NEET 2018 Questions on Electronics

“There is only one corner of the universe that you can be certain of improving, and that’s your own self.”

– Aldous Huxley

 

The following questions were included in the NEET 2018 question paper. These are of the usual type and are simple.

(1) In the circuit shown in the figure, the input voltage Vi is 20 V, V_BE = 0 and V_CE = 0. The values of I_B, I_C and β are given by

(1) I_B = 20 μA, I_C = 5 mA, β = 250

(2) I_B = 25 μA, I_C = 5 mA, β = 200

(3) I_B = 40 μA, I_C = 10 mA, β = 250

(4) I_B = 40 μA, I_C = 5 mA, β = 125

Since the emitter to base voltage of the transistor is zero, the entire input voltage 20 volt appears across the 500 kΩ base resistor. The base current is therefore given by

            I_B = 20 V/500 kΩ = (20/500) mA =0.04 mA = 40 μA

Since the emitter to collector voltage is zero, the entire 20 volt supply appears across the 4 kΩ collector resistor. The collector current is therefore given by

            I_C = 20 V/4 kΩ = 5 mA   

The current gain is given by

            β = I_C/ I_B = (5 mA)/(40 μA) = (5×10^–3)/(40×10^–6) = 125

So option (4) is the answer/

(2) In the combination of the following gates the output Y can be written in terms of inputs A and B as

The AND gate at the top has inputs A and B̅. Therefore its output is A • B̅. The bottom AND gate has inputs A̅ and B so that its output is A̅ • B.

Since the outputs of the NAND gates are applied to the two inputs of the OR gate, the final output (from the OR gate) is A • B̅ + A̅ • B as given in option (2).