A series resonant circuit contains L - 5- mH- C - 200- F and R - 100 - If a source of emf e - 200 sin 1000 - t is applied- then the rms current is-

The correct option is C 1.41 A Source emf is given as #160; e = 200 sin 1000 π t #160;Comparing it with #160; e = e_o sin wt We get ...

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Standard XII

Physics

RLC Circuit

A series reso...

Question

A series resonant circuit contains $L = \frac{5}{π} m H, C = \frac{200}{π} μ F$ and $R = 100 μ$ . If a source of emf $e = 200 s i n 1000 π t$ is applied, then the rms current is:

2 A

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200 \sqrt{2} A

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100 \sqrt{2} A

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1.41 A

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Solution

The correct option is C $1.41 A$
Source emf is given as $e = 200 s i n 1000 π t$
Comparing it with $e = e_{o} sin w t$
We get $ω = 1000 π$
$\Rightarrow f = \frac{ω}{2 π} = 500 H z$
Also, $e_{0} = 200 V$
At resonance $Z = R$
So, current flowing $i_{0} = \frac{e_{o}}{R} = \frac{200}{100} = 2 A$
Rms value of current $i_{r m s} = \frac{i_{0}}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} = 1.41 A$

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Similar questions

Q. A series resonant circuit contains

L = 5 / π m H, C = 200 / π, R = 100 Ω

. If a source of emf

e = 200 sin 1000 π t

is applied with a rms current

1.41 A

. Find the magnitude of voltage across the inductor.

In an $L - R$ series circuit $(L = \frac{175}{11} m H a n d R = 12 Ω)$ , a variable emf source $(V = V_{0} s i n ω t)$ of $V_{r m s} = 130 \sqrt{2} V$ and frequency $50 H z$ is applied. The current amplitude in the circuit and phase of current with respect to voltage are respectively $(use π = \frac{22}{7})$ .