- Home
- Standard 12
- Physics
7.Alternating Current
medium
Power factor of an $L-R$ series circuit is $0.6$ and that of a $C-R$ series circuit is $0.5$. If the element ($L, C,$ and $R$) of the two circuits are joined in series the power factor of this circuit is found to be $1.$ The ratio of the resistance in the $L-R$ circuit to the resistance in the $C-R$ circuit is
A
$6/5$
B
$5/6$
C
$\frac{4}{{3\sqrt 3 }}$
D
$\frac{{3\sqrt 3 }}{4}$
Solution
$\cos \phi_{1}=0.6=\frac{3}{5}$
$\tan \phi_{1}=\frac{4}{3}=\frac{X_{C}}{R_{2}} \dots(1)$
$\cos \phi_{1}=0.5=\frac{1}{2}$
$\tan \phi_{2}=\sqrt{3}=\frac{X_{C}}{R_{2}} \ldots(2)$
From $( 1)$ and $(2) \frac{3 \sqrt{3}}{4}=\frac{R_{1}}{R_{2}}$
Standard 12
Physics
Similar Questions
hard
