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

Step-by-step explanation:

For parallel inductors ,

`(1)/(L_R) = (1)/(L_1) +(1)/(L_2)`

`(1)/(L_R) =(1)/(25) +(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