155250
The average power dissipated in a pure capacitance $A . C$ circuit is
1 $\mathrm{CV}$
2 zero
3 $\frac{1}{2} \mathrm{CV}^{2}$
4 $\frac{1}{4} \mathrm{CV}^{2}$
Explanation:
B Average power in AC circuit is given by $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi$ For pure capacitive circuit $(\phi)=90^{\circ}$ $\mathrm{P}=\mathrm{V}_{\text {rms }} \cdot \mathrm{I}_{\text {rms }} \cdot \cos 90^{\circ}$ Since, $\cos 90^{\circ}=0$ If we put $\cos 90^{\circ}=0$, we get - Then, power will be zero.
J and K CET- 2009
Alternating Current
155251
The natural frequency $\left(\omega_{0}\right)$ of oscillations in LC-circuit is given by
1 $\frac{1}{2 \pi} \frac{1}{\sqrt{\mathrm{LC}}}$
2 $\frac{1}{2 \pi} \sqrt{\mathrm{LC}}$
3 $\frac{1}{\sqrt{\mathrm{LC}}}$
4 $\sqrt{\mathrm{LC}}$
Explanation:
C Let, natural frequency \(=\omega_0\) At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\omega_{0} \mathrm{~L}=\frac{1}{\omega_{0} \mathrm{C}}$ $\omega_{0}^{2} \mathrm{LC}=1$ $\omega_{0}^{2}=\frac{1}{\mathrm{LC}}$ $\omega_{0}=\frac{1}{\sqrt{\mathrm{LC}}}$
J and K CET- 2008
Alternating Current
155255
The resonance frequency of a series LCR circuit containing $L=12.5 \mathrm{mH}, C=500 \mu F$ and $R=160 \Omega$ is
1 $\frac{100}{2 \pi}$
2 $\frac{400}{2 \pi}$
3 $\frac{2 \pi}{300}$
4 $\frac{2 \pi}{600}$
Explanation:
B Given that, $\mathrm{L}=12.5 \mathrm{mH}=12.5 \times 10^{-3} \mathrm{H}$ $\mathrm{C}=500 \mu \mathrm{F}=500 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=160 \Omega$ We know that, resonance frequency - $2 \pi \mathrm{f}=\frac{1}{\sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{12.5 \times 10^{-3} \times 500 \times 10^{-6}}}$ $\mathrm{f}=\frac{400}{2 \pi}$
J and K CET- 1997
Alternating Current
155258
A $100 \mathrm{~V}$, AC source of frequency $500 \mathrm{~Hz}$ is connected to an $L C R$ circuit with $L=8.1 \mathrm{mH}$, $C=12.5 \mu F, R=10 \Omega$ all connected in series as shown in figure. What is the quality factor of circuit?
1 2.02
2 2.5434
3 20.54
4 200.54
Explanation:
B Given, $\mathrm{L}=8.1 \mathrm{mH}=8.1 \times 10^{-3} \mathrm{H}$, $\mathrm{C}=12.5 \mu \mathrm{F}=12.5 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=10 \Omega$ Quality factor $(\mathrm{Q})=\frac{\text { Voltage across } \mathrm{L} \text { or } \mathrm{C}}{\text { Applied voltage }}$ $\mathrm{Q}=\frac{(\omega \mathrm{L}) \mathrm{I}}{\mathrm{IR}}=\frac{\omega \mathrm{L}}{\mathrm{R}}$ Or $\quad \frac{\left(\frac{1}{\omega C}\right) I}{I R}=\frac{1}{R C \omega}$ at resonence, $\left.X_{L}=X_{C}\right]$ $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\because \quad \mathrm{Q}=\frac{\mathrm{L}}{\mathrm{R}} \times \frac{1}{\sqrt{\mathrm{LC}}}$ Or $\quad \mathrm{Q}=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$ $\therefore \quad \mathrm{Q}=\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{2 \pi \mathrm{fL}}{\mathrm{R}}$ $=\frac{2 \times \pi \times 500 \times 8.1 \times 10^{-3}}{10}$ $=\frac{8.1 \pi}{10}$ $\mathrm{Q}=2.5434$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155250
The average power dissipated in a pure capacitance $A . C$ circuit is
1 $\mathrm{CV}$
2 zero
3 $\frac{1}{2} \mathrm{CV}^{2}$
4 $\frac{1}{4} \mathrm{CV}^{2}$
Explanation:
B Average power in AC circuit is given by $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi$ For pure capacitive circuit $(\phi)=90^{\circ}$ $\mathrm{P}=\mathrm{V}_{\text {rms }} \cdot \mathrm{I}_{\text {rms }} \cdot \cos 90^{\circ}$ Since, $\cos 90^{\circ}=0$ If we put $\cos 90^{\circ}=0$, we get - Then, power will be zero.
