Answer:

Given : $R_B=270\,k\Omega,\,\,R_E=1.5\,k\Omega,\,\,V_{EE}=5V,\,\,\beta=100$

To Find: $V_E=?,\,\,I_E=?$

Solution:

Applying Kirchhoff's voltage law to the input circuit of the above figure, will result in

$-I_BR_B-V_{BE}-I_ER_E+V_{EE}=0$

but  $I_E=(\beta+1)I_B$

and  $V_{EE}-V_{BE}-(\beta+1)I_BR_E-I_BR_B=0$

with    $I_B=\dfrac{V_{EE}-V_{BE}}{R_B+(\beta+1)R_E}\,\,\,\,\,\,.....(1)$

Substituting values to the equation (1) yields,

$I_B=\dfrac{5\,V-0.7\,V}{270\,k\Omega+(101)(1.5\,k\Omega)}$

     $=\dfrac{4.3\,V}{270\,k\Omega+151.5\,k\Omega}=\dfrac{4.3\,V}{421.5\,k\Omega}$

     $=10.2\,\mu A$

Hence,

$I_E=(\beta+1)I_B=(101)(10.2\,\mu A)$

     $=1.03\,mA$

$\therefore \,\,V_E=I_ER_E=(1.03\,mA)(1.5\,k\Omega)=1.545\,V$