In This post you will get answer of Q.
The transfer ratio $beta$ of transistor is 50. The input resistance of a transistor when used in C.E. (Common Emitter) configuration is 1k$Omega$. The peak value of the collector A.C current for an A.C input voltage of 0.01V peak is
Solution:
$i_{B} = frac{V_{S}}{R_{in}}=frac{0.01}{10^{3}} = 1quad10^{5},A$
Now $beta$ of transistor is defined as
$beta_{ac}=frac{i_{c}}{i_{b}}$
or $,i_{c}=50times10^{-5}=500mu A$
Solution:
$i_{B} = frac{V_{S}}{R_{in}}=frac{0.01}{10^{3}} = 1quad10^{5},A$
Now $beta$ of transistor is defined as
$beta_{ac}=frac{i_{c}}{i_{b}}$
or $,i_{c}=50times10^{-5}=500mu A$