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Question

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


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Solution

Step 1: Given data

  1. The lower tuning frequency of the radio is f1=800KHz=800×103Hz.
  2. The upper tuning frequency of the radio is f2=1200KHz=1200×103Hz.
  3. The effective inductance of the radio is L=200μH=200×10-6H.

Step 2: The capacitance of the LCR circuit

  1. We know at the resonance of the LCR circuit the angular frequency is ω=1LC, where, L is the inductance, C is the capacitance, ω and is the angular frequency of the wave.
  2. Angular momentum, ω=2πf, where, f is the frequency of a wave.

Step 3: Finding the range of the capacitor

As we know the angular frequency of an LCR circuit is ω=1LC.

So, C=1ω2L.................(1)

Now, the capacitance of the variable capacitor for frequency f1 and angular frequency ω1 is

C1=1ω12L=12πf12×L=12×3.14×800×1032×200×10-6orC1=1.98×10-10F................(1)

And, the capacitance of the variable capacitor for frequency f2 and angular frequency ω2 is

C2=1ω22L=12πf22×L=12×3.14×1200×1032×200×10-6orC2=0.880×10-10F..............(2)

From equations 1 and 2 we get, C=C1,C2=1.98×10-10,0.880×10-10

Therefore, the range of its variable capacitor is C=1.98×10-10F,0.880×10-10F.


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