ISRO Scientist ECE Electronics Mock Test - 3 Free Online Test 2026

Concept:
Equivalent capacitance of capacitors -
Connected in series: When n capacitors C1, C2, C3, ... Cn are connected in series, the net capacitance (Cs) is given by:

Connected in parallel: When n capacitors C1, C2, C3, ... Cn are connected in parallel, the net capacitance (Cp) is given by:
⇒ Cp = C1 + C2 + C3 +... Cn
Calculation:
C₇, C₉ and C₈ are in series,
12 μF, 12 μF, 12 μF
Equivalent capacitance = (1/12) + (1/12) + (1/12) = 4 μF
4 μF is in parallel with 8 μF
Equivalent capacitance = (4 + 8) μF = 12 μF
C₄, C₅ are in parallel with 12 μF
Equivalent = 4 μF
C₃ is in series with 8 μF
Total equivalent = 12 μF

C₁ and C₂ are parallel to 12 μF
Hence equivalent = 4 μF

BFSK (Frequency shift keying):
In FSK (Frequency Shift Keying) binary 1 is represented with a high-frequency carrier signal and binary 0 is represented with a low-frequency carrier, i.e. In FSK, the carrier frequency is switched between 2 extremes.

For binary ‘1’ → S1 (A) = Acos 2π fHt
For binary ‘0’ → S2 (t) = A cos 2π fLt . The constellation diagram is as shown:

Unit impulse signal:

It is defined as a pulse compressed along the horizontal axis and stretched along the vertical axis keeping the area unity.

It is defined as,

It exists only at t = 0 and the area under impulse function is unity.

Properties:

1.

2.

3. x(t) δ(t – t0) = x(t0)

4.

5.

6.

The discrete time version of the unit impulse is defined by

Properties:

1. x[n] δ[n] = x[0] δ[n]

2.

3. x[n] δ[n – n0] = x[n0] δ[n – n0]

4.

5. δ[an] = δ[n]

R₁ → -R₁

R₂ → R₂ - 2R₁; R₃ → R₃ - 3R₁; R₄ → R₄ - 5R₁

R₄ → R₄ - 2R₂; R₃ → R₃ - 2R₂

So, r[A] = 2

Concept:
Half Wave Rectifier:

• The basic circuit of a half-wave rectifier and its waveform with a resistive load is shown in Fig.
• During the positive half-cycle of the input AC voltage, the diode is forward-biased (ON) and conducts.
• While conducting, the diode acts as a short-circuit so that circuit current flows, and hence, a positive half-cycle of the input ac voltage is dropped across RL.
• During the negative input half-cycle, the diode is reverse-biased (OFF) and so, does not conduct i.e. there is no current flow. Hence, there is no voltage drop across RL.

Consider Vm is the maximum value of Half wave rectifier,
∴ Average Value, V_av = V_m/π 
∴ RMS Value, V_R V_m/2
Whenever output is asked simply we need to find out the average output value,
Calculation:
V_m = 12.5 V
V_av = 12.5/π = 4 V

To find the value of R_th:

By applying KCL we can find that,
I₁ + 2V_ab = I --- (1)
We can see that,
V_ab = 2 I₁
now, applying the value of V_ab in equation 1, we get
0.5 V_ab + 2 V_ab = I
2.5 V_ab = I

R_th = (V_ab/I)
= 0.4 Ω

• A silicon-controlled rectifier (SCR) is a four-layer solid state current controlling device with 3 terminals
• The SCR combines the rectifying features of diodes and the on - off control features of transistors
• SCRs are generally used in power switching applications

Concept:
The propagation constant of a transmission line is a complex quantity given by:
γ = α + jβ
α = Attenuation constant, related to the line parameters as:

β = Phase constant, related to the line parameters as:

For a loss lossless line, there is no attenuation, i.e. α = 0. The propagation constant will become:
γ = jβ
|γ| = |β|
Calculation:
Given:
L = 0.35 μH/m, C = 90 pF/m and frequency = 500 MHz
ω = 2πf = 2 × 3.14 × 500 × 10⁶
ω = 3.142 × 10⁹ rad/sec
The magnitude of propagation constant for the given line will be:


|γ| ≈ 17.42
Important Point
Attenuation constant:

