Question

The self inductance of a choke coil is $10 \,mH$. When it is connected to a $10\, V \,DC$ source, the loss of power is $20 \,W$. When it is connected to a $10\, V\,AC$ source, the loss of power is $10 \,W$. The frequency of $AC$ source (in $Hz$ ) will be (Answer to be given in nearest integer)

Answer 1

For $DC : R=\frac{V^{2}}{P}\left(\therefore X_{L}=0\right)$
$R=\frac{10 \times 10}{20}=5 \, \Omega$
For $AC : P_{\text {loss }}=V I \cos \phi$
$P_{\text {loss }}=V \times \frac{V}{Z} \times \frac{R}{Z} $
$P_{\text {loss }}=\frac{V^{2} \times R}{Z^{2}} $
$Z^{2}=\frac{V^{2} R}{P_{\text {loss }}}=\frac{10 \times 10 \times 5}{10}=50\, \Omega$
Now, $Z^{2}=R^{2}+X_{L}^{2}$
$50=25+X_{L}^{2}$
$ \Rightarrow X_{L}=5 \, \Omega$
$2 \pi f L=5$
$ \Rightarrow f=\frac{5}{2 \pi L}=\frac{5}{2 \pi \times 10 \times 10^{-3}} $
$f=80 \, Hz$