Online Electronics and Communication Engineering Test - Networks Analysis and Synthesis Test 2
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- Total number of questions: 20.
- Time allotted: 30 minutes.
- Each question carries 1 mark; there are no negative marks.
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Marks : 2/20
Test Review : View answers and explanation for this test.
- 2 wattmeter
- one wattmeter
- 3 wattmeter
A minimum of 2 wattmeters is required to measure 3 phase power. Of course power can be measured by putting one wattmeter in each phase.

EI cos θ = 0.25 or I cos θ = 0.25
From above equations cos θ = 0.707 and θ = 45°.
- R = 0, I = 2.5 A
- R = 3 Ω, the value of V is given by

When R = ∞, V = 5v, Voc = 5V circuit is open
When R = 0, I = 2.5A Isc = 2.5 circuit short circuited
Hence the voltage across 3 Ω is 3 volt.

q = i dt

Convert the inner star into delta.

The 3Ω resistance is short-circuited by capacitor at t = 0.

s domain impedance of each inductance = 2s and that of each capacitor is .
As R increases, Q decreases, bandwidth = . Therefore, bandwidth increases. Hence lower half power frequency will be less than ωr.
β = - p.

Assertion (A): Charge is dual of flux.
Reason (R): Short circuit is dual of open circuit.
Assertion (A): If Z1(s) and Z2(s) are positive real then Z1(s) + Z2(s) as well as 1/Z1(s) and 1/Z2(s) are positive real.
Reason (R): The poles of a positive real function are real or occur in conjugate pairs.