• Attenuation constant gives attenuation of an electromagnetic wave propagating through a medium per unit distance from the source
• It is the real part of the propagation constant and is measured in nepers per metre

Phase Constant:

• The phase constant is the imaginary component of the propagation constant for a plane wave
• It represents the change in phase per unit length along the path travelled by the wave at any instant
• Phase constant is measured in radians per unit length

The maximum radar range is given by:

Where
P_t is the transmitted power
A₀ is capture area
S = effective area
P_min = minimum power at the receiver which can be recognized
λ = wavelength of EM radiation
r_max ∝ (P_t)⁴

The correct answer is LPF

Concept:

Z = re^jω

At low frequency ω = 0 we get Z = 1

At high frequency ω = π we get Z = -1

Solution:

Calculate the Z-transform of the given h[n]

at Z = 1 we get H(z) = 2

at Z = -1 we get H(z) = 2/3

as we can see we get a higher magnitude at low frequency hence this system will work as a Low pass filter.

Concept:

• Optical fiber is a dielectric waveguide that operates in optical frequencies, normally in cylindrical form.
• The inner cylindrical structure is core surrounded by the solid dielectric called cladding.
• Variation in material composition gives rise to two commonly used fibers.
• If the refractive index of fiber is uniform throughout and undergoes an abrupt change at the cladding boundary is called step-index fiber.
• If the refractive index of fiber is varied as a function of radial distance from the center of the fiber is called graded-index fiber.

V number
It is a dimensionless number that is related to wavelength and numerical aperture and determines how many modes fiber can support and is given by.



Where,
n₁ = refractive index of core
n₂ = refractive index of cladding
a = radius of the core
λ = operating wavelength
Δ = Relative index difference of fiber and is given by

NA = numerical aperture of the fiber and is defined as
The numerical aperture of an optical fiber is a numerical value
it is the sine of the maximum possible launching angle of the optical fiber and is given by
NA = nsin(θ)


n = refractive index of medium from which light ray enter the fiber core
For lowest order or single-mode operation, V number is given by

α = various profile parameter of the fiber
α = 1 for triangular profile
α = 2 for parabolic profile
Calculation:
Given that
Operation wavelength (λ) = 1.3 μm
Core refractive index (n₁) = 1.5
Relative index difference (Δ) = 0.01
α = 2 for parabolic profile
then


V = 2.405 × √2 = 3.401
We know that

The above equation can be rearranged as

Putting all values in the above equation

a = 3.316×10^-6 meter
a = 3.316 μm
diameter of core (d) = 2a
d = 2 × 3.316 μm
d = 6.632 μm
Explanation
The number of modes for graded-index fiber is given by

α = various profile parameter of the fiber
α = 1 for triangular profile
α = 2 for parabolic profile

Concept:
The divergence theorem states that the surface integral of a vector field over a closed surface, which is called the flux through the surface, is equal to the volume integral of the divergence over the region inside the surface.
Calculation:
is the position vector of a point inside a closed surface.

= 5 x 3 = 15V

Concept:
A continuous-time signal consisting of the sum of multiple time-varying functions is periodic, if and only if both functions are periodic and the ratio of these two periods is a rational number.
In such a case, the least common multiple(LCM) of those periods is the period of the sum signal.
Example:
x(t) = Acos(ω₁t) + Bcos(ω₂t) + Ccos(ω₃​t)
Here
T₁ = 2π/ω1
T₂ = 2π/ω₂
T₃ = 2π/ω₃
Step 1: Check whether (T₁/T₂), (T₂/T₃), (T₃/T₁) rational or not.
If any of the ratio is not rational → Signal is nonperiodic
If ratios are rational → Signal is periodic and proceed to step 2
Step 2:
Common time period = LCM( T1, T2, T3)
Note : Addition of DC signal to periodic signal does hamper the periodicity
Calculation:
GIven;
x (t) = 5 cos (2πt/3) + 9 sin (0.5π t) + 3 sin + 12
T1 = 2π/ω1 = 2π/(2π/3) = 3
T2 = 2π/ω2 = 2π/(0.5π) = 4
T3 = 2π/ω3 = 2π/(π/3) = 6
Here, (T1/T2), (T2/T3), (T3/T1) are rational
Common time period = LCM( T1, T2, T3) = LCM( 3, 4, 6)= 12 sec
Frequency = 1/ Common time period = 1/12 Hz
Hence, x(t) is periodic with frequency (1/12) Hz

