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) :

Correct answer: A. 0.93 A Capacitive Reactance aligned X_C & =(1)/(ω C)=(1)/(2 π f C)= 1 2 × 3.14 × 50 × 10 × 10^ -6 & =(1000)/(3.14) aligned aligned & V_ rms =210 V & i_ rms = V_ rms X_C =(210)/(X_C) aligned aligned Peak current & = 2 i_ ms = 2 × (210)/(1000) × 3.14=0.932 & 0.93 A aligned

NEET 2024

Exam: NEET
Year: 2024
Subject: Physics
Chapter: Alternating Current

NEET 2024 · Physics · Alternating Current

A 10 μ F capacitor is connected to a 210 ~V, 50 ~Hz source as shown...

10 \mu \mathrm{F}

capacitor is connected to a

210 \mathrm{~V}, 50 \mathrm{~Hz}

source as shown in figure. The peak current in the circuit is nearly

(\pi=3.14)

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

Options

A0.93 A
B1.20 A
C $\quad 0.35 \mathrm{~A}$
D0.58 A

Correct answer

A. 0.93 A

Step-by-step solution

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} \end{aligned}

\begin{aligned} \text{ Peak current } & =\sqrt{2} i_{\mathrm{ms}}=\sqrt{2} \times \frac{210}{1000} \times 3.14=0.932 \\ & \simeq 0.93 \mathrm{ A} \end{aligned}

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