Karnataka 2nd PUC Question Paper 2025 Answer Key Electronics

Government of Karnataka
Karnataka Secondary Education Examination Board

Question Paper
ANSWER KEY

1
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD
II PUC EXAMINATION 1 - MARCH 2025
SUBJECT: 40- ELECTRONICS Scheme of Evaluation MAX. MARKS: 70
Qn. Answer

Marks
No. PART – A 15 × 1 = 15

01. b) FET 1
02. a) Quiescent point 1
03. b) 1800 1
04. b) Decreases 1
05. d) Slew rate 1
06. b) Op-amp 1
07. d) Crystal oscillator 1
08. a) F layer 1
09. b) 455 kHz 1
10. d Thyristor 1
11. c) 0101 1001 1
12. c) Q = 1, Q̅=0 1
13. b) 2 1
14. d) != 1
15. c) Code Division Multiple Access 1
5×1=5
16. f) highest 1
17. d) comparator 1
18. b) damped 1
19. c) twice 1
20. a) combinational 1
PART – B 5 × 2 = 10
21. a. Fixed bias (base bias) 2
b. Collector to base feedback bias
c. Emitter feedback bias
d. Voltage divider bias (universal bias) (Any two)
22. Given: BW = 1 MHz, A = 100, β = 0.01
BWf = BW(1 + Aβ) 1
= 1 MHz (1 + 100 × 0.01) = 2 MHz 1
23. Given: L = 10 µH, Ceq = 1 nF
f= 1 1
2π LC eq

1 1
f = = = 1.59 MHz 1
2  3.142 10  10 -6  1  10 -9 2  3.142  10 -7
24. PT = PC(1 +
𝑚𝑎 2
) 1
2
1
Upper limit of ma = 1
25. 1. AC to DC - Rectifier 2
2. AC to AC - AC voltage controller
3. DC to DC – Chopper
4. DC to AC – Inverter Each ½ mark
IIPUC EXAM -1, MARCH 2025 KSEAB 40 - ELECTRONICS MODEL ANSWERS

2
26.

2

27.
ROM (Read Only Memory): These memory circuits permanently store binary
numbers. ROM memory cell contents are not being changed by the CPU, but they 1
may be used by CPU. ROM is also called non-volatile, because its content does not
lost when power is removed.
RAM (Random Access Memory): These memory circuits temporarily store binary
numbers. RAM memory cell contents are both read and written to by the CPU. RAM 1
is also called volatile memory, because its contents are lost when power is removed.
28. Wifi is a universal wireless networking technology that utilizes radio
frequency to transfer data. Wifi allows high speed internet connections 2
without the use of cables. The term wifi is a construction of “wireless fidelity”
and commonly used to refer to wireless networking technology.
(or any two points)
PART – C 5 × 3 = 15
29. FET BJT 3
1 Unipolar device Bipolar device
2 Current conduction is by one Current conduction is by 2 types of
type of charge carriers – charge carriers - electrons and
either electrons or holes holes
3 Voltage controlled device Current controlled device
4 Input resistance is very high Input resistance is low
5 High switching speed Low switching speed
6 Less noisy More noisy
(Any three differences)

30.
1

Consider a voltage series feedback amplifier in which, the input impedance Z i
of the basic amplifier without feedback is given by,
v
Zi = i  (1)
i
i
[Where vi is the input voltage to the internal amplifier and ii is the input
current].
IIPUC EXAM -1, MARCH 2025 KSEAB 40 - ELECTRONICS MODEL ANSWERS

3
With negative feedback, the input impedance is given by,
v
Z if = s  (2)
i
i
Output voltage of an amplifier is given by,
v0 = Avi  (3)
We know that for negative feedback amplifier, the net input vi to the internal
amplifier is
vi = vs – vf 2
i.e., vs = vi + vf
i.e., vs = vi + v0 [vf = v0]
Substituting equation (3) in the above equation we get,
vs = vi + Avi
i.e., vs = vi (1 + A).
Substituting in equation (2) we get
v (1 + Aβ)
Z if = i  (4)
ii
From equation (1) and (4),
Zif = Zi (1 + A).
Block diagram -1 mark
Derivation - 2 mark
31. 1

a) Aβ > 1

b) Aβ < 1
1

c) Aβ = 1

1
32. Noise Signal: Any unwanted electrical disturbance added to the signal in the 1
communication channel.
The maximum distance visible to the naked eye on the surface of the Earth is called
optical horizon (OH). Due to the curvature of the Earth, the sky and the Earth 2
appears to meet at a far distance and this is called optical horizon. However the
radio waves can travel far beyond the optical horizon and it is called the radio horizon
𝟒
(RH). The radio horizon is about 𝟑rd the optical horizon. For the earth’s dimensions,
the radio horizon is usually less than 100 km.

IIPUC EXAM -1, MARCH 2025 KSEAB 40 - ELECTRONICS MODEL ANSWERS

4
33.

3

Diagram – 2 mark
Waveforms – 1 mark
34.
Given: HWR 𝛼 = 600, Vrms = 230 V, Vm = Vrms×√2 = 325 V, R = 20 Ω

𝑉
Vdc = 2𝜋
𝑚
[1 + 𝑐𝑜𝑠𝛼 ] 1

𝑉
∴ Vdc = 2𝜋
𝑚
[1 + 0.5] = 77.5 V
1
𝑉𝑑𝑐
Idc = = 3.87 A
𝑅 1
35.