Calculation:
Guided wavelength is given by:
; where
f_c = Cut-off frequency
And f = Operating frequency.
For waveguide R­₁ : f₁ = 2GHz and f_c1 = 1 GHz
For waveguide R₂ : f₂ = 4 GHz_. We are to find fc­₂.
Given λ_g1 = λ_g2
Equating the wavelength for Both waveguides ;




⇒ 3 = 16 – fc₂²
fc₂² = 13
f_c2 = √13
Option (4) is the correct option.

Transmitted power = 75%

⇒ Reflected power = 25% = 1/4

Magnitude of reflection coefficient,

Also,

VSWR = 3

Concept:
The Fourier transform does not exist for all non-periodic signal.
And therefore, in order for a function x(t) to have Fourier transform, the following requirements, known as Dirichlet's conditions, must be satisfied
Function should be absolutely integrable form - to

Laplace Transform: It is an integral that converts real variable(usually time domain ) to a complex frequency domain (usually s domain)
In order to have Laplace Transform following criteria must be satisfied:

Concept:
Whenever diodes are tied together, the diode which experiences the more potential difference will ON first.

D₃ is ON and remaining OFF.

When all diodes are opposite to the previous case.

D₁ is ON and remaining OFF

If all are given the same voltage then all will be ON

Calculation:
Given voltage levels are V₁ = 0 to 5 V and V₂ = 3V
Until V₁ < 3 V diode D₂ is OFF
For V₁ = 1 V

V₀ = 1.6 V
For V₂ = 2 V

V₀ = 2.6 V
For V₁ = 3 V
Both will be ON

V₀ = 3.6 V
For V₁ = 4 V diode D₁ will be OFF.

V₀ = 3.6 V
For V₁ = 5 V diode D₁ will be OFF.

V₀ = 3.6 V
So the output is constant at 3.6 V for V₁ ≥ 3 V
Option B is correct.

Calculation:

Given, message signal x(t) = 100 cos (24 π × 10³t)

Sampling period, T_s = 50 μsec

So, we have the sampling frequency

Also, the frequency of the message signal f_m is:

Clearly, the sampling frequency (f_s) is less than the minimum Nyquist sampling frequency of 2f_m.

So, the output will contain frequency components of:

f_m ± nf_s, where n = 0, 1, 2 …

For the given signal, we have the frequency components of the sampled output as:

f_m = 12 kHz

f_s + f_m = 20 + 12k = 22 kHz

f_s – f_m = 20 – 12 k = 8 kHz

2f_s + f_m = 40 + 12 = 52 kHz

2f_s – f_m = 40 – 12 = 28 kHz and so on.

When these signal components are passed through a low pass filter with cut-off frequency f_c = 15 kHz, we get the output with frequency components:

Concept:
In FM (Frequency Modulation), the modulation index is defined as the ratio of frequency deviation to the modulating frequency.
Mathematically, this is defined as:
m_f = Δ_f/f_m
mf = Modulation index
Δf = Frequency deviation
fm = Modulating frequency
We observe that the modulation index for FM signal is inversely proportional to the modulating frequency f_m.
Important Point
A wave has 3 parameters Amplitude, Phase, and Frequency. Thus there are 3 types of modulation techniques.
Amplitude Modulation: The amplitude of the carrier is varied according to the amplitude of the message signal.
Frequency Modulation: The frequency of the carrier is varied according to the amplitude of the message signal.
Phase Modulation: The Phase of the carrier is varied according to the amplitude of the message signal.

Concept:
The total transmitted power for an AM system is given by:

Pc = Carrier Power
μ = Modulation Index
The above expression can be expanded to get:

The total power is the sum of the carrier power and the sideband power, i.e.

