Question

$\text{Assertion}\left(A\right):$ Transistor can be used as a switch.
$\text{Reason}\left(R\right):$ Both linear and non-linear voltage bias dependence occurs in it.

• A. Both $A$ and $R$ are true, and $R$ is the correct explanation of $A$.
• B. $A$ is true, but $R$ is false.
• C. Both $A$ and $R$ are true, and $R$ is not the correct explanation of $A$.
• D. Both $A$ and $R$ are false.

Answer 1

The correct option is A Both $A$ and $R$ are true, and $R$ is the correct explanation of $A$.

In the transistor, unless the current flows in the base circuit, no current can flow in the collector circuit. This property allows the transistors to work as a switch.

The operating conditions of the transistor are zero input base current $(I_B=0)$ , Zero collector current $(I_C=0)$ , and the maximum collector voltage $(V_{CE}=\text{maximum})$ which results in the large depletion layer. There is no current flowing through the device. The transistor is switched to $\text{OFF}$. In the cut-off region, the transistor works as an $\text{OFF}$ switch.

In the saturated region, the transistor will be biased so that the maximum amount of base current $(I_B=\text{maximum})$ is applied, resulting in maximum collector current $(I_C=\text{maximum})$ and then resulting in the minimum collector-emitter voltage $(V_{CE}\approx 0)$ drop. The depletion layer will be small in this condition. Hence, it allows maximum current to flow through it.
Therefore, in the saturated regions, transistors are used as an $\text{ON}$ switch.

Hence, both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

​​​​​​​Hence, $\left(A\right)$ is the correct answer.