Q1.Which of the following is an invalid state in an 8-4-2-1. Binary Coded Decimal counter
Answer : Option DExplaination / Solution:
In binary coded decimal (BCD) counter the valid states are from 0 to 9 only in binary system 0000 to 1001 only. So, 1100 in decimal it is 12 which is invalid state in BCD counter.
Q3.Electric-field magnitude E at points inside and outside a positively charged spherical conductor having charge Q and a radius R are
Answer : Option BExplaination / Solution: Electric field inside a spherical conductor is zero since all the charge resides on the surface of the conductor and there is no charge in the interior of the conductor. For all external points, the conductor behaves as though its entire charge is concentrated at its center.E=Q4πε0R2
Q4.In the following limiter circuit, an input voltage Vi = 10 sin100πt is applied.
Assume that the diode drop is 0.7 V when it is forward biased. When it is
forward biased. The zener breakdown voltage is 6.8 V
The maximum and minimum values of the output voltage respectively are
Answer : Option CExplaination / Solution: No Explaination.
Q5.what is common to both biot savart law and ampere's law
Answer : Option DExplaination / Solution: Both Biot- Savart’s lawdB⃗ =μ04πIdl⃗ ×r^r2and ampere’s circuital law∮B⃗ ⋅dl⃗ =μ0Ideal with the consequences of steady current. Ampere’s circuital law is to Biot- savart’s law as is Gauss’s law to Coulomb’s law in electrostatics.
Q6.In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal
switch S is closed at t = 0. The current i(t) for all t is
Answer : Option DExplaination / Solution:
Time constant = RC
In the given circuit, R = 0
Rise time = 0; hence capacitor charges instantaneously and the current can be represented as impulse function
Q7.A potential difference of 220 V is maintained across 12000Ωrheostat. Then voltmeter V has a resistance of 6000 Ωand point C is at one fourth the distance from a to b. Then the reading of voltmeter is
Answer : Option BExplaination / Solution: Resistance R = 1/4R = 12000/4 = 3000 parallel. . This is in series with Rbc. The current in the circuit I = 220/11000 = 0.02A The reading of the voltmeter is the potential drop across Rp. V = 0.02 * 2000 = 40V
Answer : Option DExplaination / Solution:
The term p-type refers to the positive charge of the hole. In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. P-type semiconductors are created by doping an intrinsic semiconductor with acceptor impurities (or doping an n-type semiconductor).