The natural frequency of an L-C circuit is 125000 cycles/s . When a dielectric medium of dielectric constant K is introduced between the plates of the capacitor, the frequency decreases by 25 kHz . The value of K is

Question

The natural frequency of an L-C circuit is 125000 cycles/s . When a dielectric medium of dielectric constant K is introduced between the plates of the capacitor, the frequency decreases by 25 kHz . The value of K is (A) 3.05 (B) 2.18 (C) 1.56 (D) 1.74

Tardigrade Admin · Accepted Answer

f=(1/2 π √L C) When a dielectric is introduced the capacitance becomes C ' = K â¡ C â¡ ⇒ (f '/f)=√(C â¡/C â¡') (125000 - 25000/125000)=√(C / textitKC) (100/125)=(1/√K) K=((125/100))2=1.56

Q. The natural frequency of an $L - C$ circuit is $125000 cyc l es / s$ . When a dielectric medium of dielectric constant $K$ is introduced between the plates of the capacitor, the frequency decreases by $25 k Hz$ . The value of $K$ is

$3.05$

$2.18$

$1.56$

$1.74$

$f = \frac{1}{2 π L C}$
When a dielectric is introduced the capacitance becomes
$C^{^{'}} = K C$
$\Rightarrow \frac{f ^{^{'}}}{f} = \frac{C}{C ^{^{'}}}$
$\frac{125000 - 25000}{125000} = \frac{C}{KC}$
$\frac{100}{125} = \frac{1}{K}$
$K = (\frac{125}{100})^{2} = 1.56$

Q. The natural frequency of an L−C circuit is 125000cycles/s . When a dielectric medium of dielectric constant K is introduced between the plates of the capacitor, the frequency decreases by 25kHz . The value of K is

3.05

2.18

1.56

1.74