Electronic Science - Online Test
Q1. The logic function implemented by the circuit below is (ground implies logic 0)
Answer : Option D
Explaination / Solution:
Q2. A source 
has an internal impedance of 
If a purely resistive load connected to this source has to extract the maximum power out of the source, its value inΩ should be
has an internal impedance of If a purely resistive load connected to this source has to extract the maximum power out of the source, its value inΩ should be
Answer : Option C
Explaination / Solution:
For the purely resistive load, maximum average power is transferred when
where
is the equivalent thevinin (input) impedance of the circuit. Hence,
we obtain
where
is the equivalent thevinin (input) impedance of the circuit. Hence,we obtain
Q3. The output of a 3-stage Johnson (twisted ring) counter is fed to a digital-to
analog (D/A) converter as shown in the figure below. Assume all the states
of the counter to be unset initially. The waveform which represents the D/A
converter output v0 is
Answer : Option A
Explaination / Solution:
All the states of the counter are initially unset.
State Initially are shown below in table:
Q2Q1Q0
0 0 0 0
1 0 0 4
1 1 0 6
1 1 1 7
0 1 1 3
0 0 1 1
0 0 0 0
Q4. In the circuit shown below, the Norton equivalent current in amperes with
respect to the terminals P and Q is
Answer : Option A
Explaination / Solution:
Replacing P - Q by short circuit as shown below we have
Using current divider rule the current ISC is
Q5. Two D flip-flops are connected as a synchronous counter that goes through the
following QBQA sequence 00⟶11⟶01⟶10⟶00⟶… The combination to the inputs DA and DB are
Answer : Option D
Explaination / Solution:
The sequence is QBQA
00⟶11⟶01⟶10⟶00⟶…
Q6. In the circuit shown below, the value of RL such that the power transferred to RL is maximum is
Answer : Option C
Explaination / Solution:
Power transferred to RL will be maximum when RL is equal to the thevenin
resistance. We determine thevenin resistance by killing all source as follows :
Q7. An 8085 assembly language program is given below. Assume that the carry
flag is initially unset. The content of the accumulator after the execution of the
program is
Answer : Option C
Explaination / Solution:
Initially Carry Flag, C = 0
MVI A, 07 H ; A = 0000 0111
RLC ; Rotate left without carry. A = 00001110
MVO B, A ; B = A = 00001110
RLC ; A = 00011100
RLC ; A = 00111000
ADD B ; A = 00111000
; +00001110
; 01000110
RRC ; Rotate Right with out carry, A = 0010 0011
Thus A = 23 H
Q8. In the circuit shown below, the current I is equal to
Answer : Option B
Explaination / Solution:
From star delta conversion we have
Replacing in circuit we have the circuit shown below :
Now the total impedance of circuit is
Q9. The first six points of the 8-point DFT of a real valued sequence are 5, 1 - j3, 0, 3 - 4j, 0 and 3 + 4j .....The last two points of the DFT are respectively
Answer : Option B
Explaination / Solution:
For 8 point DFT, 
and it is conjugate
symmetric about x[4], x[6] = 0; x[7] = 1 + j3
and it is conjugate
symmetric about x[4], x[6] = 0; x[7] = 1 + j3
Q10. In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with
the voltage polarity as indicated. The switch is closed at time t = 0. The current i(t) at a time t after the switch is closed is
Answer : Option A
Explaination / Solution:
Here we take the current flow direction as positive.
