A radio can tune over the frequency range of a portion of MW broadcast band : (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 `mu H`, what must be the range of its variable capacitor ?

Explanation

Hint : For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.
Given, minimum frequency
`f_(1)=800 KHz =8xx10^(5)Hz`
Inductance
`L = 200 mu H = 200xx10^(-6)H=2xx10^(-4)H`.
Maximum frequency
`f_(2)=1200 KHZ = 12xx10^(5)Hz`
for tuning, the natual frequency is equal to the frequency of oscillations that means it is the case of resonance.
Frequency of oscillations `f = (1)/(2pi sqrt(LC))`
For capacitance `C_(1), f_(1)=(1)/(2pi sqrt(LC_(1)))`
`C_(1)=(1)/(4pi^(2)f_(1)^(2)L)`
`=(1)/(4xx3.14xx3.14xx(8xx10^(5))^(2)xx2xx10^(-4))`
`= 197.7xx10^(-12)F`
`= 197.7 PF`
for capacitance `C_(2), f_(2)=(1)/(2pi sqrt(LC_(2)))`
`C_(2)=(1)/(4pi^(2)f_(1)^(2)C)`
`= (1)/(4xx3.14xx3.14xx(12xx10^(5))^(2)xx2xx10^(-4))`
`= 87.8xx10^(-12)F=87.8pF`
Thus, the range of capacitor is 87.8pF to 197.7pF.