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Alternating Current — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current3 Marks
Question
  1. Determine the value of phase difference between the current and the voltage in the given series $\text{LCR}$ circuit.
  1. 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.
✓

Answer

  1. In $\text{LCR}$ circuit: $\tan\varphi= \frac{\text{X}_{L} - \text{X}_{c}}{\text{R}} = \frac{\text{wL} - \frac{1}{\text{wC}}}{\text{R}}$
Now $\text{X}_{L} = \text{wL} = ( 1000\times100\times10^{-3})\Omega$
$= 100\Omega$
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 $
$\varphi = 45^{o}$
  1. Power Factor:
When power factor $= 1,$ we have $X_L=X_C$
$\therefore\text{X'}_{c} = \frac{1}{\omega\text{C}'} = 100 \Omega$
This gives $\text{C'} = \frac{1}{100\omega} = 10 \mu\text{F}$
We, therefore, need to add a capacitor of capacitance $(10-2) \mu F = 8\mu F$ 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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