A series LCR circuit driven by 300V at a frequency of 50 Hz contains a resistance R = $3 k Ω,$ an inductor of inductive reactances X_L = 250π $Ω$ and an unknown capacitor. The value of capacitance to maximize the average power should be:

(Take π² = 10)

Option 1 -  400 <math><mrow></mrow></math> <math> <mrow> <mi>μ</mi> <mi>F</mi> </mrow> </math>

Option 2 - 4 <math> <mrow> <mi>μ</mi> <mi>F</mi> </mrow> </math>

Option 3 - 40 <math> <mrow> <mi>μ</mi> <mi>F</mi> </mrow> </math>

Option 4 - 25 <math> <mrow> <mi>μ</mi> <mi>F</mi> </mrow> </math>

4 Views|Posted 7 months ago

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Vishal Baghel | Contributor-Level 10

7 months ago

Correct Option - 2

Detailed Solution:

Power is maximum at resonance

So, $X_{L} = X_{C} = \frac{1}{ω C}$ $_{}_{} \frac{}{}$

$\Rightarrow C = \frac{1}{ω * X_{L}}$

$= \frac{1}{250 * 100 * π^{2}}$

$= 4 μ F [π^{2} = 10]$

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A series LCR circuit driven by 300V at a frequency of 50 Hz contains a resistance R = 3 k Ω , an inductor of inductive reactances XL = 250π Ω and an unknown capacitor. The value of capacitance to maximize the average power should be: (Take π2 = 10)

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