A $10 μ F$ capacitor is connected to a $210 V, 50 Hz$ source as shown in figure. The peak current in the circuit is nearly ( $π = 3.14$ ):

Option 1 -

0.58 A

Option 2 -

0.93 A

Option 3 -

1.20 A

Option 4 -

0.35 A

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

Capacitive Reactance $X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 1 0^{- 6}}$

$= \frac{1000}{3.14}$

$V_{rms} = 210 V$

$i_{rms} = \frac{V_{rms}}{X_{C}} = \frac{210}{X_{C}}$

$Peak current = \sqrt[]{2} i_{ms} = \sqrt[]{2} \times \frac{210}{1000} \times 3.14 = 0.932$

$? 0.93 A$

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A $10 μ F$ capacitor is connected to a $210 V, 50 Hz$ source as shown in figure. The peak current in the circuit is nearly ( $π = 3.14$ ):

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A 10μ F capacitor is connected to a 210 V,50 Hz source as shown in figure. The peak current in the circuit is nearly ( π = 3 . 1 4 ):

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A $10 μ F$ capacitor is connected to a $210 V, 50 Hz$ source as shown in figure. The peak current in the circuit is nearly ( $π = 3.14$ ):