Question

The power factor of the circuit in the above figure is $1/\sqrt{2}$. The capacitance of the circuit is equal to

• A. $400\mu F$
• B. $300\mu F$
• C. $500\mu F$
• D. $200\mu F$

Answer 1

The correct option is C $500\mu F$

Here, it is given that,

power factor in the circuit, = $\frac{1}{\surd 2}$

Resistance, $R=10\Omega$

inductance, $L=0.1H$

Supply voltage, $V=2sin\left(100t\right)$

peak voltage, $V_0=2$

$\omega =100rad/s$

capacitance $=?$

$=>powerfactor=cos\varphi =\frac{R}{Z}$

$\frac{1}{\surd 2}=\frac{10}{Z}$

$=>$ By Taking the square on both the sides,

$\frac{1}{2}$ = $\frac{100}{Z^2}$

$Z^2=200$

$=>$ Now, we can use below formula to calculate the capacitance of the circuit.

$R^2+(Xc-XL{)}^2=Z^2$

${10}^2+$ $(\frac{1}{cw}$ -$L\omega {)}^2=200$

$(\frac{1}{cw}-L\omega {)}^2=100$

$\frac{1}{cw}$ $-L\omega =10$

$\frac{1}{cw}$ $=L\omega +10$

$\frac{1}{cw}$ $=0.1\times 100+10$

$\frac{1}{cw}$ $=20$

$c=$ $\frac{1}{20\omega }$

$c=$ $\frac{1}{20\times 100}$

$c=$ $\frac{1}{2000}F$

$c=\frac{{10}^6}{2000}\left(\mu \right)F$ $[\because 1F={10}^6(\mu \left)F\right]$

$=500\left(\mu \right)F$