2

(1010)2 = (1111)Gray
1
36. 1. To provide LAN connection.
2. Long distance telephone and cable TV systems. 3
3. Secure communication system at military bases.
4. Short range transmission of health sensor data from medical device to the
computer.
5. Closed circuit TV system.
6. Internet connection. (or any three)
PART – D (Section I) 3 × 5 = 15
37. VCC, R1, R2 and RE provide the necessary DC bias to the transistor.
C1 and C2 are called the coupling capacitors whose function is to block the DC and
allow ac signals.

IIPUC EXAM -1, MARCH 2025 KSEAB 40 - ELECTRONICS MODEL ANSWERS

5

3

Working: 1
During positive half cycle of the input signal,
1. The forward bias across EB junction increases hence i B and iE increases [because
iE = (1 + β) iB].
2. Thus voltage drop across RE (i.e., iE RE) which is the output voltage increases in
the same proportion of the input voltage. 1
During negative half cycle of the input signal,
1. The forward bias across EB junction decreases hence i B and iE decreases.
2. Thus voltage drop across RE (i.e., iE RE), which is the output voltage, decreases in
the same proportion of the input voltage.
Circuit diagram 2 mark
Working 2 mark
Input output waveform 1 mark
38.

2

For an ideal op-amp, open loop gain A = ∞ and Zi = ∞
By virtual ground concept, VB = VA = 0
Since input impedance Zi = ∞ → ib = 0
Applying KCL at A
ii = ib + if
ii = if (since ib = 0)
i1 + i2 + i3 = if
V1 − VA V2 − VA V3 − VA VA − VO
+ + =
R1 R2 R3 Rf
3

IIPUC EXAM -1, MARCH 2025 KSEAB 40 - ELECTRONICS MODEL ANSWERS

6

As VA = 0,
V1 − 0 V2 − 0 V3 − 0 0 − VO
+ + =
R1 R2 R3 Rf
V1 V2 V3 − VO
+ + =
R1 R2 R3 Rf

V1 V2 V3
VO = −R f ( + + )
R1 R2 R3

If R1 = R2 = R3 = Rf = R,
Vo = - (V1 + V2 + V3)

Circuit diagram 2 mark
Derivation 3 mark
39.
NOT gate using NAND gate:

1

AND gate using NAND gates:

1

OR gate using NAND gates:

1

XOR gate using NAND gates:

2

IIPUC EXAM -1, MARCH 2025 KSEAB 40 - ELECTRONICS MODEL ANSWERS

7
40. MOV A, #3FH 1
MOV B, #2AH 1
DIV AB 1

Quotient is stored in register A
1
Remainder is stored in register B
1
41. #include <stdio.h> 1
void main()
{ 1
int x,y;
printf ("Enter the two integer number \n");
scanf ("%d%d”, &x,&y); 1
if (x==y)
{
printf (“The given numbers are equal\n”); 1
}
else
{
printf ("The given numbers are not equal \n");
} 1
}
PART D (Section II) 2 x 5 = 10
42. Given: β = 150, R1 = 45 kΩ, R2 = 5 kΩ, RE = 470 Ω, RC = 2 kΩ, VCC =15 V,
26 mV
VBE = 0.7 V and re1 = IE
i) V5k = V2
V
V2= R1+R2
cc
× R2
1
15
V2 = 45k+5k × 5 k

V2 =1.5 V
V −V
ii) IE = 2 R BE 1
E
1.5−0.7
IE = = 1.7 mA
470
26 mV 26 mV
iii) re1 = = 1.7𝑚𝐴 = 15.29 Ω 1
IE
R 2k
iv) AV =r c = 15.29 = 130 1
e′
v) Ai = β = 150 1
43. (a) Inverting amplifier
𝑅
V01 = - ( 𝑅𝑓1 ) × Vi1 1
𝑖1
10 k
= -(
) × 1 = - 2.12 V 1
4.7 k
(b) Non inverting amplifier
𝑅
1
V02 = (1 + 𝑅𝑓2 ) × Vi2
𝑖2

20 k 1
= (1+ )×2
8.2 k
= 6.87 V 1

IIPUC EXAM -1, MARCH 2025 KSEAB 40 - ELECTRONICS MODEL ANSWERS

8
44. (a) Given: νm =30sin(2 × 103t) and vc = 60sin(2 ×105t)
Standard Equations are vm = Vmsin(ωmt) and vc = Vcsin(ωmt)
 Vm = 30 V and Vc = 60 V
Vm 30 1
Modulation index ma= = = 0 .5
VC 60
𝑚𝑎 𝑉𝑐 0.5×60 1
Amplitude of each sideband= = = 15 V
2 2
ω𝑚 2 π×103
fm = = =1 kHz.
2π 2π
ω𝑐 2 π×105
fc = = =100 kHz.
2π 2π
Lower sideband frequency = fLSB = fc− fm = (100-1) kHz = 99 kHz
Upper sideband frequency = fUSB= fc+fm = (100+1) kHz = 101 kHz 1

1

(b) Bandwidth = 2fm = 2 x1 kHz = 2 kHz.
1
45.

2

̅𝑪
Y=𝑨 ̅ +AC 1

2

Drawing K-map and grouping – 2 mark
Writing simplified expression – 1 mark
Drawing logic circuit – 2 mark

*****

IIPUC EXAM -1, MARCH 2025 KSEAB 40 - ELECTRONICS MODEL ANSWERS

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