The natural frequency of an LC - circuit is $1,25000$ cycles per second. Then the capacitor $C$ is replaced by another capacitor with a dielectric medium of dielectric constant \[k\]. In this case, the frequency decreases by \[25kHz\]. The value of \[k\] will be
\[\begin{align}
& A.3.0 \\
& B.2.1 \\
& C.1.56 \\
& D.1.7 \\
\end{align}\]
Answer
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Hint: The new frequency is found out by taking the difference between the natural frequency and the decreased frequency as per the question. The frequency of an LC circuit is found by the reciprocal of the product of twice the value of \[\pi \] and square root of product of the value of the inductance and the resistance. Using this, the relation between the capacitance and the frequency should be calculated. The dielectric constant is the ratio of the value of capacitance.
Complete answer:
natural frequency of this LC circuit is given as,
\[\text{natural frequency}=125000Hz\]
The new frequency can be found by reducing the decreased frequency from this. That is the decreased frequency is mentioned as,
\[f=25kHz\]
Therefore the new frequency will be,
\[\text{new frequency=}125000-25000=100000Hz\]
The frequency of an LC circuit found by using the equation,
\[f=\dfrac{1}{2\pi \sqrt{LC}}\]
Therefore from this we can obtain the relation between the capacitance and the frequency.
\[f\propto \dfrac{1}{\sqrt{C}}\]
That is,
\[\dfrac{{{f}_{1}}}{{{f}_{2}}}=\sqrt{\dfrac{{{C}_{2}}}{{{C}_{1}}}}\]
Where \[{{f}_{1}}\] be the natural frequency, \[{{C}_{1}}\] be the capacitance at the time of natural frequency, \[{{f}_{2}}\]be the new frequency.
Substituting the values in it will give,
\[\sqrt{\dfrac{{{C}_{2}}}{{{C}_{1}}}}=\dfrac{125000}{100000}=\dfrac{5}{4}\]
The square of the whole equation will give,
\[\dfrac{{{C}_{2}}}{{{C}_{1}}}=\dfrac{25}{16}\]
As we all know the value of the dielectric constant can be found by the relation,
\[k=\dfrac{{{C}_{2}}}{{{C}_{1}}}\]
Therefore the value of dielectric constant will be given as,
\[k=\dfrac{{{C}_{2}}}{{{C}_{1}}}=\dfrac{25}{16}=1.56\]
So, the correct answer is “Option C”.
Note:
Natural frequency is defined as the frequency at which a whole system begins to vibrate even in the absence of any kind of damping force. This frequency is also called Eigen frequency. The pattern of the motion of a system which is in vibration at its natural frequency is referred to as the normal mode.
Complete answer:
natural frequency of this LC circuit is given as,
\[\text{natural frequency}=125000Hz\]
The new frequency can be found by reducing the decreased frequency from this. That is the decreased frequency is mentioned as,
\[f=25kHz\]
Therefore the new frequency will be,
\[\text{new frequency=}125000-25000=100000Hz\]
The frequency of an LC circuit found by using the equation,
\[f=\dfrac{1}{2\pi \sqrt{LC}}\]
Therefore from this we can obtain the relation between the capacitance and the frequency.
\[f\propto \dfrac{1}{\sqrt{C}}\]
That is,
\[\dfrac{{{f}_{1}}}{{{f}_{2}}}=\sqrt{\dfrac{{{C}_{2}}}{{{C}_{1}}}}\]
Where \[{{f}_{1}}\] be the natural frequency, \[{{C}_{1}}\] be the capacitance at the time of natural frequency, \[{{f}_{2}}\]be the new frequency.
Substituting the values in it will give,
\[\sqrt{\dfrac{{{C}_{2}}}{{{C}_{1}}}}=\dfrac{125000}{100000}=\dfrac{5}{4}\]
The square of the whole equation will give,
\[\dfrac{{{C}_{2}}}{{{C}_{1}}}=\dfrac{25}{16}\]
As we all know the value of the dielectric constant can be found by the relation,
\[k=\dfrac{{{C}_{2}}}{{{C}_{1}}}\]
Therefore the value of dielectric constant will be given as,
\[k=\dfrac{{{C}_{2}}}{{{C}_{1}}}=\dfrac{25}{16}=1.56\]
So, the correct answer is “Option C”.
Note:
Natural frequency is defined as the frequency at which a whole system begins to vibrate even in the absence of any kind of damping force. This frequency is also called Eigen frequency. The pattern of the motion of a system which is in vibration at its natural frequency is referred to as the normal mode.
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