NEET Practice Questions (MCQs) with Answers & Solutions

In an N-P-N transistor circuit, the emitter current (I_E) is the sum of the base current (I_B) and the collector current (I_C). Therefore, the correct relationship is $$ I_E = I_B + I_C $$, which implies $$ I_B < I_C < I_E $$.

To form an AND gate using NAND gates, we need to use three NAND gates. The first two NAND gates are used to invert the inputs, and the third NAND gate combines these inverted inputs. The configuration is as follows:

1. Connect the inputs to the first two NAND gates to produce the inverted inputs.
2. These outputs are then connected to the third NAND gate to get the AND gate output.

Ripples refer to the residual periodic variation of the DC voltage within a power supply which has been derived from an AC source. Essentially, it is the unwanted AC component present in the DC output of a rectifier. Thus, ripples are AC mixed with DC.

$ I_c =10 mA =0.90I_E $ $ \therefore I_E \approx 11 mA $ $ \therefore I_B \approx 1mA $

When a P-type semiconductor is heated, the thermal energy supplied increases the number of electron-hole pairs generated. This means that the number of electrons and holes increases equally due to the intrinsic carrier generation.

The depletion layer in a PN junction diode is caused by the diffusion of charge carriers. When a PN junction is formed, electrons from the N-region diffuse into the P-region and recombine with holes, and holes from the P-region diffuse into the N-region and recombine with electrons. This diffusion results in the creation of a region devoid of free charge carriers, known as the depletion layer.

In a solar cell, we aim to maximize the power output. A larger active junction area allows for the absorption of more light, which in turn generates more electron-hole pairs and increases the current produced by the solar cell. Hence, the active junction area in a solar cell is large to achieve more power.

The current gain ( $\beta$ ) of a transistor is given by the ratio of the change in the collector current ( $\Delta I_C$ ) to the change in the base current ( $\Delta I_B$ ). Given the parameters: $\beta = 100$ and $\Delta I_B = 100 \mu A$ . We can calculate the change in the collector current as follows:

$$\Delta I_C = \beta \times \Delta I_B$$

Substituting the values:

$$\Delta I_C = 100 \times 100 \mu A$$

$$\Delta I_C = 10000 \mu A$$

$$\Delta I_C = 10 \text{ mA}$$

Thus, the change in collector current is $10 \text{ mA}$ .