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Electrical Oscillations

A choke coil ...

Question

A choke coil of resistance $5 Ω$ and inductance $0.6 H$ is in series with a capacitance of $10 μ F$ . If a voltage of $200 V$ is applied and the frequency is adjusted to resonance, the current and voltage across the inductance and capacitance are $I_{0}, V_{0}$ and $V_{1}$ respectively. We have

I_{0} = 40 A

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V_{0} = 9.8 k V

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V_{1} = 9.8 k V

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V_{1} = 19.6 k V

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Solution

The correct options are
A $I_{0} = 40 A$
B $V_{0} = 9.8 k V$
C $V_{1} = 9.8 k V$
At resonance current, $I_{0} = \frac{V}{R} = \frac{200}{5} = 40 A$
At resonance frequency voltage across L $= V_{0}$ , and voltage across C $= V_{1}$ are always equal and out of phase by 180 degree.
Therefore, if $f = 65$ Hz , $V_{0} = j I_{0} L ω = I_{0} L 2 π f = j 9800 V$
$V_{1} = - j \frac{I_{0}}{C ω} = - j \frac{I_{0}}{C 2 π f} = - j 9800 V$

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A choke coil of resistance 5Ω and inductance 0.6 H is in series with a capacitance of 10 μF. If a voltage of 200 V is applied and the frequency is adjusted to resonance, the current and voltage across the inductance and capacitance are I0,V0 and V1 respectively. We have

The correct options are A I0=40 A B V0=9.8 kV C V1=9.8 kVAt resonance current, I0=VR=2005=40AAt resonance frequency voltage across L =V0, and voltage across C =V1 are always equal and out of phase by 180 degree.Therefore, if f=65 Hz , V0=jI0Lω=I0L2πf=j9800V V1=−jI0Cω=−jI0C2πf=−j 9800V

A choke coil of resistance $5 Ω$ and inductance $0.6 H$ is in series with a capacitance of $10 μ F$ . If a voltage of $200 V$ is applied and the frequency is adjusted to resonance, the current and voltage across the inductance and capacitance are $I_{0}, V_{0}$ and $V_{1}$ respectively. We have