Courses
Courses for Kids
Free study material
Offline Centres
More

Find the capacity of a capacitor, which when put in series with a $10\,ohm$ resistor, makes the power factor equal to $0.5$. Assume an $80 V - 100 Hz$ a.c. supply.

seo-qna
Last updated date: 22nd Feb 2024
Total views: 303k
Views today: 7.03k
Rank Predictor
Answer
VerifiedVerified
303k+ views
Hint: To answer this question, we first need to understand what is a capacitor. A capacitor (also called a condenser) is a two-terminal passive electrical component that stores energy electrostatically in an electric field. Practical capacitors come in a variety of shapes and sizes, but they all have at least two electrical conductors (plates) separated by a dielectric (i.e., insulator).

Complete step by step answer:
As given in the equation, R=10 ohm, $\cos \Phi = 0.5$(power factor), ${V_{rms}}$= 80V, $f = 50Hz$.As we know that
$\cos \Phi = \dfrac{R}{Z}$
(Here $R$ is the resistance and $Z$ is the impedance)
Substituting values
$0.5 = \dfrac{{10}}{Z}$ so, Z = 20
As we know that $Z = \sqrt {{R^2} + X_C^2} $(where R is the resistance and ${X_c}$is the capacitive impedance)
Putting values of Z and R,
$20 = \sqrt {100 + X_C^2} $
We get ${X_C} = 10\sqrt 3 $
And we know that ${X_C} = \dfrac{1}{{wc}}$
Substituting value of ${X_C}$ and value of $w = 2\pi f$
$10\sqrt 3 = \dfrac{1}{{2 \times 3.14 \times 50 \times C}}$
$\Rightarrow C = \dfrac{1}{{2 \times 50 \times 3.14 \times 10\sqrt 3 }}$
$\Rightarrow C = \dfrac{1}{{100 \times 3.14 \times 10 \times 1.732}}$(putting value of $\sqrt 3 $= 1.732)
$\therefore C = \dfrac{1}{{1000 \times 3.14 \times 1.732}}$
On further calculating we get $C = 9.2 \times {10^{ - 5}}F$

So, the final answer is $C = 9.2 \times {10^{ - 5}}\,F$.

Note: Impedance is the active resistance of an electrical circuit or component to AC current, which is determined by the interaction of reactance and ohmic resistance. In other words, impedance in AC circuits is simply an extension of resistance concepts. We also describe it as any obstruction or measurement of an electric current's resistance to energy flow when voltage is applied.
Recently Updated Pages