An electrical device draws 2 kW power form AC mains [voltage 223 V (rms) `= sqrt(50,000) V`]. The current differs (lags) in phase by `phi (tan = (-3)/(4))` as compared to voltage. Find (i) R, (ii) `X_(C ) - X_(L)`, and (iii) `I_(M)`. Another device has twice the values for R, `X_(C )` and `X_(L)`. How are the answers affected ?

Answer

A

Explanation

Given, powe drawn = P = 2kW= 2000 W
`tan phi= -3/4, I_(M) = I_(0)=?, X_(C)-X_(L)=?`
`V_("rms")= V= 223V`
Power P=`V^(2)/Z`
`Z= V^(2)/P = (223 xx 223)/(2 xx 10^(3))=25`
Impedance Z=`25Omega`
Impedance z = `sqrt(R^(2) + (X_(L)-X_(C))^(2)`
`25 = sqrt(R^(2) + (X_(L)-X_(C))^(2)`
`625 = R^(2) + (X_(L)^(2)-X_(C))6(2)`.............(i)
Agian, `tanphi= (X_(L)-X_(C))/(R)=3/4`
or `X_(L)-X_(C) = (3R)/4`.........(ii)
From Eq. (ii), we put `X_(L)-X_(C)=(3R)/4` in Eq. (i), we get
`625 = R^(2)+((3R)/4)^(2) = R^(2) +(9R)^(2)/(16)`
or 625= `(25R^(2))/(16)`
a) Resistance R = `sqrt(25 xx 16)= sqrt(400) = 20 Omega`
b) `X_(L) - X_(C) = (3R)/(4) = 3/4 xx 20 = 15 Omega`
c) Main current, `I_(M) =sqrt(2)I=sqrt(2)V/Z = 223/25 xx sqrt(2)=12.6`A
As R, `X_(C), X_(L)` are all doubled, `tanphi` does not change. Z is doubled, current is halved, So, power is also halved.