In the circuit shown in figure , find the value of `R_C`.
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Answer

A

Explanation

Consider the figure to solve this question,
` I_E = I_C + I_B and I_C = beta I_B`
` I_C R_C + V_(CE) + I_E R_E = V_(C C) `
` RI_B + V_(BE) + I_E R_E = V_(C C) `
` I_E = I_C = beta I_B`

From Eq. (iii) ,
` (R + beta R_E) I_B = V_(C C) - V_(BE)`
` rArr I_B = (V_(C C) - V_(BE)) / (R + beta R_E )`
` = (12 -0.5) / (80 + 1.2 xx 100)`
` = 11.5/ 200 mA `
` From Eq. (ii),
` (R_C + R_E ) = (V_(CE) - V_(BE) )/ I_C `
` = (V_(C C) - V_(CE) )/ (beta I_B)` (`because I_C = beta I_B) `
` (R_C + R_E ) = 2 / 11.5 (12-3) k Omega `
` = 1.56 k Omega`
` R_C + R_E = 1.56`
` R_C = 1.56 - 1 = 0.56 k Omega`