Question #167867

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?

Expert's answer

"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.

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