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

Answer.1. 10

Explanation

The power triangle or Impedance triangle of the AC circuit is shown Below

Cosφ = Base ⁄ Hypotenuse = Active Power  ⁄ Apparent Power
 Active Power = Apparent power × Cosφ
Sinφ = Perpendicular ⁄ Hypotenuse = Reactive Power  ⁄ Apparent Power

Reactive Power = Apparent power × Sinφ

or

Reactive Power = (Active Power × Sinφ) ⁄ Cosφ

cosφ = 0.8

Sin²φ = 1 − cos²φ

Sin²φ = 1 − (0.8)²

Sinφ = 0.6

Reactive Power =(20 × 0.6) ⁄ 0.8

Reactive Power = 15 VAR

Answer.1. 40 W

Explanation

Comparing the 60W bulb to a 40W bulb means that the 60W bulb consumes 20W more power than the 40W bulb. Now note the question its a 40W and 60W bulbs we talking about, not resistors! For the bulbs, wattage is brightness. The 40W and 60W bulbs will be rated for a particular (I mean same) voltage, not different. So considering power P=V²/R power is inversely proportional to resistance (bulbs rated for the same voltage). Which means that the 40W bulb has more resistance across it compared to the 60W bulb. ( Compare in terms of brightness, a 0W bulb has to glow so dim just because it has high resistance while a 100W bulb glows very much bright because its resistance is low)

Connecting the 2 bulbs in series implies that the current is the same, hence the power consumption totally depends on how voltage gets divided between the bulbs. Naturally, the 40W bulb will have more voltage across it since it has more resistance. Hence, the 40W bulb will glow brighter.

Answer.1. M⁻¹ L⁻² T³ A²

Explanation

The conductance is given as

Conductance = 1 ⁄ Resistance

Firstly consider the resistance.

Ohm’s law states that electric current flowing through the conductor is directly proportional to the potential difference between its two ends when the temperature and other physical parameters of the conductor remain unchanged.

$V=IR$

$\Rightarrow R=\frac{V}{I}$

Now $V$ has units of (electric field)*(distance).

But the electric field has units (force)/(charge).

Also, charge has dimensions of (current)(time) and force has dimensions (mass)(length)/(time)².

Thus, dimensions of $V$ is,

$\left[V\right]=\frac{LML{T}^{-\mathrm{2 ⁄}}}{AT}$

$\Rightarrow \left[V\right]=\; M\; L²\; T ⁻³\; A⁻¹$

Dimensional formula for I = A

Put these values in equation (1), we get:

The dimensional formula for R = M L² T⁻³ A⁻¹/ A

Dimensional formula for R = [M L² T ⁻³ A⁻²]

Now the Dimensional formula of conductance will be

G = 1 ⁄ [M L² T ⁻³ A⁻²]

G = [M⁻¹ L⁻² T³ A²]

Answer.1. εA/d

Explanation

The capacitance C of a capacitor is the ratio of the magnitude of the charge on either conductor (plate) to the magnitude of the potential c^lif^rerence between the conductors (plates).

C = Q/V

THE PARALLEL-PLATE CAPACITOR

The figure shows the structure of a parallel plate capacitor. It is made up of two parallel plates of the area of cross-section A, kept in an insulating medium and separated from each other by a distance of d metros. Then, if the two plates are applied with potential, one of the plates will be positively charged and the other negatively charged.

Let the charge on the plates were given by Q coulombs. 

The capacitance of a device depends on the geometric arrangement of the conductors. The capacitance of a parallel-plate capacitor with plates separated air can be easily calculated from three facts.

The magnitude of the electric field between two plates is given E = σ/ε_o, where σ is the magnitude of the charge per unit area on each plate and ε_ο = Permittivity of free space = 8.85 × 10^-12

Second, the potential difference between the two plates is ΔV = Ed, where d is the distance between the plates.

Third, the charge one plate is given by q = σA, where A is the area of the plate. Substituting the three facts into the definition of capacitance gives the desired result:

[latex]\begin{array}{l}C = \dfrac{Q}{{\Delta V}} = \frac{{\sigma A}}{{Ed}} = \dfrac{{\sigma A}}{{\left( {\sigma /{\varepsilon _o}} \right)d}}\\\\C = {\varepsilon _o}\dfrac{A}{d}\end{array}[/latex]

Answer.4. Greater than the largest inductance of the combination

Explanation

Let us Suppose that the three Resistance L₁, L₂, L₃ of 1 Henry, 2 Henry, and 3 Henry are connected in series respectively.

