A variable capacitor with a range from 10 to 365pF is used with a coil to form a variable-frequency LC circuit to tune the input to a radio. (a) What is the ratio of maximum frequency to minimum frequency that can be obtained with such a capacitor? If this circuit is to obtain frequencies from 0.54 MHz to 1.6 0MHz, the ratio computed in (a) is too large. By adding a capacitor in parallel to the variable capacitor, this range can be adjusted. To obtain the desired frequency range,(b) What capacitance should be added ? (c) What inductance should the coil have?
We can use the concept of frequency of the oscillator, which is inversely proportional to the capacitance. So from the minimum value of the capacitor, we get the maximum frequency. Now, we can take the ratio of the maximum frequency to the minimum frequency. Adding the additional capacitor, which is parallel to the original capacitor, and by using the condition that the frequency range must be the same, we can first take the ratio of frequency and find the capacitance of the additional capacitor. Finally, we can find the inductance of the coil.
The frequency of an LC circuit, (i)
From equation (i), we conclude that the frequency is inversely proportional to the square root of the capacitor. So, the ratio of the maximum frequency to the minimum frequency range using equation (i) can be written as follows:
Hence, the value of the ratio is 6.0.
An additional capacitor is added to the capacitor so that we take the ratio of the frequency as follows:
Now, by using equation (i) and the given condition, we can get the added capacitor as follows:
Hence, the value of the added capacitance is 36 pF.
Now we have to solve the frequency equation to find out the inductance. For the minimum frequency, the new capacitance can be given as:
Now, using the given data in equation (i), we can get the inductance value as follows:
Hence, the value of the inductance is .
An alternating emf source with a variable frequency is connected in series with a resistor and a capacitor. The emf amplitude is . (a) Draw a phasor diagram for phasor (the potential across the resistor) and phasor (the potential across the capacitor). (b) At what driving frequency do the two phasors have the same length? At that driving frequency, what are (c) the phase angle in degrees, (d) the angular speed at which the phasors rotate, and (e) the current amplitude?
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