In a study of a transistor as an amplifier,th | vedclass

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(D) Given: The ratio of collector current $(i_c)$ to emitter current $(i_e)$ is $\alpha = \frac{i_c}{i_e} = 0.98$.The collector current is $i_c = 3

Vedclass · Accepted Answer

$60 \mu	ext{A}$

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(D) Given: The ratio of collector current $(i_c)$ to emitter current $(i_e)$ is $\alpha = \frac{i_c}{i_e} = 0.98$.The collector current is $i_c = 3 	ext{ mA}$.We know that the emitter current is the sum of collector current and base current: $i_e = i_c + i_b$.Substituting this into the ratio: $\frac{i_c}{i_c + i_b} = 0.98$.Rearranging the equation: $\frac{i_c + i_b}{i_c} = \frac{1}{0.98} = \frac{100}{98}$.$1 + \frac{i_b}{i_c} = \frac{100}{98} \implies \frac{i_b}{i_c} = \frac{100}{98} - 1 = \frac{2}{98} = \frac{1}{49}$.Therefore,$i_b = \frac{i_c}{49} = \frac{3 	ext{ mA}}{49} = \frac{3000 \mu	ext{A}}{49} \approx 61.22 \mu	ext{A}$.Rounding to the nearest given option,the base current is approximately $60 \mu	ext{A}$.

In a study of a transistor as an amplifier,the ratio of collector current to emitter current is $0.98$. If the collector current is $3 \text{ mA}$,then the base current will be approximately:

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