Now add all the inductance of the circuit we get

L_eq = L₁ + L₂ + L₃
L_eq = 1 + 2 + 3 = 5H

Now from the above result, we can conclude that the equivalent Inductance in the series combination is greater than the largest Inductance in the combination (since the largest Inductance was 3H).

Therefore, the Option.4. is correct.

Answer.3. Conductivity

Explanation

Resistivity:-

The resistance of a material having unit length and the unit cross-sectional area is known as Specific Resistance or Resistivity.

In the S.I. system of units

Conductivity

The electrical resistance of an electrical conductor is a measure of the difficulty to pass an electric current through that conductor. The inverse quantity is electrical conductance or conductivity and is the ease with which an electric current passes. Electrical conductance is measured in siemens (S) σ and denoted by (G).

The material having a high value of conductivity is a good conductor of electricity while the material having the low value of conductivity is a good insulator.

Answer.2. L/4

Explanation

Inductors are sometimes connected in parallel in a circuit. Two or more circuit elements are said to be in parallel when they share the same pair of nodes. In a parallel circuit, the voltage across each of the parallel circuit elements is the same, but the current through each of the elements might be different. Let us consider four inductors that are connected in a parallel to a source voltage V as shown in Fig. In these cases, the voltage across each of these resistors will be V.

Note:- Inductor connected in parallel combine the same way as the resistor in parallel.

According to Ohm’s law, the currents the current I₁, I₂, I₃, I₄ through the corresponding inductors L₁, L₂, L₃ & L₄ can be written as

I₁ = V ⁄ L₁

I₂ = V ⁄ L₂

I₃ = V ⁄ L₃

I₄ = V ⁄ L₄

Applying KCL to the circuit we get

I = I₁ + I₂ + I₃ + I₄

[latex]\begin{array}{l}I = \dfrac{V}{{{L_1}}} + \dfrac{V}{{{L_2}}} + \dfrac{V}{{{L_3}}} + \dfrac{V}{{{L_4}}}\\\\I = V\left( {\dfrac{1}{{{L_1}}} + \dfrac{1}{{{L_2}}} + \dfrac{1}{{{L_3}}} + \dfrac{1}{{{L_4}}}} \right)\\\\{L_1} + {L_2} + {L_3} + {L_4} = L\\\\\dfrac{I}{V} = \left( {\dfrac{{1 + 1 + 1 + 1}}{L}} \right)\\\\R = \dfrac{L}{4}\end{array}[/latex]

Answer.1. 0.1

Explanation

Conductance “G” is measured in Siemens and it is the reciprocal the resistance 

G = 1/R = G = 1/V/I

G = I/V

Where

I = current = 3A

V = voltage = 30V

G = 3/30 = 0.1 Siemens

Answer.4. 90

Explanation

Given

Resistance R = 10Ω

Current I = 3 A

Power dissipated by the resistor is

P = I²R

P = 3² × 10

P = 90 watts

Answer.4. 26

Explanation

Resistance Temperature Coefficient:

The change in resistance of a material with the increase in temperature can be expressed b means of the temperature coefficient of resistance. Consider a conductor having resistance R_o at 0°c and R_t at t°c. From the above discussion, we can conclude that the change in the resistance i.e (R_t – R_o) is

1. Directly proportional to the initial resistance R_o
2. Directly proportional to the rise in temperature t°c.
3. Depends on the nature of the material for conductor metals and alloy

Hence

(R_t – R_o) ∝ R_ot

(R_t – R_o) = αR_ot

R_t  = R_o(1 + α_ot)

Where α_o is constant and called as the temperature coefficient of resistance at 0°c and its value depends upon the nature of material and temperature.

R_o = 10 ohms

T = 40°

α =.04

R_T = ?

Hence the value of resistance at 40°C is

R_T = 10(1 + .04 × 40)

R_T = 26 Ω

Answer.2. 90

The energy stored in the magnetic field of an inductor can be expressed as

W = 1/2 L I²

where

W = energy stored (joules, J)

L = inductance (henrys, H) = 5 H

I = current (amps, A) = 6A

W = (5 × 6²) ⁄ 2

W = 90 Joules