AFCAT EKT EE Electrical Mock Test - 5 Free Online Test 2026

cosec²y (dy/dx) + x cot y = x³

Let cot y = t ⇒ -cosec²y (dy/dx) = dt/dx

Therefore, -dt/dx + x t = x³

⇒ dt/dx - x t = -x³

Integrating Factor: IF = e^(∫(-x)dx) = e^-x²/2

In trimetric projection, trimetric scales are used in which three perpendicular edges of the object make different angles with POP

A MOSFET is a voltage controlled device and requires only a small input current. This is because of the fact that gate circuit impedance in MOSFET is extremely high, of the order of 10⁹ ohm.

A NOR gate is same as combination of NOT and OR because NOR is same as N+OR. N means NOT whereas OR means OR gate.

For a noiseless channel, the number of input symbols is equal to number of output symbols. Thus, m = n.

When a signal with energy E is fed to a matched filter, the output signal also has the same energy as the input waveform. Thus, the output signal wave also has energy E.

In a conductor, the forbidden band is missing and the conduction band and valance band are overlapped. Therefore, no holes can be created in a metal.

Cathode rays are stream of electrons observed in vacuum tubes. Rest all are electromagnetic waves in nature.

ASK signaling has the highest probability of error. And BPSK has the lowest probability of error.

NOR gate is logically equivalent to an OR gate followed by an inverter. The logic or Boolean expression given for a logic NOR gate is that for Logical Multiplication which it performs on the complements of the inputs.

Boolean expression for A NOR B which gives output Q is

Mutual inductance of the pair of coils depends on the distance between two coils and geometry of two coils. The rate of change of current impacts the emf in coil.

The radiation resistance of a loop antenna is Rrad = (320π⁴S²) / λ⁴.

Now S is area.
If the area of a single turn is A, then for n turns S = nA.

Thus, Rrad = (320π⁴n²A²) / λ⁴.

⇒ Rrad ∝ n².

Now if Rrad = 0.04 and n₁ = 1, then for Rrad₂ = 1Ω, we have n₂:

Rrad₁ / Rrad₂ = n₁² / n₂²

⇒ 0.04 / 1 = 1 / n₂²

⇒ n₂ = √(1 / 0.04) = √25 = 5.

A reverse anode to cathode voltage will tend to interrupt the normal cycle and turn off the SCR. This way is also known as forced commutation.

Let,
I = ∫[0 to π/2] [(sin x - cos x) / (1 + sin x cos x)] dx

Now, we know that,
∫[0 to π/2] f(x) dx = ∫[0 to π/2] f(π/2 - x) dx

Thus, the integral can also be written as
I = ∫[0 to π/2] [(cos x - sin x) / (1 + cos x sin x)] dx

Adding the above two integral equations, we get
2I = 0
∴ I = 0

In a comparator if V ⁺ > V ^- then output is V_sat+ and if V ⁺ > V ^- then output is V_sat ^-. In the question V ^- is set to be ground, hence when during positive cycle of input sinusoid we get output as V_sat + and V_sat ^- during negative cycle. The overall output is a square wave.

The feedback amplification factor is given by A_f = A / (1 + βA), where A is open loop gain and βA is loop gain.

As feedback increases, the gain decreases, thereby bandwidth increases since the gain bandwidth product of the Op-amp is constant. Also, feedback increases input impedance while decreasing output impedance.

In this problem, compared to a non-inverting amplifier with unity gain, the 100 gain amplifier has less negative feedback, leading to lesser bandwidth and input impedance.

The sensitivity of the feedback amplifier to Op-amp amplification factor variances is given by:
ΔA_f / A_f = (1 / (1 + βA)) (ΔA / A) where (1 / (1 + βA)) = A_f / A

Therefore, compared to a unity gain amplifier, a 100 gain amplifier is more sensitive to changes in open loop Op-amp gain.

A general expression of FM signal is given by:

Thus, instantaneous frequency, f^_i = f_c + k_f m(t)

Now, maximum frequency deviation, Δf_max = k_f m(t)_max

Thus, maximum frequency deviation at any time is the function m(t) simply scaled by k_f = 25.

For Binary addition use the following rule:

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 0 and a carry = 1

So, using these rules the binary addition of 1001 and 1101 gives 10110.

Piggybacking is a bi-directional data transmission technique in the network layer (OSI model). It makes the most of the sent data frames from receiver to emitter, adding the confirmation that the data frame sent by the sender was received successfully (ACK acknowledge). This practically means, that instead of sending an acknowledgement in an individual frame it is piggy-backed on the data frame.

The bandwidth and raise time of a system is defined by relation Bt_r = 0.35
Hence as bandwidth is increased the raise time will decrease i.e. system response is faster.

In induction motor, the rotor has fan blades to circulate cooling air through the motor frame.

A satellite transponder consists of a receiver and a transmitter to relay microwave transmission between one point on earth to another point. It uses separate range of frequencies for data transmission known uplink and downlink frequencies for each direction of transmission.

In ideal voltage source the load voltage is independent of load current and it is always equal to source voltage.
Whereas In ideal current source the load current is independent of load voltage  and it is always equal to source current.

The bit rate of the signal is R_b = 36 kHz × log_₂256 = 36 kHz × 8 = 288 kHz
Thus, bit duration, T_b = 1 / R_b = 1 / (288 × 10³) = 3.47 µsec

Equilibrium is the status of the body when it is subjected to a system of forces. If the resultant force is equal to zero it implies that the net effect of the system of forces is zero this represents the state of equilibrium.

In series the equivalent capacitance will be C = 1.6 μF. The charge will be same on each capacitor. Its value will be Q = CV = 1.6 × 10^-6 × 300 = 4.8 × 10^-4 C. The potential difference across C₁ and C₂ will be 240 V and 60 V respectively. The energy stored in the system will be 1/2CV² = 7.2 × 10^-2 J.

In TE₃₀ mode, the electric field is transverse to the guide axis. Thus, the component  does not exist.

A step up transformer is one in which the voltage in secondary side is more than that of primary side. However, the KVA remains the same.

In a P-type semiconductor the Fermi level is close to valence band while in n-type the Fermi level is closer to conduction band. Only in intrinsic semiconductor it is at center of conduction and valence band.

The main advantage of delta modulation over PCM is its simple circuit. The simple circuit is a direct consequence of a quantizer having only two quantisation levels.