Physics NCERT Exemplar Solutions Class 12th Chapter Seven

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

If we consider a L-C circuit analogous to a harmonically oscillating spring block system. The electrostatic energy $\frac{1}{2}$ CV2 is analogous to potential energy and energy associated with moving charges (current) that is magnetic energy $\frac{1}{2}$ LI2 is analogous to kinetic energy.

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

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

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

Total current i=i1+i2

Vmsinwt=Ri1

I1= $\frac{V_{m}sinwt}{R}$

If q2 is charge on the capacitor at any time t then for series combination of C and L

Applying KVL in below circuit

So q2=qmsin(wt $\frac{q_2}{C}+\frac{Ld{i}_2}{dt}-V_{m}sinwt=0$ )……….1

qm[ $+\varnothing$ sin(wt+ $\frac{q_2}{C}+\frac{L{d}_2{q}_2}{d{t}^2}=V_{m}sinwt$ )]= Vmsinwt

if $\frac{d{q}_2}{dt}=-q_m{w}^2\mathrm{s}\mathrm{i}\mathrm{n}?(wt+\varnothing )$

qm= $usethisvalueineqn1$

from above equation i2= $\frac{1}{C}+L(-w^2)$

i2= $\varnothing$ when $\varnothing =0$

so total current I = i1+i2

I= $\frac{V_m}{(\frac{1}{c}-L{w}^2)}$

I= i1+i2= Ccos $\frac{d{q}_2}{dt}=w{q}_m\cos ?\left(wt+\varnothing \right)$ +Csin $\frac{w{V}_m\mathrm{c}\mathrm{o}\mathrm{s}?(wt+\varnothing )}{\frac{1}{c}-L{w}^2}$

I= Csin(wt+ $\varnothing =0,i_2=\frac{V_{m}coswt}{\frac{1}{wC}-Lw}$ )

C= $\frac{V_{m}sinwt}{R}+\frac{V_{m}coswt}{\frac{1}{wC}-Lw}$

$\varnothing sinwt$ 1/2

$\varnothing coswt$ = tan-1 $\varnothing$

$\sqrt[]{A^2+B^2}$ 1/2

This is the expression for impedance.2

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

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

Power drawn, p= 2KW= 2000W

tan $\varnothing$ =-3/4

P=V2/Z

Z=V2/P= $\frac{223*223}{2*{10}^3}$ =25

Z = $\sqrt[]{R^2+({X_L-X_C)}^2}$

25= $\sqrt[]{R^2+({X_L-X_C)}^2}$

625 = $R^2+({X_L-X_C)}^2$ ………… (1)

tan $\varnothing$ = $\frac{X_L-X_C}{R}$ = ¾

XL-XC= 3R/4

Use this in eqn 1

625= R2+ (3R/4)2 = R2+9R2/16

625 = 25R2/16, R= 20ohm

XL-XC=15ohm

Im= $\sqrt[]{2}$ I= $\sqrt[]{2}V/Z$ = $\frac{223*\sqrt[]{2}}{25}$ = 12.6A

If R, XL and Xc are all doubled, tan $\varnothing$ does not change . Z is doubled, current is halved. So power is also halved.