Calculation:
Given Pt = 600 W, Pc = 400 W
We can write:


(1.5 - 1) × 2 = μ²
μ = 1

Sequential circuit
This is represented when the process has the clock signal in its sensitivity list else we can’t call that a sequential circuit.
Since the given process doesn’t have any clock signal. So, it can’t be a sequential circuit and hence the process is a combinational circuit.
WAIT statement also only one and this represents the given circuit is sequential.
An example of sequential circuit code is represented below:
entity d_ff is port ( CLK, d: in bit; q: out bit ) ;
end d_ff ;
architecture rtl of d_ff is
begin
process(CLK)
begin
if clk’event and clk = ’1’
then q <= d ;
end if ;
end process ;
end rtl ;

Conclusion:
The given circuit is a combinational circuit.

Concept:

• Shift Registers are sequential logic circuits, capable of storage and transfer of data.
• They are made up of Flip Flops which are connected in such a way that the output of one flip flop could serve as the input of the other flip-flop, depending on the type of shift registers being created.
  
• Shift registers are a type of register that has the ability to transfer (“shift”) data.
• Registers are storage devices that are created by connecting a specific number of flip flops in series.
• The number of bits that can be stored by the register is always directly proportional to the number of flip flops ( as each flip flop is capable of storing only one bit at a time).
• The number of clock pulses needed to load and unload an n-bit shift register will be (2n – 1).

∴ To load a 4-bit binary number 1101 in a register, it will take 4 clock pulses.

Example: Let 1101 is the input to a shift register. The working of the register after each clock pulse is as shown:

After the 4^th clock pulse, all the bits are stored in each flip-flop of the register. We can observe that one bit is already at the output terminal.

∴ It will require another 3 clock pulses to get the other three bits at the output terminal, i.e. it requires 7 clock pulses to load and unload the 4-bit register.

This is as shown:

Concept:

According to Carson rule signal, BW is given as:

B.W = (β + 1) 2f_m

B.W. = 2[Δf + fm]

If multiple frequencies are available in modulating signal then,

B.W. = 2(β + 1) fmax

Calculation:

Given:

m(t) = 10 cos (2π 100t)

The resulting FM signal is given as:

x(t) = 20 cos {2π 106 t + 15 sin (2π 100 t)}

The instantaneous frequency is given by:

Maximum instantaneous frequency

Maximum frequency deviation

= 1.5 kHz

Given, f_m = 100 Hz = 0.1 kHz

So, Approximate bandwidth according to Carson’s rule

⇒ 2(Δf + fm)

= 2(1.5 + 0.1)

= 2 (1.6 k)

= 3.2 kHz

Concept:
In Digital Electronics, Logic 1 means High and Logic 0 means low.
Whenever in diode, if 1 is applied to anode and 0 to cathode then Diode acts as a short circuit i.e. ON.
And if 0 is applied to anode and 1 to cathode Diode acts as open circuit i.e. OFF.

Explanation:
The given logic circuit is

For different logic of A and B,4 cases are there and according to that logic of C will vary.

Case 1
When A is logic 0 and B is logic 0

Then the logic of C will be 0.

Case 2
When A is logic 0 and B is logic 1

Then the logic of C will be 1.

Case 3
When A is logic 1 and B is logic 0

Then the logic of C will be 1.

Case 4
When A is logic 1 and B is logic 1

Then the logic of C will be 1.
According to Result, we make a table

This Table is of Logic OR gate.
∴ C = A + B

Important PointsLogic Circuit for AND gate is C = AB

Under equal load conditions, the bridge rectifier needs four diodes, two of which conduct in alternate half cycles.
Because of this, the total voltage drop in diodes becomes double that in the case of the center tap rectifier, losses are increased, and rectification efficiency is somewhat reduced.
The efficiency of a rectifier is defined as the ratio of the dc output power to input power.
The efficiency of Bridge rectifier = Efficiency of the full-wave rectifier

For Half wave rectifier maximum efficiency = 40.6%
For Full wave rectifier maximum efficiency = 81.2%
A half-wave rectifier conducts when the alternating current is positive, which happens during 180° of the A.C cycle, and blocks current during the other 180 degrees
A full-wave rectifier conducts during both 180° periods of an A.C cycle; So we can see, it conducts current for twice the amount of time
The full-wave rectifier has twice the efficiency of a half-wave rectifier is, which utilizes both half-cycles of the input.

