Answer to Question #165814 in Electrical Engineering for Enthada Nokkunne

Question #165814

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.


1
Expert's answer
2021-02-25T04:10:51-0500

The frequency is 60 Hz, so, the reactance of the second coil is


X2=2πfL2=1.5 Ω.X_{2}=2\pi fL_2=1.5\space\Omega.

So, since the total voltage is 110 V, the connection is series, and the active power output is 2400 W:


P=V2R1+R2, R2=V2PR1=3.04 Ω.P=\frac{V^2}{R_1+R_2},\\\space\\ R_2=\frac{V^2}{P}-R_1=3.04\space\Omega.

The voltage across the first one:


V1=IR1=VR1+R2R1=43.7 V, V2=IR2=VR1+R2R2=66.3 V.V_1=IR_1=\frac{V}{R_1+R_2}R_1=43.7\text{ V},\\\space\\ V_2=IR_2=\frac{V}{R_1+R_2}R_2=66.3\text{ V}.

Find the power factor:


 cosϕ=PP2+Q2=0.8.\text{ cos}\phi=\frac{P}{\sqrt{P^2+Q^2}}=0.8.

The reactance of the first coil can be found from the following:


cosϕ=R1+R2(R1+R2)2+(X1+X2)2, X1=(R1+R2cosϕ)2(R1+R2)2X2=2.28 Ω, L1=X12πf=0.006 H.\text{cos}\phi=\frac{R_1+R_2}{\sqrt{(R_1+R_2)^2+(X_1+X_2)^2}},\\\space\\ X_1=\sqrt{\bigg(\frac{R_1+R_2}{\text{cos}\phi}\bigg)^2-(R_1+R_2)^2}-X_2=2.28\space\Omega,\\\space\\ L_1=\frac{X_1}{2\pi f}=0.006\text{ H}.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment