The figure shows a series LCR circuit with L = 5.0 H, C = 80 $\mu$F, R = 40 W connected to a variable frequency 240V source. Calculate 1. The angular frequency of the source which drives the circuit at resonance. 2. The current at the resonating frequency. 3. The rms potential drop across the capacitor at resonance.

1. $\omega = \frac{1}{\sqrt{\text{LC}}}$ $ = \frac{1}{\sqrt{5\times80\times10^{-6}}} = 50\text{radian / s }$ 2. Current at resonance $\text{I}_{rms} = \frac{\text{V}_{rms}}{\text{R}} = \frac{240}{40}\text{A} = 6\text{A}$ 3. $V_{rms}$ across capacitor $\text{V}_{rms} = \text{I}_{rms}\text{X}_{c}$ $ = 6\times\frac{1}{50\times80\times10^{-6}}\text{V}$ $ = \frac{6\times10^{6}}{4\times10^{3}}\text{V} = \frac{6000}{4}\text{V} = 1500\text{V}.$

The figure shows a series LCR circuit with L = 5.0 H, C = 80 $\mu$F, R = 40 W connected to a variable frequency 240V source. Calculate 1. The angular frequency of the source which drives the circuit at resonance. 2. The current at the resonating frequency. 3. The rms potential drop across the capacitor at resonance.

1. $\omega = \frac{1}{\sqrt{\text{LC}}}$ $ = \frac{1}{\sqrt{5\times80\times10^{-6}}} = 50\text{radian / s }$ 2. Current at resonance $\text{I}_{rms} = \frac{\text{V}_{rms}}{\text{R}} = \frac{240}{40}\text{A} = 6\text{A}$ 3. $V_{rms}$ across capacitor $\text{V}_{rms} = \text{I}_{rms}\text{X}_{c}$ $ = 6\times\frac{1}{50\times80\times10^{-6}}\text{V}$ $ = \frac{6\times10^{6}}{4\times10^{3}}\text{V} = \frac{6000}{4}\text{V} = 1500\text{V}.$

The figure shows a series LCR circuit with L = 5.0 H, C = 80 $\mu$F, R = 40 W connected to a variable frequency 240V source. Calculate 1. The angular frequency of the source which drives the circuit at resonance. 2. The current at the resonating frequency. 3. The rms potential drop across the capacitor at resonance.

1. $\omega = \frac{1}{\sqrt{\text{LC}}}$ $ = \frac{1}{\sqrt{5\times80\times10^{-6}}} = 50\text{radian / s }$ 2. Current at resonance $\text{I}_{rms} = \frac{\text{V}_{rms}}{\text{R}} = \frac{240}{40}\text{A} = 6\text{A}$ 3. $V_{rms}$ across capacitor $\text{V}_{rms} = \text{I}_{rms}\text{X}_{c}$ $ = 6\times\frac{1}{50\times80\times10^{-6}}\text{V}$ $ = \frac{6\times10^{6}}{4\times10^{3}}\text{V} = \frac{6000}{4}\text{V} = 1500\text{V}.$

The figure shows a series LCR circuit with L = 5.0 H, C = 80 muF, R = 40 W connected to a variable frequency

3 Marks Question

GSEB - English Medium

The figure shows a series LCR circuit with L = 5.0 H, C = 80 $\mu$F, R = 40 W connected to a variable frequency 240V source. Calculate

The angular frequency of the source which drives the circuit at resonance.
The current at the resonating frequency.
The rms potential drop across the capacitor at resonance.

CBSE DELHI - SET 1 2012

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Solution

$\omega = \frac{1}{\sqrt{\text{LC}}}$

$ = \frac{1}{\sqrt{5\times80\times10^{-6}}} = 50\text{radian / s }$

Current at resonance

$\text{I}_{rms} = \frac{\text{V}_{rms}}{\text{R}} = \frac{240}{40}\text{A} = 6\text{A}$

$V_{rms}$ across capacitor

$\text{V}_{rms} = \text{I}_{rms}\text{X}_{c}$

$ = 6\times\frac{1}{50\times80\times10^{-6}}\text{V}$

$ = \frac{6\times10^{6}}{4\times10^{3}}\text{V} = \frac{6000}{4}\text{V} = 1500\text{V}.$

STD 12 Science
Physics
Alternating Current
NCERT

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