Alternating Current

This is a short answer type question as classified in NCERT Exemplar

P_s=60W, I_s= 0.54 A

P=VI, V= 60/0.54= 110V

Voltage in the secondary is less than voltage in primary so transformer is step down transformer

As transformation ratio is, k = E_s/E_p= I_p/I_s

110/220=I_p/0.54A

I_p= 0.27A

This is a short answer type question as classified in NCERT Exemplar

L= 0.01H

R= 1ohm

Voltage =200V

Frequency = 50Hz

Z= $\sqrt[]{R^2+X_L^2}$ = $\sqrt[]{1^2+({2*3.14*50*0.01)}^2}$ = $\sqrt[]{10.86}$ =3.3ohm

tan $\varnothing =\frac{wL}{R}$ = $\frac{2\pi fL}{R}=\frac{2*3.14*50*0.01}{1}$ = 3.14

$\varnothing$ = tan-13.14=72

$\varnothing$ =72 $\frac{\pi }{180}rad$

$?t$ = $\varnothing$ /w= $\frac{72\pi }{180*2\pi *50}$ = $\frac{1}{250}$ s

This is a short answer type question as classified in NCERT Exemplar

For a Direct Current (DC),

1ampere= 1coulomb/sec

An AC current changes direction with the source frequency and the attractive force would average to zero. Thus, the AC ampere must be defined in terms of some property that is independent of the direction of current.

Joule's heating effect is such property and hence it is used to define rms value of AC.

This is a short answer type question as classified in NCERT Exemplar

(a) We know that P= VI

that is curve of power will be having maximum amplitude, equals to multiplication of amplitudes of voltage (V)and current (I) curve. So, the curve will be represented by A.

(b) As shown by shaded area in the diagram, the full cycle of the graph consists of one positive and one negative symmetrical area. Hence average power over a cycle is zero.

(c) As the average power is zero, hence the device may be inductor (L) or capacitor (C) or the series combination of L and C.

This is a short answer type question as classified in NCERT Exemplar

This is a short answer type question as classified in NCERT Exemplar

I_rms= $\sqrt[]{\frac{1^2+2^2}{2}}$ = $\sqrt[]{\frac{5}{2}}$ = 1.58A = 1.6 approx

This value is indicating in graph

This is a short answer type question as classified in NCERT Exemplar

In the figure

Band width = w₂-w₁ where these two corresponds to frequencies at which magnitude of current is 1/ $\sqrt[]{2}$ times of maximum value

I_rms= $\frac{I_{max}}{\sqrt[]{2}}$ = 1/ $\sqrt[]{2}$ = 0.7A

From the diagram the corresponding frequencies are 0.8 and 1.2rad/s

This is a short answer type question as classified in NCERT Exemplar

E=E₀sinwt

And current I=I₀sin (wt $\frac{+}{-}\varnothing$ )

Power, P= EI = E₀sinwt = E₀I₀sinwtsin (wt+ $\varnothing$ )

= $\frac{E_0{I}_0}{2}$  [cos $\varnothing -cos(2wt+\varnothing )$ ]………… (1)

Pav= $\frac{V_0}{\sqrt[]{2}}\frac{{\mathrm{I}}_0}{\sqrt[]{2}}$ cos $\varnothing$ =V_rmsI_rmscos $\varnothing$ ………. (2)

From eqn 1 ……But when cos $\varnothing$ < cos (2wt+$\mathrm{\varnothing )}$

P<0 power can be nagative of an AC source

From eqn 2……… P_av>0

cos $\varnothing$ =R/Z>0 average power can not be negative