SSC JE Basic Electrical Questions (2009 – 2018) Solved | MES Electrical

Answer.2. Coulomb

Explanation:-

The S.I unit of electric charge is Coulomb and it is equivalent to ampere per second. The amount of charge transported through any section of an electric circuit in one second, when the current in the circuit is one international ampere. 

Electric charge Q = Current (I) × Time taken (t)

Note:- 

The Henry (symbolized H) is the Standard International ( SI ) unit of inductance.

The Tesla (symbolized T) is the standard unit of magnetic flux density. 

The Weber (symbolized Wb) is the standard unit of magnetic flux. 

Answer.3. 42.4

Explanation

The r.m.s. value or effective value or virtual value of alternating current is 0.707 times the peak value of alternating current or voltage for the half-cycle as well as for full cycle.

I_RMS = 0.707 I_Peak

I_peak = I_RMS ⁄ 0.707 

I_peak = 30 ⁄ 0.707 = 42.4 Ampere

I_peak = 42.4 Ampere

Answer.1. 50

Explanation:-

As we know that the frequency is inversely proportional to the time period

F = 1/t = F = 1 ⁄ 20 × 10⁻³

F = 50 Hz

Answer.3. 8.66

Let the expression be as follows:-

I(t) = I_m sin(ωt ± φ)

Where φ represent the concerned phase-shift

I_m = Maximum value of the current

Time t = 3 sec

I(t) = 10 sin(25t − 15)A

I(t) = 10 sin(25 × 3 − 15)

I(t) = 10sin(75 − 15)

I(t) = 10sin(60)

I(t) = 8.66 A

Answer.4. Minimum

An RLC series circuit the total impedance of the series LCR circuit is given as

Z² = R² +  (X₁ – X₂)²

where X₁ is inductive reactance

and X₂ is capacitive reactance.

At a particular frequency (resonant frequency), we find that X₁=X₂ because the resonance of a series RLC circuit occurs when the inductive and capacitive reactances are equal in magnitude but cancel each other because they are 180 degrees apart in phase. Therefore, the phase angle between voltage and current is zero and the power factor is unity.

Thus, at the resonant frequency, the net reactance is zero because of X₁=X₂. The circuit impedance Z becomes minimum and is equal to the resistance R. “Since the impedance is minimum, the current will be maximum”.

Hence electrical resonance is said to take place in a series LCR circuit when the circuit allows maximum current for a given frequency of alternating supply at which capacitive reactance becomes equal to the inductive reactance.

Answer.2. 30

Explanation

The phase angle of the series RC circuit is given as

φ = tan⁻¹X_C/R

φ = tan⁻¹11.56/20

φ = tan⁻¹0.57

φ = 30°

Answer.3. 15.59

Explanation:-

The power consumed by the three-phase circuit is

P = 3 V_ph I_ph cosφ_ph

In delta connected system

Line voltage is equal to the phase voltage i.e

V_L = V_PH = 200 V

Phase current I_PH = 30 A

phase angle φ = 30°

∴ P = 3 × 200 × 30 × cos30°

P = 15.58 watts

Answer.1. 0.126

Explanation

Susceptance (B) is an expression of the ease with which alternating current ( AC ) passes through capacitance or an Inductance.

Capacitive Susceptance B_C is given as

B_C = 1/X_C Siemens

Capacitive reactance X_C = 1/ωC = 1/2πfC

X_C = 1/2 × π × 50 × 0.4 × 10⁻³

X_C = 7.96Ω

B_C = 1/X_C = 1/7.96

B_C = 0.126 Siemens

Answer.4. 19.6

Explanation

Given

Line Voltage V_L = 400 V

Line current I_L = 40A

Phase angle φ = 45

The reactive power of a three-phase AC circuit is given as

Q = √3 × V_L × I_L × sinφ

Q = √3 × 400 × 40 × sin 45°

Q = √3 × 400 × 40 × 0.707

Q = 19592.9 VAR

or

Q = 19.59 KVAR

Answer.1. 0.5

Explanation

Given

Resistance R = 40 Ω

Inductance L = 8mH = 8 × 10⁻³ H

Capacitance C = 0.02 mF = 0.02 × 10⁻³ H

The Quality factor of the RLC circuit is given as

[latex]\begin{array}{l}Q = \dfrac{1}{R}\sqrt {\dfrac{L}{C}} \\\\Q = \dfrac{1}{{30}}\sqrt {\dfrac{{27 \times {{10}^{ – 3}}}}{{0.03 \times {{10}^{ – 3}}}}} \\\\Q = \dfrac{1}{{40}}\sqrt {400} \\\\Q = 0.5\end{array}[/latex]

Answer.3. 0.4

Explanation

The time taken by the RL circuit to reach steady state is 5 times constant of RL circuit i.e

Total time = 5τ

Also, the time constant of the RL circuit is given as τ = R/L

Total time = 5 × R/L

= (5 × 0.8) ⁄ 10

= 0.4 sec

Answer.2. ML²/T²

Explanation

The dimension of energy is the same as that of work.

Finding the dimension of work:

Work = Force × Displacement

Force = Mass × Acceleration

Acceleration = Velocity/Time

Velocity = Distance/Time

Now we will start solving all the from the bottom i.e from the velocity

Displacement, Length = L & Time = T , Mass = M 

⇒ Velocity = Displacement / Time

Velocity = [ M L T⁻¹]

⇒ Acceleration = Velocity ⁄ Time = [ M⁰ L T⁻¹] ⁄ [T]

Acceleration = [M⁰ L T⁻²] 

⇒ Force = Mass × Acceleration = [M] × [M⁰ L T⁻²] 

Force = [M L T⁻²] 

⇒ Work = Force × Displacement = [M L T ⁻²] × [L]

Work = [M L² T ⁻²] or [M L² ⁄ T²]

Answer.2. 10

Explanation:-

We can determine the relation between apparent power and reactive power by the power triangle

Reactive Power in an A.C circuit is given as

Q = V.I.Sinθ

or

V.I = Q/Sinθ———-(1)

Apparent power in an A.C circuit is

S = V.I

or

S = Q/Sinθ——-(From equation 1)

Power factor cosφ = 0.8

cosφ = 0.8

Sin²φ = 1 − cos²φ

Sin²φ = 1 − (0.8)²

Sinφ = 0.6

S = 60/0.6

S = 10 watt