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Alternating Current question

2024 · Q196
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Alternating Current question

2024 · Q196

NEETPhysicsAlternating CurrentMCQ+4 / −1

A 10μF10 \mu \mathrm{F}10μF capacitor is connected to a 210 V,50 Hz210 \mathrm{~V}, 50 \mathrm{~Hz}210 V,50 Hz source as shown in figure. The peak current in the circuit is nearly (π=3.14)(\pi=3.14)(π=3.14) :

NEET 2024 Physics - Alternating Current Question 5 English

  1. A
    0.58 A
  2. B
    0.93 A
  3. C
    1.20 A
  4. D
    0.35 A
View written solutionFree

Correct answer: B

Capacitive Reactance

$$\begin{aligned} X_C =\frac{1}{\omega C}=\frac{1}{2 \pi f C}=\frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} \\ =\frac{1000}{3.14}\end{aligned}$$

$$ \begin{aligned} & V_{\mathrm{rms}}=210 \mathrm{~V} \\ & i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}=\frac{210}{X_C} \\ & \text { Peak current }=\sqrt{2} i_{\mathrm{ms}}=\sqrt{2} \times \frac{210}{1000} \times 3.14=0.932 \\ & \qquad =0.93 \mathrm{~A} \end{aligned} $$

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