TS EAMCET 5 May 2018 Shift 2 Solved Paper

Consider the following circuit.Use KVL to the circuit.
VCC=iCRC+VCE+iERE
VCC=iCRC+VCE+iCRE (∵iE≈iC)
VCE=VCC−iC(RC+RE) …… (1)
Calculate the current i1 by ohm's law.
i1=

VCC

R1+R2

=

10V

15kΩ

=

2

3

mA
The value of V2 is calculated as follows.
V2=i1R2
=

2

3

mA×5kΩ
=

10

3

=3.3V
Apply KVL to base circuit to obtain,
V2=VBE+iERE
iE=

V2−VBE

RE

=

3.3−0.7

526

=5mA
Therefore, from equation (1) VCE is calculated as,
VCE=VCC−iC(RC+RE)
=10−5mA(1kΩ+526Ω)
=2.4V