In the BJT circuit shown, beta of the PNP transistor is 100. Assume V_BE =−0.7V. The voltage across R_C will be 5 V when R₂ is ________ kΩ . (Round off to 2 decimal places)

The correct answer is 18.099.

Given parameters from the circuit diagram:
Supply voltage, $V_{CC}=12\text{V}$
Base resistor, $R_1=4.7\text{k}\Omega$
Emitter resistor, $R_E=1.2\text{k}\Omega$
Collector resistor, $R_C=3.3\text{k}\Omega$
Transistor current gain, $\beta =100$
Base-emitter voltage, $V_{BE}=-0.7\text{V}$ (which implies $V_{EB}=0.7\text{V}$)
Voltage across collector resistor, $V_{R_C}=5\text{V}$

Step 1: Calculate the Collector Current ($I_C$)
Using Ohm's law across the collector resistor $R_C$:

$I_C=\frac{V_{R_C}}{R_C}=\frac{5\text{V}}{3.3\text{k}\Omega }\approx 1.515\text{mA}$

Rounding $I_C$ to two decimal places for intermediate calculations gives:

$I_C=1.51\text{mA}$

Step 2: Calculate the Base Current ($I_B$) and Emitter Current ($I_E$)
The base current is related to the collector current by:

$I_B=\frac{I_C}{\beta }=\frac{1.51\text{mA}}{100}=0.0151\text{mA}$

The emitter current is:

$I_E=I_C+I_B=1.51\text{mA}$ + $0.0151\text{mA}=1.5251\text{mA}$

Step 3: Calculate the Emitter Voltage ($V_E$) and Base Voltage ($V_B$)
The emitter terminal is connected to the supply rail via $R_E$:

$V_E=V_{CC}-I_E{R}_E=12\text{V}-(1.5251\text{mA}\times 1.2\text{k}\Omega )=12\text{V}-1.8301\text{V}=10.1699\text{V}$

Since $V_{EB}=0.7\text{V}$ for a PNP transistor, the base voltage is:

$V_B=V_E-V_{EB}=10.1699\text{V}-0.7\text{V}=9.4699\text{V}$

Step 4: Solve for the Voltage-Divider Resistor ($R_2$)
Applying Kirchhoff's Current Law (KCL) at the base node:

$\frac{V_{CC}-V_B}{R_1}-I_B=\frac{V_B}{R_2}$

Substituting the known values into the equation:

$\frac{12-9.4699}{4.7}-0.0151=\frac{9.4699}{R_2}$

$\frac{2.5301}{4.7}-0.0151=\frac{9.4699}{R_2}$

$0.5383-0.0151=\frac{9.4699}{R_2}$

$0.5232=\frac{9.4699}{R_2}$

$R_2=\frac{9.4699}{0.5232}\approx 18.099\text{k}\Omega$