GATE Practice Test: Electrical Engineering (EE)- 1 - Electrical Engineering (EE) MCQ

The sentence talks about a past incident and a condition has been mentioned. Thus option 2 fits here correctly. 'Would have' is the correct tense to be used here. Option 1 is illogical. 'Would' must be used here and not 'will' as it is in the past tense. Past perfect tense is incorrect here.

The word that fits here is a noun thus options 3 and 4 are eliminated. The form should be singular as for abstract nouns we do not use the plural form. Option 1 is thus the correct answer.

The word powerful is an adjective and 'empowering' is a verb.

The radius of outer circle = r

∴ Diagonal of square = 2r

Side of square (a) = 

Now, the radius of the inner circle = 

Average speed = Total distance/Total time

Total distance = a + a + a = 3a

Total time = 

Average speed  = 

The sentence mentions a person doing something as per her wishes thus the first word must be a positive reaction. 'Restive' which means 'restless' and 'jinxed' which means 'cursed' are incorrect here. Between options 1 and 3,the word 'elated' which means 'too happy' and the word 'venture' which means 'undertake a risky or daring journey or course of action' fit here correctly. The word 'cease' means 'stop' and does not convey a proper meaning. Thus option 3 is the correct answer.

As, the faster runner is twice as fast as the slowest runner, the faster runner would have complete two rounds by the time the slowest runner completes one round.

Their first meeting takes place after the fastest runner takes 6 min to complete two rounds and the slowest runner completes one round.

Time taken to complete one round by fastest runner = 6/2 = 3 minutes

Rounds needed to complete the race = 1600/400 = 4

Time taken to complete the race = 4 × 3 = 12 min

The young one of a horse is called a foal. Similarly, the young one of a goose is known as gosling. Thus, option 2 contains the pair that best expresses a relationship similar to that expressed in the given pair. In all the other pairs, the first word expresses the male form of the latter.

Since teenagers here have tender age and commit crimes unknown to its consequences, clearly they should not be treated as any criminal. They need to be taught so that they can make difference between right and wrong and do not commit these crimes in future. Hence inference in option 1 is correct.

450 y = n³ implies that 450 y is the cube of an integer. 

When we prime-factorize the cube of an integer, we get 3 (or a multiple of 3) of every prime factor.

8 is the cube of an integer because 8 = 2³ = 2*2*2.

Thus, when we prime-factorize 450 y, we need to get at least 3 of every prime factor. 

Here's the prime factorization of 450 y:

450 y = 2 * 3² * 5² * y 

Since 450 provides only one 2, two 3's, and two 5's, and we need at least 3 of every prime factor, the missing prime factors must be provided by y. Thus, y must provide at at least two more 2's, one more 3, and one more 5. 

Smallest possible case: 

y = 2² * 3 * 5

In this problem, cumulative frequency is given, so converting into frequency

From the table, it is clear that option b, i.e.

60% of the scholars published at least 2 papers is correct.

A matrix is singular if and only if its determinant is zero.

x(3+1)−2(x²+1) = 0−2x²+4x−2 = 0

x²−2x+1 = 0

(x−1)² = 0

x = 1

According to formula, L(tⁿ f(t)) =​

Here, f(t) = sin (2t) ⇒ F(s) =

∴ L(t²sin (2t)) =  = 

ii) (4 - x)_x=2 = 2

Hence at x = 2, f(x) is continuous

∴ f(x) is not differentiable.

The given limit is in 0/0 form, So

Applying L hospitals rule

⇒ log b - 1 = -1 - log b

⇒ 2 log b = 0

⇒ b = 1

y log y dx + (x – log y) dy = 0

It is a Leibnitz’s equation in x.

Integrating factor =

By applying Laplace transform,

PMMC meter reads average value if the output of the XOR gate.

The output of XOR gate will be high when both the inputs are different.

For a phase difference of T/2 both X and Y will be different as shown below.

Equations of hybrid parameter are,

When, V₂ = 0

V₁ = -2I₂

By applying KCL at 0,

⇒ h₂₁ = -2

V₁ = -2I₂ = -2 (-2I₁) = 4I₁

⇒ h₁₁ = 4

When, I₁ = 0

⇒ h₂₂ = 0.5

By applying KVL,

⇒ -V₁ + 2V₂ – I₂ = 0

⇒ -V₁ + 2V₂ – 0.5 V₂ = 0

⇒ h₁₂ = 1.5

For H(z) to be stable the poles of H(z) must be inside the unit circle.

For the inverse of H(z) to be stable the poles inverse of H(z) must be inside the unit circle.

The poles of inverse of H(z) are the zeros of H(z)

Hence, both poles and zeros of H(z) must be inside the unit circle.

By voltage division

By voltage division,

Given circuit is Buck Boost converter

Voltage across inductor would be, V_s = 200 V

V_o = 200 V

⇒ δ = 1 – δ

⇒ δ = 0.5

Sampling period, T_s = 2 × 5 = 10 μs

Sampling rate, 100k samples/sec

V_GS = 0.65 V_p

The drain current of JFET is given as

 (1−0.65)² = 0.1225 = 12.25% 

Electric field at a point P(x, y, z) is given as

In π model, the shunt admittance at each end of the line is 

= j300×10⁻⁶  

= 300×10⁻⁶ ∠90^∘ mho

x[n] = 3 sin (1.3 πn + 0.5 π) + 5 sin (1.2 π n)

Time period = LCM (20, 5) = 20

To turn off the SCR anode current should go less than holding current

⇒ V = 4 × 10^-3 × 4 × 10³

⇒ V = 16 V

Form the options, V = 15 V is correct. for the remaining options, holding current is less than anode current.

Fourier transform of

⇒ g(t) = δ(t) + 2.e^-3|t|

From the given polar plot,

From the options, 5(1 + 0.1s) satisfies this condition.

From the given circuit diagram,

P = A⊕B⊕C

Q = AB + BC + CA

The above two expressions presents sum (P) of a full adder and carry (Q) of a full adder