SSC JE Basic Electrical Questions (2009 – 2018) Solved | MES Electrical
Ques.1. The S.I unit of Electric energy is (SSC 2018 -S-1)
Watt
Volts
Ampere
Joul
Answer.4. Joul
Explanation:-
Electrical Power
The rate at which the electric work is done in an electric circuit is called electric power. The SI unit of power is joule per second or watt.
Note:-
Electrical Energy
Electric energy is the total amount of electrical work done in an electrical circuit. The electric energy can also be defined as the product of power and time. The S.I Unit of Electrical- Energy is joule or watt-sec.
Ampere is the Unit of Current
Volt is the S.I unit of Potential Difference.
Ques.2. What is the value of current I4 (in A) for the given junction? (SSC 2018 -S-1)
4
−4
6
−6
Answer.3.6.
Explanation:-
According to Kirchhoff’s Current Law: At any point in an electrical circuit, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.
From the above Diagram
Current Flowing towards the Point: I2, I6, I4
Current Flowing Away from the Point: I1, I3, I5
Hence I2 + I6 + I4 = I1 + I3+ I5
Putting the value of the current
2A + 7A + I4 = 4A + 3A + 8A
I4 = 15A − 9A = 6A
I4 = 6A
Ques.3. Which of the following statement is TRUE? (SSC 2018 -S-1)
The equivalent resistance in series combination is larger than the largest resistance in the combination
The equivalent resistance in series combination is smaller than the largest resistance in the combination
The equivalent resistance in series combination is equal to the smallest resistance in the combination
The equivalent resistance in series combination is equal to the largest resistance in the combination
Answer.1. The equivalent resistance in series combination is larger than the largest resistance in the combination
Let us Suppose that the three Resistance R1, R2, R3 of 1 ohm, 2 ohms, and 3 ohms are connected in series respectively.
Now add all the resistance of the circuit we get
Req = R1 + R2 + R3 Req = 1 + 2 + 3 = 5Ω
Now from the above result, we can conclude that the equivalent resistance in the series combination is larger than the largest resistance in the combination (since the largest resistance was 3Ω).
Therefore, Option.1. is correct.
Ques.4. The resistivity of a conductor depends upon (SSC 2018 -S-1)
Pressure
Temperature
Degree of Illumination
Shape of cross-section
Answer.2. Temperature
Explanation
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
Hence, the unit of resistivity is ohm-meter (Ω-m).
Specific Resistance depends only on temperature and material of the conductor but not on its dimensions of the conductor, on which resistance depends, and mechanical deformation such as stretching, etc. As ρ depends only on the material of a conductor at a given temperature, hence it is a characteristic constant.
Material with the highest value of resistivity is the best insulator while the material with the low value of resistivity is a good conductor.
Ques.5. What will be the equivalent capacitance (in mF) of three capacitors connected in a series having the capacitance of 0.04 mF, 0.08 mF, and 0.02 mF respectively? (SSC 2018 -S-1)
0.026
0.032
0.065
0.011
Answer.2.0.011
Explanation
The three capacitor C1, C2, C3 is connected in a series as shown in the figure
Now the Equivalent capacitance connected in the series will be
1/Ceq = 1/C1 + 1/C2 + 1/C3
1/Ceq = 1/0.04 + 1/0.08 + 1/0.02
1/Ceq = 25 + 12.5 + 50
1/Ceq = 87.5
Ceq = 1/87.5
Ceq = 0.011
Ques.6. Determine the voltage (in V) of a battery connected to a parallel plate capacitor (filled with air) when the area of the plate is 10 square centimeters, the separation between the plate is 5 mm and the charged stored on the plate is 20 pC. (SSC 2018 -S-1)
12.3
10.3
11.3
14.3
Answer.3.11.3
Explanation
Consider the parallel plate capacitor connected to the battery as shown in the figure
The capacitance “C” of the parallel plate capacitor is given as
C = εοA/d
Where A = Area = 10 square centimeter = 10 × 10−4 meter εο = Permittivity of free space = 8.85 × 10-12 d = distance between the parallel plate capacitor = 5mm = 5 × 10−3 meter
q = Charge stored in the capacitor = 20 pC = 20 × 10−12
C = (8.85 × 10-12 × 10 × 10−4) ⁄ 5 × 10−3
C = 1.77 × 10−12 F
The self-capacitance of a conductor is defined by
C = q ⁄ V
Hence the voltage across the capacitor
V = q ⁄ V = (20 × 10−12) ⁄ (1.77 × 10−12)
V = 11. 3 Volts
Ques.7. Determine the value of the current (in A) through both the resistors of the given circuit. (SSC 2018 -S-1)
−2, −1.5
2, 1.5
−2, 1.5
2, −1.5
Answer.2. 2, 1.5
Explanation:-
Current through the 10Ω resistance
I1 = V/R = 20/10 = 2A
I1 = 2A
Now current through the 20Ω resistance
I2 = V − (-10)/R = (20 + 10)/20 = 1.5 A
I2 = 1.5 A
Ques.8. Which one of the following statements is TRUE? (SSC 2018 -S-1)
Kirchhoff’s Law is not applicable to the circuit with the passive elements
Kirchhoff’s Law is not applicable to circuits with the distributed elements
Answer.4.Kirchhoff’s Law is not applicable to circuits with the distributed elements
Explanation
Types of Elements in a Network Circuit
Passive Network/Element:- Passive network is one, which contains no source of emf in it i.e., R, L, C.
An element that is capable of receiving power is called passive element Ex: Resistor, Inductor, Capacitor.
Active Network/Element:- Active network is one that contains one or more than one source of emf, then the network is called an active network.
An element that is capable of supplying power is called an active element.
The network element can further be classified into more different groups
Linear or Non-Linear
Unilateral or Bilateral
Time-variant or Time invariant
Lumped or distributed Network
Lumped Elements:- When it is possible to separate the elements of a network physically like resistors, inductors, capacitors, etc. the elements are called as lumped elements. Kirchhoff’s law is only applicable to the circuit with lumped elements.
Distributed Elements:- When it is not possible to separate the elements of a network physically the elements are known as distributed elements. The best example of such a network is a transmission line where resistance, inductance, and capacitance of a transmission line are distributed all along its length and cannot be shown as a separate element, anywhere in the circuit.
Ques.9. Which one of the following is a passive element to an electric circuit? (SSC 2018 -S-1)
Current source
Voltage source
Resistors
Battery
Answer.3. Resistors
Explanation
Passive Network/Element:- Passive network is one, which contains no source of emf in it i.e., R, L, C.
An element that is capable of receiving power is called passive element Ex: Resistor, Inductor, Capacitor.
Active Network/Element:- Active network is one that contains one or more than one source of emf, then the network is called an active network.
An element that is capable of supplying power is called an active element. Examples are the current source, voltage source, battery, cell, Transistor, etc.
Ques.10.Which one of the following is the mathematical expression of the Ohm’s Law? (SSC 2018 -S-1)
V = I
V = R/I
V = I.R
V = R
Answer.3.V = I.R
Explanation
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.