An alternating voltage- of peak value 283 V and of varying frequency- is applied across a series LCR combination in which- R- 3 - L-25 mH and C-400 -F- The frequency of the source for which the current and voltage are in phase- is -

The current and voltage will be in phase at resonance. So, frequency, f = nbsp;12 π√(LC) = 1 2 × 3.14×√(25 × 10^ 3× 400 × 10^ 6) ⇒ f ≈ 50 Hz nbsp; Hence ...

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Standard XII

Physics

Dimensional Analysis

An alternatin...

Question

An alternating voltage, of peak value $283 V$ and of varying frequency, is applied across a series LCR combination in which, $R = 3 Ω, L = 25 mH$ and $C = 400 μ F$ . The frequency of the source for which the current and voltage are in phase, is -

90 Hz

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70 Hz

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30 Hz

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50 Hz

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Solution

The correct option is D $50 Hz$
The current and voltage will be in phase at resonance.

So, frequency,

$f = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 \times 3.14 \times \sqrt{25 \times 10^{- 3} \times 400 \times 10^{- 6}}}$

$\Rightarrow f \approx 50 Hz$

Hence, option $(D)$ is the correct answer.

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An alternating voltage, of peak value 283 V and of varying frequency, is applied across a series LCR combination in which, R=3 Ω,L=25 mH and C=400 μF. The frequency of the source for which the current and voltage are in phase, is -

The correct option is D 50 HzThe current and voltage will be in phase at resonance. So, frequency, f=12π√LC=12×3.14×√25×10−3×400×10−6 ⇒f≈50 Hz Hence, option (D) is the correct answer.

An alternating voltage, of peak value $283 V$ and of varying frequency, is applied across a series LCR combination in which, $R = 3 Ω, L = 25 mH$ and $C = 400 μ F$ . The frequency of the source for which the current and voltage are in phase, is -

The correct option is D $50 Hz$
The current and voltage will be in phase at resonance.

So, frequency,

$f = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 \times 3.14 \times \sqrt{25 \times 10^{- 3} \times 400 \times 10^{- 6}}}$

$\Rightarrow f \approx 50 Hz$

Hence, option $(D)$ is the correct answer.