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An alternating EMF of e=200 sin 100πt is applied across a capacitor of capacitance 2×10^-6F . The capacitive reactance and the peak current is?
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An alternating EMF of e=200 sin 100πt is applied across a capacitor of...
E = Eo/√2. E = 200/√2. E = 100√2.
Xc = 1/wc. Xc= 1/100π*2*10-6. Xc=5000/π
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An alternating EMF of e=200 sin 100πt is applied across a capacitor of...
Capacitive Reactance:
The capacitive reactance (Xc) is a measure of the opposition offered by a capacitor to the flow of alternating current (AC). It depends on the frequency (f) of the AC signal and the capacitance (C) of the capacitor. The formula to calculate the capacitive reactance is given by:

Xc = 1 / (2πfC)

where Xc is the capacitive reactance in ohms, f is the frequency in hertz, and C is the capacitance in farads.

Peak Current:
The peak current (I) is the maximum value of the alternating current that flows through the circuit. It can be calculated using Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the total impedance (Z) of the circuit. In the case of a purely capacitive circuit, the total impedance is equal to the capacitive reactance (Xc). The formula to calculate the peak current is given by:

I = V / Xc

where I is the peak current in amperes, V is the peak voltage in volts, and Xc is the capacitive reactance in ohms.

Given:
EMF (e) = 200 sin(100πt)
Capacitance (C) = 2×10^-6F

Calculating Capacitive Reactance:
The frequency (f) can be obtained from the given EMF equation. In this case, the frequency is 100π Hz.

Using the formula for capacitive reactance, we can calculate Xc:

Xc = 1 / (2πfC)
= 1 / (2π(100π)(2×10^-6))
= 1 / (200π^2×10^-6)
≈ 7.96 ohms

The capacitive reactance is approximately 7.96 ohms.

Calculating Peak Current:
The peak voltage (V) can be obtained from the given EMF equation. In this case, the peak voltage is 200 volts.

Using the formula for peak current, we can calculate I:

I = V / Xc
= 200 / 7.96
≈ 25.13 amperes

The peak current is approximately 25.13 amperes.

Conclusion:
The capacitive reactance of the capacitor is approximately 7.96 ohms, and the peak current flowing through the circuit is approximately 25.13 amperes. The capacitive reactance is determined by the frequency and capacitance, while the peak current is determined by the peak voltage and capacitive reactance.
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An alternating EMF of e=200 sin 100πt is applied across a capacitor of capacitance 2×10^-6F . The capacitive reactance and the peak current is? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An alternating EMF of e=200 sin 100πt is applied across a capacitor of capacitance 2×10^-6F . The capacitive reactance and the peak current is? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An alternating EMF of e=200 sin 100πt is applied across a capacitor of capacitance 2×10^-6F . The capacitive reactance and the peak current is?.
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