Question

A capacitor, a 101.5 mH inductor and a 5 ohm resistor are connected in series with a 50 Hz a.c. source. If current and voltage in the circuit are in phase, what is the capacitance of the capacitor ?​

Answer 1

Given :

• Indicator =101.5mH

•Resistor =5ohm

• Current and voltage in the circuit are in phase

To Find :

• The capacitance of the capacitor =?

Formula :

$\\ \blue\bigstar\boxed{\bf{tan\theta =\dfrac{\omega_{o}L- \dfrac{1}{\omega_{o} C}}{R}}}$

Solution :

• In a LCR circuit , the current and the voltage are in phase $\rm{ \theta=0}$ , when

$\\ \implies\sf{tan\theta=\dfrac{\omega_{o}L-\dfrac{1}{\omega_{o}C}}{R}=0}$

$\\ \\ \implies\sf{\omega_{o}L=\dfrac{1}{\omega_{o}C}}$

$\\ \\ \implies\sf{C=\dfrac{1}{(\omega_{o})^{2}L}}$

Here, $\boxed{\bf{\omega=2 \pi f}}$

$\\ \\ \implies\sf{\omega= 2 \times 3.14 \times 50s^{-1}=314s^{-1}}$

$\\ \bullet\bf{L=101.5mH=101.5 \times 10^{-3} H}$

$\\ \\ \therefore \implies\sf{C=\dfrac{1}{(314s^{-1})^{2} \times (101.5 \times 10^{-3}H)}}$

$\\ \\ \implies\sf{C=\dfrac{10^{3}}{(314s^{-1})^{2} \times (101.5)H}}$

$\\ \\ \implies\sf{C= \approx 1000 uF}$