A 40mH inductive coil has a resistance of 10Ω. - How much current will it draw if connected across a $100V, 60H_Z$ supply? - What is the power factor of the coil? - Determine value of capacitance that must lie connected across the coil to make the power factor of overall circuits units? -

Mumbai University > FE > Sem 1 > Basic Electrical and Electronics Engineering

Marks: 4 M

Year: May 2013

Given:

$L = 46mH \\ r = 10Ω \\ V = 100V \\ f = 60H_Z$

Solution:

$X_L$ $=2πfL \\ =2π×60×46×10^{-3} \\ =17.34Ω$

$Z= \sqrt{r^2+X_L^2 }= \sqrt{10^2+17.34^2} =20.02Ω \\ p F = cos⁡∅ = r/Z= 10/20.02 \\ = 0.5 (lagging)$

When a capacitor is connected across a coil, it is resonance.

$f_0 = \dfrac{1}{2π} \sqrt{\dfrac{1}{LC}-\dfrac{λ^2}{L^2}} \\ 60=\dfrac{1}{2π} \sqrt{\dfrac{1}{46×10^{-3}×C}-\dfrac{10^2}{(46×10^{-3})^2}} \\ ∴C=114.79μF$