Question

In an Hartley oscillator ,if L1=0.2mH,L2=0.3mH and C=0.003 μF, calculate the frequency of its oscillation in KHz ,neglect mutual inductance

Answer 1

Answer:

130.208Khz

Explanation:

given

L1=0.2mH

L2=0.3mH

C=0.003uh=3x10 power -3 x 10 power -6

= 3x10 power -9

we know

f=1/2π√LtC

here LT = L1+L2

= 0.5mH= 5x10 power -4

substitute in above formula

1/6.28x√5x3x 10 power -13

= 1/6.28x√1.5x10 power -12

if we calculate it

10 power 6 divided by 7.68

ans is :130208.33hz

in khz divided by 1000

:: 130.208Khz