Question

Two inductors, L1 and L2, are in parallel. L1 has a value of 25 mH and L2 a value of 50 mH. The parallel combination is in series with L3, a 20 mH coil. The entire combination is across an ac source of 60 Vrms at 300 kHz. The total rms current is

Answer 1

Answer:

Explanation:

For parallel inductors ,

$\frac{1}{L_R} = \frac{1}{L_1} +\frac{1}{L_2}$

$\frac{1}{L_R} =\frac{1}{25} +\frac{1}{50}$

$L_R=16.67 mH.$

For series combination

Total inductance

= 16.67 + 20

= 36.67 mH .

reactance of total inductance at 300 kHz

= ω$L_{total}$ where ω is angular frequency

= 2πf$L_{total}$

= 2 x 3.14 x 300 x 10³ x 36.67 x 10⁻³

= 69.1 x 10³ ohm

Total rms current = Vrms / reactance

= 60 / 69.1 x 10³ A

= .87 x 10⁻³ A

= .87 mA