Correct Answer - A
Given , `i_c = 4 mA `
Applying Kirchhoffs second law in loop 1, we have
`V_(CE) = 8 - i_c R_L`
`:. R_L = 8 - V_(CE) / i_c = (8 - 4)/ (4xx 10^(-3)) `
` = 1000 Omega `
` = 1 k Omega `
Further, ` beta = i_c / i_b `
` :. i_b = i_c / beta = (4 xx 10^(-3)) / 100 A = 40 mu A `
Now, ` V_(BE) = 8 - i_b R_B `
`:. R_B = (8 - V_(BE)) / i_b `
`= (8 -0.6 )/( 40 xx 10^(-6) )`
` = 1.85 xx 10^(5) Omega ` .