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New answer posted

a year ago

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P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) inductive reactance XL= 2 πϑ L

 
For an LCR circuit Z= R2+ (XL-XC) 2

As frequency (ν) increases, Z decreases and at certain value of frequency know as resonant frequency (ν0), impedance Z is minimum that is Zmin= R current varies inversely with impedance and at Zmin current is maximum.

New answer posted

a year ago

0 Follower 11 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Secondary voltage Vs= 24V

Power associated with secondary Ps = 12W

Is= PSVS = 12/24= 0.5A

I0= Is 2 = 1/ 2 A

New answer posted

a year ago

0 Follower 11 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Average power dissipated is ErmsIrmscos ∅

Z= R2+XL2

= 1+4 = 5

Irms= 6/ 5 A

cos ∅ =R/Z=2/ 5

pav= 6 *65*25 = 14.4W

New answer posted

a year ago

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P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) As we know that Q= 1R√LC to make Q high R should be low, L should be high and C should be low.

New answer posted

a year ago

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P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

As we know resonant frequency is v0= 12πLc

To increase capacitance we must connect it to another capacitor parallel to the first.

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a year ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) The voltmeter connected to AC Mains reads mean value (2>) and is calibrated in such a way that it gives value of, which is multiplied by form factor to give rms value.

New answer posted

a year ago

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P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) For delivering maximum power from the generator to the load, total internal reactance must be equal to conjugate of total external reactance.

XL=-Xg

New answer posted

a year ago

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P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) I0= 2 Irms= 2  (5)A

I=I0sinwt = 5 2 sin π/3 = 5 32 A

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b)

Explanation- Due to the point source light propagates in all directions symmetrically and hence, wavefront will be spherical as shown in the diagram.

If power of the source is P, then intensity of the source will be I= p/4 π r 2

 where, r is radius of the wavefront at anytime.

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b)

Explanation- (a) When a decreases w increases. So, size decreases.

(b) Now, light energy is distributed over a small area and intensity?1/Area  is decreasing so intensity increases

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