Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  Test: Network Elements - 1 - Electronics and Communication Engineering (ECE) MCQ

Test: Network Elements - 1 - Electronics and Communication Engineering (ECE) MCQ


Test Description

10 Questions MCQ Test - Test: Network Elements - 1

Test: Network Elements - 1 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Test: Network Elements - 1 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Network Elements - 1 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Network Elements - 1 below.
Solutions of Test: Network Elements - 1 questions in English are available as part of our course for Electronics and Communication Engineering (ECE) & Test: Network Elements - 1 solutions in Hindi for Electronics and Communication Engineering (ECE) course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt Test: Network Elements - 1 | 10 questions in 30 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
Test: Network Elements - 1 - Question 1

Find the energy (power) drawn by the circuit shown in the figure below.

Detailed Solution for Test: Network Elements - 1 - Question 1

Sign convention for Power delivered and absorbed:
When current is leaving the positive voltage terminal of the element or current is entering the negative voltage terminal of the element then the particular element is delivering power.
∴ P = +Vi

  • When current is entering the positive voltage terminal of the element or current is leaving the negative voltage terminal of the element then the particular element is absorbing power.

∴ P = -Vi

  • The power delivered to an element is denoted by the positive sign, whereas, power supplied by an element is denoted by the negative sign.
  • In figure (i), when current enters through the positive terminal of the circuit, the power has a positive sign.
  • In figure (ii) when the current leaves the positive terminal, the power has a negative sign

Calculation:
Given that, 
V = 2 volts, 
i = 3 A

In the above circuit, the current is leaving the positive voltage terminal of the element or the current is entering the negative voltage terminal of the element then the particular element is delivering power.
Thus, the power drawn by the circuit shown in the above figure can be calculated as
P = + Vi
=> P = +(2 × 3)
∴ P = + 6 W

Test: Network Elements - 1 - Question 2

For the circuit shown in the figure, V1 =8 V, DC and I1 =8A, DC. The voltage Vab  in Volts is ___ (Round off to 1 decimal place).

Detailed Solution for Test: Network Elements - 1 - Question 2

Reraw the circuit:

Now, using voltage division,

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Network Elements - 1 - Question 3

Three resistors of 6 Ω are connected in parallel. So, what will be the equivalent resistance?

Detailed Solution for Test: Network Elements - 1 - Question 3

When resistances are connected in parallel, the equivalent resistance is given by

When resistances are connected in series, the equivalent resistance is given by

Calculation:
Given that R1 = R2 = R= 6 Ω and all are connected in parallel

​​​​​​​

Test: Network Elements - 1 - Question 4

In the given circuit, the value of capacitor C that makes current I = 0 is __________μF.

 

Detailed Solution for Test: Network Elements - 1 - Question 4

Test: Network Elements - 1 - Question 5

If 5 A of electric current flows for a period of 3 minutes, what will be the amount of charge transferred?

Detailed Solution for Test: Network Elements - 1 - Question 5

Electric current: If the electric charge flows through a conductor, we say that there is an electric current in the conductor.
If Q charge flow through the conductor for ‘t’ seconds, then the current given by that conductor is

Q = I × t
I = current
t = times

Calculation:
Given I = 5 amp
t = 3 min = 180 sec
Q = I × t
Q = 5 × 180 = 900 C

Test: Network Elements - 1 - Question 6

The power supplied by the 25 V source in the figure shown below is ________W.

Detailed Solution for Test: Network Elements - 1 - Question 6

Using KCL at node , we get
I + 0.4I = 14
I= 10A
Now, Power supplied,
P = 25 x 10  = 250W

Test: Network Elements - 1 - Question 7

In the circuit shown below, the voltage and current sources are ideal. The voltage (Vout) across the current source, in volts, is

Detailed Solution for Test: Network Elements - 1 - Question 7

So, Vout = (5 x 2) + 10 = 20V

Test: Network Elements - 1 - Question 8

In the given circuit, the current supplied by the battery, in ampere, is _______.

Detailed Solution for Test: Network Elements - 1 - Question 8

Applying KCL at node A,
-I1 + I2 + I2 = 0
2I2 = I1  .... (i)
and applying KVL in loop ABCD,
1 - I1 - I2 -  I2 = 0
I1 + 2I2  = 1 .... (ii)
From equation (i) and (ii),

Test: Network Elements - 1 - Question 9

When two identical resistors are connected in series across a battery, the power dissipated is 10 W. If these resistors are connected in parallel across the same battery, the total power dissipated will be

Detailed Solution for Test: Network Elements - 1 - Question 9

When two or more electrical devices of the same voltage are connected in parallel connection, their equivalent power rating can be calculated as
Peq = P1 + P2 + P3 + ……
When ‘n’ number of devices of the same voltage and same power rating (P) are connected in parallel connection, their equivalent power rating can be calculated as
Peq = nP
When two or more electrical devices of the same voltage rating are connected in a series connection, their equivalent power rating can be calculated as

When ‘n’ number of devices of the same voltage and same power rating (P) are connected in series connection, their equivalent power rating can be calculated as

Calculation:
When two identical bulbs are resistors are connected in series across a battery, their equivalent power (Peq) is ‘10 W’

∴ The power rating of each bulb, P = 20 W

When these two resistors are connected in parallel, then the equivalent power dissipated is
Peq = 2 × P
Peq = 40 W

Alternate Approach:
Series circuit: 
In the series circuit, the voltage drop across each resistor = V/2
The total current through the circuit = V/2R amps.
The power dissipated by each resistor = V/2 × V/2R = V2/4R
Which must be equal to 5 Watts in order to make a total of 10 W.

Parallel circuit:
The voltage drop across each resistor = V.
The current through each resistor = V/R.
Multiplying these gives V2/R which is 4 times the series value of V2/4R
∴ The total power dissipated in parallel circuit = 4 × 10 = 40 W

Test: Network Elements - 1 - Question 10

Consider an element represented by the relationship between current i (t) and voltage v (t) as follows: v(t) = i2(t). This device is classified as:

Detailed Solution for Test: Network Elements - 1 - Question 10

Linear Element: An element is linear if :

  • It follows additivity and superposition
  • V-I graph is a straight line passing through the origin


Time variant Element: An element is said to be time-variant if the V-I graph varies with time.

Application:
Given V-I characteristic : v(t) = i2(t)
The characteristics can be drawn as,

It is not a straight line so non-linear
It is time-invariant because in any network analysis problem we take the circuit parameters V and I to be time-invariant.

Alternate Method:

Shifting the input, we get:
v1(t) = i12 (t - t1)
Now, shifting the output with the same amount, we can write:
v2 (t) = v1(t - t1) = i12 (t - t1) = v1 (t)

Since v1(t) = v2(t), the system is time invariant.

Information about Test: Network Elements - 1 Page
In this test you can find the Exam questions for Test: Network Elements - 1 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Network Elements - 1, EduRev gives you an ample number of Online tests for practice

Top Courses for Electronics and Communication Engineering (ECE)

Download as PDF

Top Courses for Electronics and Communication Engineering (ECE)