Question

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 1

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 =$IZ_B=52.72V$