Question

Two coils A and B are connected in series across a 110V, 60Hz supply. The resistance of coil A is 2  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

The kVA triangle and the circuit is shown above. The circuit kVA is given by, kVA=$\sqrt{(2.4)^2+(1.8)^2}=3$ or VA = $3000$ voltmeters

Circuit current = $\dfrac{3000}{110}=27.27$ A

$\therefore27.27^2(R_A+R_B)=2400$

$\Rightarrow R_A+R_B=\dfrac{2400}{742.38}=3.22\Omega$

$\Rightarrow R_B=1.22\Omega$

Impedance of the whole circuit is given by $Z=110/27.27=4.033\Omega$

$X_A+X_B=\sqrt{Z^2-(R_A+R_B)^2}=2.42\Omega$

Now, $X_B=2\pi\times60\times0.004=1.50\Omega$

$\therefore X_A=0.912\Omega$

$\Rightarrow2\pi\times60\times L_A=0.912$

$\Rightarrow L_A=0.00241H$

Now, $Z_A=\sqrt{R_A^2+X_A^2}=2.198\Omega$

P.d across coil A = $I\times Z_A=59.9V$

Now, $Z_B=\sqrt{R_B^2+X_B^2}=1.933\Omega$

P.d across coil B = $I\times Z_B=52.72\Omega$