- Determine the value of phase difference between the current and the voltage in the given series LCR circuit.
- Calculate the value of the additional capacitor which may be joined suitably to the capacitor C that would make the power factor of the circuit unity.
CBSE OUTSIDE DELHI - SET 3 DEHRADUN 2015
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- In LCR circuit: $\tan\varphi= \frac{\text{X}_{L} - \text{X}_{c}}{\text{R}} = \frac{\text{wL} - \frac{1}{\text{wC}}}{\text{R}}$
$= 100\Omega$
and $\text{X}_{c} = \frac{1}{\text{wc}} = \bigg(\frac{1}{100\times2\times10^{-6}}\bigg)\Omega$
$\therefore\text{X}_{c} = 500 \Omega$
$\therefore\tan\varphi = \frac{500 - 100}{400} = 1 $
$\tan\varphi = 1 $
$\varphi = 45^{o}$
- Power Factor:
$\therefore\text{X'}_{c} = \frac{1}{\omega\text{C}'} = 100 \Omega$
This gives $\text{C'} = \frac{1}{100\omega} = 10 \mu\text{F}$
We, therefore, need to add a capacitor of capacitance (10-2) μF = 8μF in parallel with the given capacitor.
Alternate Answer
Let addition capacitance $C_1$ be connected:
$\text{X}'_{c} = \frac{1}{1000(2 + \text{C}_{1})\times10^{-6}}$
$\therefore100 = \frac{1}{1000(2 + \text{C}_{1})\times10^{-6}}$
$\therefore2 + \text{C}_{1} = 10 $
$\text{C}_{1} = 8 \mu\text{F}$.
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