Concept:
1. Every branch of a root locus diagram starts at a pole (K = 0) and terminates at a zero (K = ∞) of the open-loop transfer function.
2. The root locus diagram is symmetrical with respect to the real axis.
3. A number of branches of the root locus diagram are:
N = P if P ≥ Z
= Z, if P ≤ Z
4. Number of asymptotes in a root locus diagram = |P – Z|
5. Centroid: It is the intersection of the asymptotes and always lies on the real axis. It is denoted by σ.

ΣPi is the sum of real parts of finite poles of G(s)H(s)
ΣZi is the sum of real parts of finite zeros of G(s)H(s)
6. The angle of asymptotes:
l = 0, 1, 2, … |P – Z| – 1
7. On the real axis to the right side of any section, if the sum of the total number of poles and zeros are odd, the root locus diagram exists in that section.
8. Break-in/away points: These exist when there are multiple roots on the root locus diagram.
At the breakpoints gain K is either maximum and/or minimum.
So, the roots of dK/ds are the breakpoints.

Observation:
n root-locus plot, the breakaway points, must lie on the root loci.

• A blocking oscillator used to produce narrow pulses or trigger pulses it is also called Saw-tooth wave oscillator
• When we take feedback from the output signal, it blocks the feedback, after a cycle, for certain predetermined time

Concept:
The polarization of a plane wave is the “figure traced by the tip of the electric field vector as a function of time, at a fixed point in space.”

Calculation:
The given wave can be written as;
⇒ E(x, t) = a_y 100 cos (ωt - βx) + a_z 200 sin (ωt - βx)
Fixing a point, i.e. x = 0
⇒ E (0, t) = a_y 100 cos (ωt) + a_z 200 sin (ωt)
When, ωt = 0
E = a_y (100)

E = a_z (200)
This can be explained as:
Since the magnitude of ‘y’ and ‘z’ components are unequal and the tip of the field is rotating in the counter-clockwise direction. The polarization is right-hand elliptical polarization. So, option (B).

The pattern followed here is:

Code = Sum of positional values of letters ÷ Number of letters in the word.

• In SHINE → 11

• Code = Sum of positional values of letters ÷ Number of letters in the word
• Code = (19 + 8 + 9 + 14 + 5) ÷ 5
  = 55 ÷ 5
  = 11​​​
• In ENTRY → 16.4 ​

• Code = Sum of positional values ÷ Number of letters in the word
• Code = (5 + 14 + 20 + 18 + 25) ÷ 5
  = 82 ÷ 5
  = 16.4
• Similarly, In VIDEO → ​

• Code = Sum of positional values ÷ Number of letters in the word
• Code = (22 + 9 + 4 + 5 + 15) ÷ 5
  = 55 ÷ 5
  = 11.​​​

We get '11'.
Hence, "Option 3" is the correct answer.

The logic followed here is → (First digit of the first number × Second digit of the first number) + (Sum of the digits of first number) = Second number

So,
Option - (1) → 68 - 62
⇒ (6 × 8) + (6 + 8) = 62
⇒ 48 + 14 = 62
⇒ 62 = 62 (LHS = RHS)

Option - (2) → 72 - 23
⇒ (7 × 2) + (7 + 2) = 23
⇒ 14 + 9 = 23
⇒ 23 = 23 (LHS = RHS)

Option - (3) → 35 - 22
⇒ (3 × 5) + (3 + 5) = 22
⇒ 15 + 8 = 22
⇒ 23 ≠ 22 (LHS ≠ RHS)

Option - (4) → 91 - 19
⇒ (9 × 1) + (9 + 1) = 19
⇒ 9 + 10 = 19
⇒ 19 = 19 (LHS = RHS)
Here, "35 - 22" is different from rest of the alternatives.
Hence, "Option - (3)" is the correct answer.