I have attached the images for the question and the way i attempted it please refer to it and tell me have i made any mistake.I have also attached the image for the solution provided by book, and my answer is not matching to the answer given in the book. Question has asked to calculate the output voltage and it is (Q.28) for solution given in my book.
Lazyaditya, Your work was correct until the last line of the second picture you attached. You substituted 430 uA for Ic which is not correct. 430uA is the total current = Ic + I(4 Kohm resistor). The solution in the book is correct, you don't need to substitute each current value, you just calculate the voltage on 2K resistor, & substitute from 12 V -> you will get Vo.
^^^^ That does not seem right. If the transistor was not connected then 2.0 milli-amps would be flowing through the 4k Ω resistor making 8.0 volts at V_{o}. Once the transistor is connected in parallel to the the 4k Ω resistor voltage at V_{o} will only get lower than 8.0 volts.
I computed Vo = 7.4133 volts. The Vbe forward junction drop is 0.60V at such low currents. Ib is 5.0V - 0.60V, divided by 1.0 megohm, which is 4.4 uA. Multiplying by beta value of 100 gives 0.440 mA for Ic, the collector current. This Ic equals the current in the 2.0 kohm resistor minus that in the 4.0 kohm resistor. Using algebra and solving for Vo gives 7.4133V. The 2.0 kohm resistor current is 2.2933 mA, and the 4.0 kohm current is 1.8533 mA. The difference is 0.440 mA, exactly equal to collector current Ic. Claude
Posts 1-8 have too many significant digits. Vbe is known to about 0.65 +/- 0.10 V. So post 9 is the best.