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Solution :: Ans: Given that \beta=100 and V_{A}=100 \mathrm{~V}Each transistor is biased at I_{E}=0.5 \mathrm{~mA}\begin{array}{l}\therefore r_{e l}=r_{e 2}=\frac{V_{T}}{I_{E}}=\frac{25 \mathrm{mV}}{0.5 \mathrm{~mA}}=50 \Omega, g_{\mathrm{m}}=\frac{1}{r_{e}}=20 \mathrm{~mA} / \mathrm{V} \\\therefore V_{A}=100 \mathrm{~V} \text { and } I_{E}=0.5 \mathrm{~mA} ; Y_{0}=\frac{V_{A}}{I_{E}}=\frac{100}{0.5}=200 \mathrm{k} \Omega \\\end{array}Differential Half ciraut * Common mode Half circuitGain A_{V}= Total Resistance in the collectorsTotal Resistance in the EmittersDifferential gain A_{d}=\frac{2\left(R_{c} \| R_{L / 2}\right)}{2\left(r_{e}+R_{E / 2}\right)}=\frac{2(10 \| 10) \mathrm{K}}{2(0.05+0.15) \mathrm{K}}\therefore \quad A_{d}=25 \mathrm{~V} / \mathrm{V}Differential Input resistance R id:\begin{aligned}R_{\text {id }} & =2\left((\beta+1)\left(r_{e}+R_{E} / 2\right)\right)=2(101)(50+150)=40.4 \mathrm{kn} \\& \therefore \quad R_{i d}=40.4 \mathrm{k} \Omega .\end{aligned}Common mode gain with R_{c} have 1 \% tolerance\begin{array}{l}\therefore \quad A_{\mathrm{cm}}=5 \times 10^{-4} \mathrm{~V} / \mathrm{V} \quad\left[\because \frac{\Delta R_{c}}{R_{c}}=1 \%=0.01\right] \\\end{array}common mode input resistance Ricm:\left.\begin{array}{rl}R_{i c m} & =(B+1)\left[\frac{r_{e}}{2}+\left(\frac{R_{E E}}{2} \| \frac{r_{0}}{2}\right)\right]_{V_{i c m}} \\& =(101)\left(25+\left(\frac{200 \mathrm{k}}{2} \| \frac{200 \mathrm{k}}{2}\right)\right. \\\therefore R_{i c m} & =50.525 \mathrm{M} \Omega\end{array}\right\} PLEASE RATE THE ANSWER POSITIVELY BY GIVING THUMBS UP ...