my Fundamentals Of Physics By Halliday, Resnick and Walker , I faced several confusions. My book says that when the switch S1 is closed, the collector current starts flowing , its value slowly increases, as shown by the segment XY and thus emitter current is induced in the input circuit.

My question is why the collector current increases slowly and why Emitter current increases with increase in collector current? Also what is the significance of the cell between base and emitter when it is receiving voltage through mutual inductance? Please help me in the stand.

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The cell between the base and emitter is responsible for biasing the transistor on when the switch is open. If this cell was not present, then the collector current would not flow at all when the switch is closed. The voltage coupled through the mutual inductance will be superimposed on top of the DC bias current.

Now, as for why the current increases slowly. The inductor and capacitor in parallel form what is called a resonant tank circuit. This circuit has an impedance that varies with frequency. In this case, the impedance well be lowest at the resonant frequency. The variable capacitor can be adjusted to tune the resonant frequency as desired. Since the impedance is frequency dependent, when you apply a step function by closing the switch, the tank circuit will 'ring' at is resonant frequency. This ringing will decay unless there is something to keep it going - in this case, the transistor.

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    \$\begingroup\$ It's a parallel tuned circuit so its impedance will be highest at resonance. \$\endgroup\$ – Brian Drummond Feb 28 '15 at 11:11

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