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Standard XII

Physics

AC Applied to a Resistor

An LCR circui...

Question

An LCR circuit contains resistance of $110 Ω$ and a supply of $220 V$ at $300 \frac{r a d}{s}$ angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by $45^{0}$ . If on the other hand, only inductor is removed the current leads by $45^{0}$ with the applied voltage. The $r m s$ current flowing in the circuit will be:

$2.5 A$

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$2 A$

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$1 A$

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$1.5 A$

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Solution

The correct option is B
$2 A$

Step1: Given data.
Resistance, $R = 110 Ω$
Supply voltage, $V = 220 V$
Angular frequency, $ω_{r} = 300 \frac{r a d}{s}$
Current lags behind the voltage, $θ = 45^{0}$
Current leads the voltage by removing inductor, $θ = 45^{0}$
Step2: Find the $r m s$ current flowing in the circuit.
Since we know that when $L$ and $C$ are connected with $R$ in series the circuit will come in resonance.
Therefore,
$I_{r m s} = \frac{V_{r m s}}{R}$
Where $I_{r m s}$ and $V_{r m s}$ is $r m s$ current and voltage through the circuit.
$I_{r m s} = \frac{220}{110}$
$I_{r m s} = 2 A$
Hence, option B is correct.

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