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Alternating Current — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current3 Marks
Question
  1. Determine the value of phase difference between the current and the voltage in the given series 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 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) 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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