Question

When a voltage of 100 volts at 50 Hz is applied to a series connection of a resistor A

and an inductor A, the current taken is 8 A and the power is 120 watts. When applied

to another series connection of a resistor B and an inductor B, the current is 10 A and

the power is 500 watts. What current and power will be taken when 100 V is applied

to the two coils in series?

Answer 1

$V=100V\newline f=50Hz$

CaseA:

$Ia=8A\newline Pa=120W$

$Ra=Pa/Ia^2=120/8^2=1.875\Omega$

$Za=V/Ia=100/8=12.5\Omega$

$Xa=\sqrt{Za^2-Ra^2} =\sqrt{12.5^2-1.875^2} =12.36\Omega$

$\implies La\omega=12.36, La=12.36/2\pi50=39.34mH$

CaseB:

$Ib=10A\newline Pb=500W$

$Rb=Pb/Ib^2=500/10^2=5\Omega$

$Zb=V/Ib=100/10=10\Omega$

$Xb=\sqrt{Zb^2-Rb^2} =\sqrt{10^2-5^2} =8.66\Omega$

$\implies Lb\omega=8.66, Lb=8.66/2\pi50=27.56mH$

When La and Lb are connected in series current $Iab=V/(Xa+Xb) =100/(12.36+8.66) =4.76A$

Power= 0 , as both are inductors there is no power loss.