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

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Answer:

Explanation:

For parallel inductors ,

[tex]\frac{1}{L_R} = \frac{1}{L_1} +\frac{1}{L_2}[/tex]

[tex]\frac{1}{L_R} =\frac{1}{25} +\frac{1}{50}[/tex]

[tex]L_R=16.67 mH.[/tex]

For series combination

Total inductance

= 16.67 + 20

= 36.67 mH .

reactance of total inductance at 300 kHz

= ω[tex]L_{total}[/tex] where ω is angular frequency

= 2πf[tex]L_{total}[/tex]

= 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

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