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
(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.
1 comment:
Thanks good to make ones basic clear here
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