In the circuit Fig. the value of beta is 200. Find `I_(B), V_(CE), V_(BE)` and `V_(BC)`, when `I_(C)=2.5mA`. The transistor is in active, cut off or saturation state.

Explanation

`beta=I_C/I_B or I_B=I_C/beta=(2.5mA)/200=0.0125mA`
In the collector-emitter circuit, collector-emitter voltage,
`V_(CE)=V_C-I_CR_(L)`
`=20-(2.5xx10^(-3))xx(5xx10^(3))=7.50V`
In the collector-base circuit, base emitter voltage,
`V_(BE)=V_C-I_BCxxR_B`
`=20-(0.0125xx10^(-3))xx(120xx10^(3))`
`=18.50V`
Now, base collector voltage
`V_(BC)=V_(BE)-V_(CE)=18.50-7.50`
`=11.0V`
Since both the junction are forward biased and voltage are high therefore, the transistor is said to be in saturation state.