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# In an n-p-n transistor, ${10^{10}}$ electrons enter the emitter in ${10^{ - 6}}s$. If $2\%$ of the electrons are lost in the base, then the current transfer ratio and the current amplification factor are:$(A)0.98,49$ $(B)0.49,49$$(C)0.98,98$$(D)0.49,98$

Last updated date: 22nd Feb 2024
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Hint: First, we have to find the emitter current from the given number of electrons and time using the charge of an electron.
Since $2\%$ of current is lost in the base, so we may calculate the base current also.
The sum of the collector current and base current is called the emitter current. So find the collector current also.
Note that, the current transfer ratio means the ratio of the collector current to the emitter current. And, the current amplification factor is the ratio of the collector current to the base current.

Formula used:
The emitter current ${I_e} = \dfrac{{ne}}{t}$ , $n =$ the number of electrons, $e =$ the charge of an electron, and $t =$ the time.
If ${I_b}$ and ${I_c}$ are the base current and the collector current respectively,
${I_e} = {I_b} + {I_c}$
The current transfer ratio = $\dfrac{{{I_c}}}{{{I_e}}}$
The current amplification factor = $\dfrac{{{I_c}}}{{{I_b}}}$

Current is defined as the rate of change of total charge at a certain time. From this concept, we may calculate the emitter current in the given n-p-n transistor.
The emitter current ${I_e} = \dfrac{{ne}}{t}$, $n =$ the number of electrons, $e =$ the charge of an electron, and $t =$ the time.
Given, $n = {10^{10}}$
And, $t = {10^{ - 6}}\sec$
We know $e = 1.6 \times {10^{ - 19}}coulomb$
Putting these values we get, ${I_e} = \dfrac{{{{10}^{10}} \times 1.6 \times {{10}^{ - 19}}}}{{{{10}^{ - 6}}}}$
$\Rightarrow {I_e} = 1.6 \times {10^{ - 3}}A = 1.6mA$
Given that $2\%$ of current is lost in the base, so the base current ${I_b} = 1.6 \times \dfrac{2}{{100}} = 0.032mA$
The sum of the collector current and base current is called the emitter current.
Hence, If ${I_b}$ and ${I_c}$ are the base current and the collector current respectively,
$\Rightarrow {I_e} = {I_b} + {I_c}$
$\Rightarrow {I_c} = {I_e} - {I_b}$
$\Rightarrow {I_c} = 1.6 - 0.032 = 1.568mA$
Now, the current transfer ratio means the ratio of the collector current to the emitter current I.e
The current transfer ratio = $\dfrac{{{I_c}}}{{{I_e}}} = \dfrac{{1.568}}{{1.6}}$
$\Rightarrow \dfrac{{{I_c}}}{{{I_e}}} = 0.98$
And, the current amplification factor is the ratio of the collector current to the base current i.e
The current amplification factor = $\dfrac{{{I_c}}}{{{I_b}}} = \dfrac{{1.568}}{{0.032}}$
$\Rightarrow \dfrac{{{I_c}}}{{{I_b}}} = 49$

So the right answer is in option $(A) \Rightarrow 0.98,49.$

Note: At which ratio, The collector current is changed with the change of emitter current in a transistor – this ratio is called the current transfer ratio of the transistor. This term is denoted by $\alpha$. Since ${I_c} < {I_e}$ the value of $\alpha < 1$.here in the problem, this criteria is satisfied.
The current amplification is also known as the current gain. This is defined by the ratio of the output ac collector current to the input ac base current.