EAMCET 2010 Multiple Choice Questions on Electronics

Here are the two questions on electronics which were included in EAMCET 2010 Agriculture-Medicine and Engineering question papers respectively:

(1) In the figures shown below

(1) In both Fig. (a) and Fig. (b) the diodes are forward biased

(2) In both Fig. (a) and Fig. (b) the diodes are reverse biased

(3) In Fig. (a) the diode is forward biased and in Fig. (b) the diode is reverse biased

(4) In Fig. (a) the diode is reverse biased and in Fig. (b) the diode is forward biased

In Fig. (a) the anode of the diode is at a positive potential of 10 V while the cathode is at zero volt and hence it is evidently forward biased. In Fig. (b) the anode of the diode is at a higher negative potential compared to the cathode and hence it is reverse biased [Option (3)].

(2) A transistor having a β equal to 80 has a change in base current of 250 μA. Then the change in collector current is

(1) 20,000 mA

(2) 200 mA

(3) 2000 mA

(4) 20 mA

The current gain β is given by

β = ∆I_c/∆I_b at constant collector voltage where ∆I_c is the change in collector current and ∆I_b is the change in base current.

Therefore, ∆I_c = β∆I_b = 80×250 μA = 20,000 μA = 20 mA.

Electronics- Multiple Choice Questions involving Diodes

I am enough of an artist to draw freely upon my imagination. Imagination is more important than knowledge. Knowledge is limited. Imagination encircles the world.

– Albert Einstein

Here are three questions involving diodes:

(1) In the circuit shown in figure the inputs at A and B are respectively 7 V and 0.8 V. If the potential drops across the forward biased diodes are negligible and the breakdown voltage of the zener diode is 5V, the output voltage is

(a) 0.8 V

(b) 7 V

(c) 3.9 V

(d) 0 V

(e) 5 V

The diode connected to input A will be forward biased and the diode connected to input B will be reverse biased. The input of 7 V will be impressed at the junction between the diodes and the resistor. The zener diode will breakdown and the output voltage will be the breakdown voltage of the zener diode, which is equal to 5 V. (Two volts will be dropped across the resistor).

(2) If the logical 0 is 0 volt and logical 1 is + 5 volt, the circuit shown in the above question can function as

(a) AND gate

(b) NOT gate

(c) EXOR gate

(d) OR gate

(e) none of the above

If both inputs are 0 volt the output will be zero. If at least one input is 5V or all inputs are 5 V, the output will be 5 V and the circuit can function as an OR gate.

(3) The current through the 40 Ω resistor in the circuit shown in the adjoining figure is

(a) 50 mA

(b) 80 mA

(c) 40 mA

(d) 92.3 mA

(e) 44.4 mA

The diode D₂ is reverse biased and hence no current flows through its branch. The branches containing diodes D₁ and D₃ are in parallel and the diodes are forward biased. The effective resistance of these two branches in parallel is (20×20)/(20+20) = 10 Ω. Therefore, the total resistance in series with the 4 volts battery is 40 Ω +10 Ω = 50 Ω.

Therefore,the current through the 40 Ω resistor is 4 V/50 Ω = 0.08 A = 80 mA.

You will find some useful multiple choice questions (with solution) in electronics here as well as here.