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In the circuit, the value of beta is 100...

In the circuit, the value of `beta` is 100. find `I_B,V_(CE),V_(BE)` and `V_(BC)`, when `I_C`=1.5mA. The transistor is in active, cut off or saturation state?

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`beta=100,I_C=1.5mA=1.5xx10^(-3)A,V_CC=24V`
`beta=(I_C)/(I_B)`
`I_B=(1.5xx10^(-3))/(100)=15muA`
To calculate `V_CE` we apply kirchhoff.s rule CEFDC, therefore
`V_CC=I_Cxx4.7kOmega+V_CE`
`24=1.5xx10^(-3)xx4.7xx10^(3)+V_(CE)`
`V_CE=24-7.05=16.95V`

Again applying Kirchhoff.s rule to loop ABEFDCA, we get,
`V_CC=I_Bxx220kOmega+V_BE`
`V_BE=24-3.3`
`V_BE=20.7V`
Going along loop ABCA,we get
`I_Bxx220kOmega+V_BC=I_Cxx4.7kOmega`
`15xx10^(-6)xx220xx10^(3)+V_BC=1.5xx10^(-3)xx4.7xx10^(3)`
`V_BC=7.05-3.3=3.75V`
As `V_CEltV_BE`, both the juction are forward biased.So, the transistor is in the saturation state.
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