Physics NCERT Exemplar Solutions Class 12th Chapter Seven Alternating Currents

1. An electrical device draws 2KW power from AC mains (voltage 223V(rms)=) $\sqrt[]{50000}\mathit{V}$ . The current differs (lags) in phase by $\mathit{\varnothing }$ (tan $\mathit{\varnothing }=-3/4$ ) as compared to voltage . find

(a) R

(b) X_L-X_C

(c) I_m. Another device has twice the values for R,X_c and X_L. how the answer affected?

Explanation- Power drawn , p= 2KW= 2000W

tan  $\varnothing$ =-3/4

P=V2/Z

Z=V2/P=  $\frac{223\times 223}{2\times {10}^3}$ =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=  $\sqrt[]{2}V/Z$ =  $\frac{223\times \sqrt[]{2}}{25}$ = 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.

2. 1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if

(i) Power is transmitted at 220V. Comment on the feasibility of doing this.

(ii) A step-up transformer is used to boost the voltage to 11000V, power transmitted, then a step-down transformer is used to bring voltage to 220 V. (ρ_cu= 1.7×10^-8 SI unit)

3. Consider the LCR circuit shown in figure find the net current I and the phase of i. show that i=V/Z. find the impedance Z for this circuit.

Explanation- 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