Ncert Exercises & Solutions

In the circuit shown in the figure below, find the value of R_C.
(Given: \(R_E\) =1 k\(\Omega\))

Hint:

I_{C} R_{C} + V_{CE} + I_{E} R_{E} = V_{CC} and, I_{B} R + V_{BE} + I_{E} R_{E} = V_{CC}

Step 1: Use Kirchoff's voltage law.

Consider the fig. (b) to solve this question.

I_{E} = I_{C} + I_{B} and I_{C} = {βI}_{B} . . . (i)

I_{C} R_{C} + V_{CE} + I_{E} R_{E} = V_{CC} . . . (ii)

I_{B} R + V_{BE} + I_{E} R_{E} = V_{CC} . . . (iii)

∴ I_{E} \approx I_{C} = {βI}_{B}

Step 2: Find the base current.

From Eq. (iii),

(R + βR) I = V_{CC} - V_{BE}

I_{B} = \frac{V_{CC} - V_{BE}}{R + β \cdot R_{E}}

= \frac{12 - 0.5}{80 + 1.2 \times 100} - \frac{11.5}{200} mA

Step 3: Find the emitter resistance.

From Eq. (ii),

(R_{C} + R_{E}) = \frac{V_{CC} - V_{BE}}{I_{C}} = \frac{V_{CC} - V_{CE}}{{βI}_{B}} (∴ I_{C} = {βI}_{B})

(R_{C} + R) = \frac{2}{11.5} (12 - 3) kΩ = 1.56 kΩ

R_{C} + R = 1.56

R = 1.56 - 1 = 0.56 kΩ