In the circuit shown in figure, when the input voltage of the base resistance is 10V, `V_(BE)` is zero and `V_(CE)` is also zero. Find the values of `I_(B), I_(C ) and beta`.

Explanation

Here, `I_(C )=1mA`
`beta=(I_(C ))/(I_(B)) rArr I_(B)=(I_(C ))/(beta)=(1mA)/(100)=0.01mA`
`I_(E )=I_(B)+I_(C )=0.01+1=1.01mA`
In output circuit,
`V_(C C)-I_(C )R_(C )-V_(CE)-I_(E )R_(E )=0`
`therefore 12-(1xx10^(-3)xx7.8xx10^(3))-3-(1.01xx10^(-3)R_(E ))=0`
`therefore 1.2=0.00101R_(E )`
`therefore R_(E )=(1.2)/(0.00101)=1188.12Omega`
In input circuit,
`V_(B)-V_(BE)-I_(E )R_(E )+(20xx10^(3))I=V_(B)`
`therefore (20xx10^(3))I=V_(BE)+I_(E )R_(E )`
`therefore (20xx10^(3))I=0.5+(1.01xx10^(-3))(1188.12)`
`therefore (20xx10^(3))I =1.7`
`therefore I = 8.5xx10^(-5)A`
Also,
`V_(C C)-(I+I_(B))R_(B)-V_(BE)-I_(E )R_(E )=0`
`therefore (I+I_(B))R_(B)=V_(C C)-B_(BE)-I_(E )R_(E )`
`therefore {(8.5xx10^(-5))+(10^(-5))}R_(B)=12-0.5-(1.01-10^(-3))(1188.12)`
`therefore (9.5xx10^(-5))R_(B)=10.3`
`therefore R_(B)=1.084xx10^(5)Omega`