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NEET Exam  >  NEET Tests  >  Test: Phasors (January 30) - NEET MCQ

Test: Phasors (January 30) - NEET MCQ


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10 Questions MCQ Test - Test: Phasors (January 30)

Test: Phasors (January 30) for NEET 2024 is part of NEET preparation. The Test: Phasors (January 30) questions and answers have been prepared according to the NEET exam syllabus.The Test: Phasors (January 30) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Phasors (January 30) below.
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Test: Phasors (January 30) - Question 1
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If a capacitor of capacitance 9.2F has a voltage of 22.5V across it. Calculate the energy of the capacitor.

Detailed Solution for Test: Phasors (January 30) - Question 1

We know that,
 ω=(1/2)CV2
After putting the values,
=(1/2)x9.2x22.5x22.5
=2328.75J
Hence option B is the answer.

Test: Phasors (January 30) - Question 2
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Alternating current is represented by

Detailed Solution for Test: Phasors (January 30) - Question 2

Alternating current is an electric current which periodically reverses direction, as opposed to direct current which flows only in one direction. And it can be easily represented by the periodic function.
So, I = Io sin wt or I = lo cos wt.

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Test: Phasors (January 30) - Question 3
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Virtual value or effective value of a.c. is​

Detailed Solution for Test: Phasors (January 30) - Question 3

 Answer :- d

Solution :- as we know that H= (I02 R)/2 * T/2 -----------------------(1)

If I(rms) be the rms value of ac then

H = I2(rms)R T/2 ---------------------(2)

From eq (1)&(2)

I2(rms)R T/2 = (I02R)/2 * T/2

I2(rms) = (I02)/2

= I(rms)= (I0)/(2)½

= I(rms) = 0.707 I0

Test: Phasors (January 30) - Question 4
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Which is more dangerous?​
Detailed Solution for Test: Phasors (January 30) - Question 4

220 volt a.c. means the effective or virtual value of a.c. is 220 volt, i.e., Ev=220
As peak value E0=√2Ev
∴E0=1.414×220=311 volt
But 220 volt d.c. has the same peak value (i.e., 220 volt only).
Moreover, the shock of a.c. is attractive and that of d.c. is repulsive.
Hence 220 volt a.c. is more dangerous than 220 volt d.c.
 

Test: Phasors (January 30) - Question 5
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What is time constant
Detailed Solution for Test: Phasors (January 30) - Question 5

Time constant is a measure of delay in an electrical circut resulting from either an inductor and resistor or capacitor and resistor. I will discuss rhe most common case which is resistor and capacitor, however the inductor resistor combination behaves in a similar manner. The time constant is equal to the value of the resistance in ohms multiplied by the value of capacitance in Farads. The time constant is measured in seconds . It represents the time for the voltage to decay to 1/2.72.

Test: Phasors (January 30) - Question 6
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What is the relationship between Em and E0​
Detailed Solution for Test: Phasors (January 30) - Question 6

peak value Em=2E0/π
so 2/π value is 0.637
therefore,
Em=-0.637 E0

Test: Phasors (January 30) - Question 7
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The only component that dissipates energy in ac circuit is:
Detailed Solution for Test: Phasors (January 30) - Question 7

The only component that dissipates energy in ac circuit is the resistor because  Pure Inductive and pure capacitive circuits have no power loss.

Test: Phasors (January 30) - Question 8
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The average power dissipation in pure resistive circuit is:
Detailed Solution for Test: Phasors (January 30) - Question 8

Power=IV
Where I=Vrms value of current=IV
And V=Vrms value of voltage=EV
Therefore, P=EVIV

Test: Phasors (January 30) - Question 9
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Given the instantaneous value of current from a.c. source is I = 8 sin 623t. Find the r.m.s value of current​
Detailed Solution for Test: Phasors (January 30) - Question 9

Compare the given eqn. with the standard from I=I0sinωt
I0=8, Irms=I0/√2=8/√2=5.656A

Test: Phasors (January 30) - Question 10
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Find the instantaneous voltage for an a.c. supply of 200V and 75 hertz​
Detailed Solution for Test: Phasors (January 30) - Question 10

Answer :- b

Solution :-  f = 75hz

w=2πf

= 2 * π * 75

= 150π

E(max) = (2)^½ E(rms)

E(max) = 1.414 * 200

= 282.8V

E(ins) = E(max)sinwt

E(ins) = 282.8 sin 150πt

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