In a series LCR circuit with an A C source R = 300 Ω , C = 20 μ F , L = 1 H , E r m s = 50 V and f = 50 π H z . The potential difference across the capacitor is -

X C = 1 C ω = 1 20 × 1 0 - 6 × 2 π × 50 π = 500 Ω X L = L ω = L × 2 π × 50 π = 100 Ω I r m s = E r m s Z = 50 R 2 + X C - X L 2 = 50 300 2 + 500 - 100 2 = 50 500 = 1 10 A V C = I r m s × X C = 1 10 × 500 V = 50 V

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PhysicsEMI/ACSeries AC CircuitsEasy2 minPYQ_2020

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In a series LCR circuit with an $A C$ source $R = 300 Ω, C = 20 μ F, L = 1 H, E_{r m s} = 50 V$ and $f = \frac{50}{π} H z$ . The potential difference across the capacitor is -

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Answer:

Solution:

$X_{C} = \frac{1}{C ω} = \frac{1}{20 \times 1 0^{- 6} \times 2 π \times \frac{50}{π}} = 500 Ω$

$X_{L} = L ω = L \times 2 π \times \frac{50}{π} = 100 Ω$

$I_{r m s} = \frac{E_{r m s}}{Z} = \frac{50}{\sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}}$

$= \frac{50}{\sqrt{{(300)}^{2} + {(500 - 100)}^{2}}}$

$= \frac{50}{500} = \frac{1}{10} A$

$V_{C} = I_{r m s} \times X_{C}$

$= \frac{1}{10} \times 500 V$

$= 50 V$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits

⏱ 2mℹ️ Source: PYQ_2020

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