Two coils A and B are connected in series across a 110V, 60Hz supply. The resistance of coil A is 2Ω and inductance of coil B is 0.004H. If the input from the supply is 2.4kW and 1.8kVAR, find the inductance of coil A and resistance of coil B. Also calculate the voltage across each coil.
Answer
The kVA triangle and the circuit is shown above. The circuit kVA is given by, kVA="\\sqrt{(2.4) ^2+(1.8) ^2}" =3
VA=3000volt -meter
Current="\\frac{3000}{110}=27.27A"
So
"R_A+R_B=\\frac{2400}{742.38}=" 3.22
"R_B=1.22\\Omega"
Impedance of the whole circuit is given by
"Z=\\frac{110}{27.27}=4.033\\Omega"
So
"X_A+X_B=\\sqrt{Z^2-(R_A+R_B) ^2}"
"X_A+X_B=2.42\\Omega"
"X_B=2\\pi \\times60\\times0.004=1.5\\Omega"
So "X_A=0.912\\Omega"
"2\\pi \\times60\\times L_A=0.912"
"L_A=0.00241H"
So
"Z_A=\\sqrt{R_A^2+X_A^2}=2.198\\Omega"
P.d across coil A ="IZ_A=59.9V"
"Z_B=\\sqrt{R_B^2+X_B^2}=1.933\\Omega"
P.d across coil B ="IZ_B=52.72V"
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