In an n-p-n common emitter (CE) transistor,th | vedclass

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(A) The current gain $(\beta)$ in a common emitter $(CE)$ transistor is defined as the ratio of the change in collector current $(\Delta I_C)$ to the

Vedclass · Accepted Answer

$110$

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(A) The current gain $(\beta)$ in a common emitter $(CE)$ transistor is defined as the ratio of the change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$.Given:$\Delta I_C = 16\,mA - 5\,mA = 11\,mA = 11 	imes 10^{-3}\,A$$\Delta I_B = 200\,\mu A - 100\,\mu A = 100\,\mu A = 100 	imes 10^{-6}\,A$Using the formula:$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{11 	imes 10^{-3}\,A}{100 	imes 10^{-6}\,A} = \frac{11 	imes 10^{-3}}{10^{-4}} = 11 	imes 10^1 = 110$Thus,the current gain of the transistor is $110$.

In an $n-p-n$ common emitter $(CE)$ transistor,the collector current changes from $5\,mA$ to $16\,mA$ for a change in base current from $100\,\mu A$ to $200\,\mu A$,respectively. The current gain of the transistor is:

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