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NEET

Exam: NEET
Subject: Physics
Chapter: Alternating Current

NEET · Physics · Alternating Current

When an A C source of emf E with frequency ω=100 ~Hz is connected...

When an

A C

source of emf

E

with frequency

\omega=100 \mathrm{~Hz}

is connected across a circuit, the phase difference between

E

and current I in the circuit is observed to be

\frac{\pi}{4}

as shown in the figure. If the circuit consist of only

\mathrm{RC}

or

\mathrm{RL}

in series, then When an A C source of emf E with frequency ω=100 ~Hz is connected across a circuit, the phase difference between E and

When an A C source of emf E with frequency ω=100 ~Hz is connected across a circuit, the phase difference between E and

Options

A $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=5 \mu \mathrm{F}$
B $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=10 \mathrm{H}$
C $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=1 \mathrm{H}$
D $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}$

Correct answer

D. $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}$

Step-by-step solution

.. Since supply voltage lags the current

\operatorname{By}(\phi=\pi / 4)

\begin{aligned}&\tan \phi=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}=1, \mathrm{X}_{\mathrm{C}}=\mathrm{R} \ldots \ldots(1) \\ &\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 10 \times 10^{-6}}=1 \mathrm{k} \Omega=\mathrm{R} \end{aligned}

$$

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