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Question: The turning circuit of a radio receiver has a resistance of \(50\Omega \) , an inductor of \(10mH\) ...

The turning circuit of a radio receiver has a resistance of 50Ω50\Omega , an inductor of 10mH10mH 10mH10mH and a variable capacitor. A 1MHz1MHz radio wave produces a potential difference of 0.1mV0.1mV 0.1mV0.1mV.The values of the capacitor to produce resonance Take\pi _{}^2 = 10$$$$\pi _{}^2 = 10
a.2.5pF b.5.0pF c.25pF d.50pF  a.2.5pF \\\ b.5.0pF \\\ c.25pF \\\ d.50pF \\\

Explanation

Solution

Hint So here we can see that for taking out resonance required values are given that is resistance inductance and frequency given so by the formula and putting their respective value we can get that

Complete Step by step solution
1. A circuit is a closed path that allows electricity to flow from one point to another.
2. Tuned circuit combines an inductor and capacitor to make a circuit that is responsive to a frequency.
3. A capacitor is a passive two-terminal electrical component used to store energy electrostatically in an electric field.
4. An inductor is a passive electronic component which is capable of storing electrical energy in the form of magnetic energy.
5. Difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.
Resistance is the hindrance to the flow of electrons in material. While a potential difference across the conductor encourages the flow of electrons . The rate at which charge flows between two terminals is a combination of these two factors.
6. A vibration of large amplitude in a mechanical or electrical system caused by a relatively small periodic stimulus of the same or nearly the same period as the natural vibration period of the system is called resonance
So resonance here,
wL=1wcwL = \dfrac{1}{{wc}} wL=1wcwL = \dfrac{1}{{wc}}
c=1w2L =1(2πf)2L c = \dfrac{1}{{w_{}^2L}} \\\ = \dfrac{1}{{(2\pi f)_{}^2L}} \\\
=14π2(106)×10×103 =2.5×1012F = \dfrac{1}{{4\pi _{}^2(10_{}^6) \times 10 \times 10_{}^{ - 3}}} \\\ = 2.5 \times 10_{}^{ - 12}F \\\
=2.5pF2.5pF

So the correct option is A.

Note
So tuning circuit is given so by this inductor and capacitor can be determined. And what quantities are given and what we need to be found out and by putting the formulae we can easily find out.