The specifications of transistor 2N525 are

$P_{D(max)}=225mW$

$h_{fe}=45(typ)$

$h_{c}=1.4 K\Omega \ \ \ \ h_{re}=3.2\times 10^{-4}, \ \ \ \ \ \ \ h_{oe}=25\mu mho$

i)Calculation of $R_{C}$

The voltage gain is given as

$A_{v}=\frac{h_{fe}R_{c}}{h_{fe}+\Delta hR_{c}}$

and $\Delta h=h_{ie}h_{oe}-h_{fe}h_{re}$

=$(1.4k\times 25 \mu=mho)-(45\times 3.2 \times 10^{-4})$

$\Delta h=0.0206$

40=$\frac{45\times R_{c}}{1.4k+0.0206\times R_{C}}$

$R_{C}=1.25K\Omega$

select $R_{C}=1.3k\Omega$

ii)Selection of Q-point

$V_{CEQ}\geq V_{p}+V_{CE(gat)}$

$\geq 4V+0.3V$

$\geq 4.3V$

Select $V_{CEQ}=6V,V_{RC}-7.5V, V_{RE}=1.5V and V_{CC}$=15V

$I_{c}=\frac{V_{R_{c}}}{R_{C}}=\frac{7.5V}{1.3k\Omega}$

=5.77mA(Select I_{C}=6mA)

Q-point will be (6V.6mA) .........(ii)

iii)Calcualtion of $R_{E}$

As $I_{E}=I_{C}=6mA$

$R_{E}=\frac{V_{RE}}{I_{C}}=\frac{1.5V}{6mA}=250 \Omega$

Select $R_{E}=270\Omega \ \ \ \ \ \ ........(iii)$

iv)Selection of $R_{1}and R_{2}$

If S=10 then

$R_{B}=9 R_{E}=9\times2700=2430\Omega$

$V_{B}=V_{E}+V_{BE}=1.5+0.7=2.2V$

$R_{i}=\frac{R_{B}V_{CC}}{V_{B}}\frac{2430\times 15V}{2.2V}=16.6k\Omega$

and $R_{B}=R_{1}||R_{2}$ substituting values of R_{B} and R_{1} in this equation we get $R_{2}=2.874k\Omega$

Select $R_{1}=16k\Omega$ and $R_{2}=3k\Omega \ \ \ \ \ .......(iv)$

v)Calculation of by-pass capacitor $C_{E}$

The $X_{CE}$ should be very much less than $R_{E}$ so as to bypass the dc signal to the ground

$X_{CE}=\frac{R_{E}}{10}=\frac{270\Omega}{10}=27\Omega$

$C_{E}=\frac{1}{2\pi fX_{CE}}$

As lower cutoff frequency is not given let us take f=20HZ

=$\frac{1}{2\pi 20Hz 27\Omega}$=$295\mu F$

Select $C_{E}=330\mu F \ \ \ \ \ .........(v)$

vi) Calculation of coupling capacitor, $C_{c}$

The coupling capacitor is used to allow only ac signal to go to next stage and to block all dc

The output impedance of first stage is

$R_{01}=R_{L}=R_{C}=1.3k\Omega$

and the input impedance of second stage

$R_{in1}=R_{B}||R_{in2}=0.888k\Omega$

The capacitace reactance

$X_{CC}=R_{01}+R_{in2}$

=1.3k+0.888k=2.188k$\Omega$

$C_{C}=\frac{1}{2\pi fX_{CC}}=\frac{1}{2\pi \times 20Hz\times 2.188k\Omega}=3.63\mu F$

Select $C_{C}=10\mu F \ \ \ \ ........(vi)$