In the circuit shown in the figure, the input voltage `V_(i)` is `20V,V_(BE)=0` and `V_(CE)=0`. The values of `I_(B),I_(C)` and `beta` are given by:

Options

A. `I_(B)=20 muA , I_(c )=5 mA ,beta =250`
B. `I_(B)=25 mu A ,I_(c )=5 mA ,beta=200`
C. `I_(B)=40 muA ,I_(c )=10 mA ,beta=250`
D. `I_(B)=40 muA ,I_(C )=5 mA ,beta=125`

Explanation

Voltage across base and emitter `V_(EE) = 0`
Voltage across collector and emitter `V_(CE) = 0`

Applying Kirchoff law between C and E
`20 - I_( c) R_(c ) - V_( b) = 0`
`I_(c ) = ((20-0)/( 4 xx 10^(3))) rArr I_( c) = 5 xx 10^(-3) = 5 mA`
Applying Kirchoff .s law between B and E
`V_(1) - I_(B) - V_(b) =-0`
`I_(B) = (20)/( 500 xx 10^(3)) = 40 muA`
`beta=(I_(c ))/( I_(b)) = (25 xx 10^(3))/( 40 xx 10^(-4)) = 125`
Hence option (d ) is correct