For a resistance R and capacitance C in series, the impedance is twice that of a parallel combination of the same element, What is the frequency of applied emf?  

As shown in figure, in case of series combination,

$Z_s=\sqrt{R^2+X_C^2}={\left[R^2+(1/\omega C{)}^2\right]}^{1/2}$

In case of Parelle combinations,

$i_R=\frac{V}{R}\sin \omega t\text{ and }{i}_C=\frac{V}{X_C}\sin \left(\omega t+\frac{\pi }{2}\right)$
So, $I=i_R+i_C=\frac{V}{R}\sin \omega t+\frac{V}{R_C}\cos \omega t$
i.e., $i=I_M\sin (\omega t+\varphi )$
with $I_M\cos \varphi =\frac{V}{R}$ and $I_M\sin \varphi =\frac{V}{X_C}$
So, $I_M={\left[{\left(\frac{V}{R}\right)}^2+{\left(\frac{V}{X_c}\right)}^2\right]}^{1/2}=\frac{V}{Z_p}$
i.e. $\frac{1}{Z_p}={\left[\frac{1}{R^2}+{\left(\frac{1}{X_c}\right)}^2\right]}^{1/2}$ i.e., $Z_p=\frac{R}{\sqrt{1+{\omega }^2{C}^2{R}^2}}$
and as according to given problem,
$Z_{\varphi }=2Z_{Dn}$ i.e., $Z_x^2=4Z_p^2$ 

i.e., $\frac{\left(R^2{\omega }^2{C}^2+1\right)}{{\omega }^2{C}^2}=4\frac{R^2}{\left(1+R^2{\omega }^2{C}^2\right)}$
i.e., ${\left(1+R^2{\omega }^2{C}^2\right)}^2=4R^2{\omega }^2{C}^2$
or $1+R^2{\omega }^2{C}^2=2R\omega C$
or $(R\omega C-1{)}^2=0$
or $\omega =\frac{1}{RC}$ i.e., $f=\frac{1}{2\pi RC}$ Ans.