Test: Basic & Electric Circuits- 2

# Test: Basic & Electric Circuits- 2

Test Description

## 20 Questions MCQ Test Topicwise Question Bank for Electrical Engineering | Test: Basic & Electric Circuits- 2

Test: Basic & Electric Circuits- 2 for Electrical Engineering (EE) 2023 is part of Topicwise Question Bank for Electrical Engineering preparation. The Test: Basic & Electric Circuits- 2 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Basic & Electric Circuits- 2 MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Basic & Electric Circuits- 2 below.
Solutions of Test: Basic & Electric Circuits- 2 questions in English are available as part of our Topicwise Question Bank for Electrical Engineering for Electrical Engineering (EE) & Test: Basic & Electric Circuits- 2 solutions in Hindi for Topicwise Question Bank for Electrical Engineering course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Basic & Electric Circuits- 2 | 20 questions in 60 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Topicwise Question Bank for Electrical Engineering for Electrical Engineering (EE) Exam | Download free PDF with solutions
 1 Crore+ students have signed up on EduRev. Have you?
Test: Basic & Electric Circuits- 2 - Question 1

### A current of 3 A flows through a resistor of 20 ohms. The energy dissipated in the resistor per minute is

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 1

Power dissipated by the resistor is
P = I2 R = 32 x 20 = 180 W
Now, energy dissipated by the resistor in one hour (or 60 min)
= [180 x 60] watt-min
∴ Energy dissipated by the resistor in one minutes Test: Basic & Electric Circuits- 2 - Question 2

### A long uniform coil of a inductance L henries and associated resistance R ohms is physically cut into two exact halves which are then rewound in parallel. The resistance and inductance of the combination are

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 2

We know that, and when coil is cut into two equal halves, (Area of cross - section = constant)
So, Also,   So, Hence, new values of resistance and inductance which are reconnected in parallel is and Test: Basic & Electric Circuits- 2 - Question 3

### In the circuit shown, the power dissipated in the resistor R is 1 W when only source ‘1’ is present and ‘2’ is replaced by short circuit. The power dissipated in the same resistor R is 4 W when only source '2’ is present and '1' is replaced by a short circuit When both the sources ‘1’ and ‘2’ are present, the power dissipated in R will be Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 3

We have; or,  or, When both the sources are present, net current through R will be
I = (I2 - I1
[as polarity of |V1| is reverse]
So, power loss in R is   Test: Basic & Electric Circuits- 2 - Question 4

If v, w, q stand for voltage, energy and charge, then v can be expressed as:

Test: Basic & Electric Circuits- 2 - Question 5

In the figure shown below, φ = power-factor angle, W = watts, VA = volt ampere and VAr = volt-ampere .reactive for an ac circuit. The correct figure is

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 5 Test: Basic & Electric Circuits- 2 - Question 6

A circuit possesses resistance R and inductive reactance XL in series, its susceptance is given by

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 6

Susceptance is the imaginary part of admittance,  or, Here, G = Conductance and S = Susceptance Test: Basic & Electric Circuits- 2 - Question 7

If V = a + jb and I = c + jd, then the power is given by

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 7

P = VI* = (a + jb).(c + jd)*
= (a + jb) (c - jd)
= (ac + bd) + j(bc - ad)
= P + jQ
Here, P = Active power
and Q = Reactive power

Test: Basic & Electric Circuits- 2 - Question 8

A 230 V, 100 W bulb has resistance RA and a 230 V, 200 W bulb has resistance RB. Here,
1. RA > RB
2. RB > RA
3. RA = 2 RB
4. RB = 2 RA
5. RA = 4 RB
From these, the correct answer is

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 8

Power, Since voltage ratings of both bulbs are same, therefore
PR = constant
or, PA RA = PB RB
or, 100 x RA = 200 x RB
or, RA = 2 RB
Thus, RA > RB

Test: Basic & Electric Circuits- 2 - Question 9

The voltage phapor of a circuit is 10∠15° V and the current phasor is 2∠- 45° A. The active and the reactive powers in the circuit are

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 9

S = VI* = (10∠15°) (2∠ - 45°)*
= 20∠15 + 45° = 20∠60°
= 20 (cos 60° + j sin 60°) = (10 + j10√3) = (10 + j17.32)
= P+ jQ

Test: Basic & Electric Circuits- 2 - Question 10

In the circuit shown below, if the power consumed by the 5 Ω resistor is 10 W, then power factor or the circuit will be Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 10

Given, power consumed is
P = IReq
or, Also, or, |Z| = 25Ω
or, or, Hence, p.f. of given circuit is Test: Basic & Electric Circuits- 2 - Question 11

The minimum requirements for causing flow of current are.

Test: Basic & Electric Circuits- 2 - Question 12

Ohm’s law is applicable to

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 12

Carbon resistor and semiconductors have nonlinear relationship between V and I. Hence, Ohm’s law is not applicable. Also, these are not bilateral.

Test: Basic & Electric Circuits- 2 - Question 13

For a fixed supply voltage, the current flowing through a conductor will increase when its

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 13 When l is reduced. I will be increased and vice-versa.

Test: Basic & Electric Circuits- 2 - Question 14

Two registors R1, and R2 give combined resistance of 4.5 Ω when in series and 1 Ω when in parallel. The resistances are

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 14

R1 + R2 = 4.5 ...(i)
and ∴ (R1-R2)2 = (R1 + R2)- 4R1R2 or, ...(ii)
On solving equations (i) and (ii), we get
R1 = 3 Ω and R2 = 1.5 Ω
or, R1 = 1.5 Ω and R2 = 3Ω.

Test: Basic & Electric Circuits- 2 - Question 15

Which of the following is not equivalent to watts?

Test: Basic & Electric Circuits- 2 - Question 16

Two heaters, rated at 1000 W, 250 V each are connected in series across a 250 V, 50 Hz ac mains. The total power drawn from the supply would be

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 16

For series connection,
Req = R1 + R2
or, or, Given,
P1 = P= 1000 W
∴ Peq = 500 Watt

Test: Basic & Electric Circuits- 2 - Question 17

A 100 watt light bulb burns on an average of 10 hours a day for one week. The weekly consumption of energy will be

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 17

Energy consumption per week is
W= 100 x 10 x 7
= 7000 kW - hr
= 7 units

Test: Basic & Electric Circuits- 2 - Question 18

Assertion (A): The direction of flow of conventional current is taken opposite to that of electrons.
Reason (R): Electrons have negative charge.

Test: Basic & Electric Circuits- 2 - Question 19

Match List-I (Materials) with List-lI (Range of resistivity) and select the correct answer using the codes given below the lists:
List-I
A. Conducting materials
B. Semiconductor materials
C. insulating material
List-II
1. 100 to 102 Ω-m
2. 10-8 to 10-6 Ω-m
3. 1012to 1018 Ω-m
4. 1020 to 1030 Ω-m
Codes: Test: Basic & Electric Circuits- 2 - Question 20

A constant current source supplies a current of 200 mA to a load of 2 kΩ When the load is changed to 100 Ω, the load current is

Detailed Solution for Test: Basic & Electric Circuits- 2 - Question 20

For a constant C.S., current will remain constant for all values of loads.

## Topicwise Question Bank for Electrical Engineering

207 tests
Information about Test: Basic & Electric Circuits- 2 Page
In this test you can find the Exam questions for Test: Basic & Electric Circuits- 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Basic & Electric Circuits- 2, EduRev gives you an ample number of Online tests for practice

## Topicwise Question Bank for Electrical Engineering

207 tests (Scan QR code)