J and K CET- 2009
Alternating Current
155251
The natural frequency $\left(\omega_{0}\right)$ of oscillations in LC-circuit is given by
1 $\frac{1}{2 \pi} \frac{1}{\sqrt{\mathrm{LC}}}$
2 $\frac{1}{2 \pi} \sqrt{\mathrm{LC}}$
3 $\frac{1}{\sqrt{\mathrm{LC}}}$
4 $\sqrt{\mathrm{LC}}$
Explanation:
C Let, natural frequency \(=\omega_0\) At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\omega_{0} \mathrm{~L}=\frac{1}{\omega_{0} \mathrm{C}}$ $\omega_{0}^{2} \mathrm{LC}=1$ $\omega_{0}^{2}=\frac{1}{\mathrm{LC}}$ $\omega_{0}=\frac{1}{\sqrt{\mathrm{LC}}}$
J and K CET- 2008
Alternating Current
155255
The resonance frequency of a series LCR circuit containing $L=12.5 \mathrm{mH}, C=500 \mu F$ and $R=160 \Omega$ is
1 $\frac{100}{2 \pi}$
2 $\frac{400}{2 \pi}$
3 $\frac{2 \pi}{300}$
4 $\frac{2 \pi}{600}$
Explanation:
B Given that, $\mathrm{L}=12.5 \mathrm{mH}=12.5 \times 10^{-3} \mathrm{H}$ $\mathrm{C}=500 \mu \mathrm{F}=500 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=160 \Omega$ We know that, resonance frequency - $2 \pi \mathrm{f}=\frac{1}{\sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{12.5 \times 10^{-3} \times 500 \times 10^{-6}}}$ $\mathrm{f}=\frac{400}{2 \pi}$
J and K CET- 1997
Alternating Current
155258
A $100 \mathrm{~V}$, AC source of frequency $500 \mathrm{~Hz}$ is connected to an $L C R$ circuit with $L=8.1 \mathrm{mH}$, $C=12.5 \mu F, R=10 \Omega$ all connected in series as shown in figure. What is the quality factor of circuit?
1 2.02
2 2.5434
3 20.54
4 200.54
Explanation:
B Given, $\mathrm{L}=8.1 \mathrm{mH}=8.1 \times 10^{-3} \mathrm{H}$, $\mathrm{C}=12.5 \mu \mathrm{F}=12.5 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=10 \Omega$ Quality factor $(\mathrm{Q})=\frac{\text { Voltage across } \mathrm{L} \text { or } \mathrm{C}}{\text { Applied voltage }}$ $\mathrm{Q}=\frac{(\omega \mathrm{L}) \mathrm{I}}{\mathrm{IR}}=\frac{\omega \mathrm{L}}{\mathrm{R}}$ Or $\quad \frac{\left(\frac{1}{\omega C}\right) I}{I R}=\frac{1}{R C \omega}$ at resonence, $\left.X_{L}=X_{C}\right]$ $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\because \quad \mathrm{Q}=\frac{\mathrm{L}}{\mathrm{R}} \times \frac{1}{\sqrt{\mathrm{LC}}}$ Or $\quad \mathrm{Q}=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$ $\therefore \quad \mathrm{Q}=\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{2 \pi \mathrm{fL}}{\mathrm{R}}$ $=\frac{2 \times \pi \times 500 \times 8.1 \times 10^{-3}}{10}$ $=\frac{8.1 \pi}{10}$ $\mathrm{Q}=2.5434$
155250
The average power dissipated in a pure capacitance $A . C$ circuit is
1 $\mathrm{CV}$
2 zero
3 $\frac{1}{2} \mathrm{CV}^{2}$
4 $\frac{1}{4} \mathrm{CV}^{2}$
Explanation:
B Average power in AC circuit is given by $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi$ For pure capacitive circuit $(\phi)=90^{\circ}$ $\mathrm{P}=\mathrm{V}_{\text {rms }} \cdot \mathrm{I}_{\text {rms }} \cdot \cos 90^{\circ}$ Since, $\cos 90^{\circ}=0$ If we put $\cos 90^{\circ}=0$, we get - Then, power will be zero.
J and K CET- 2009
Alternating Current
155251
The natural frequency $\left(\omega_{0}\right)$ of oscillations in LC-circuit is given by
1 $\frac{1}{2 \pi} \frac{1}{\sqrt{\mathrm{LC}}}$
2 $\frac{1}{2 \pi} \sqrt{\mathrm{LC}}$
3 $\frac{1}{\sqrt{\mathrm{LC}}}$
4 $\sqrt{\mathrm{LC}}$
Explanation:
C Let, natural frequency \(=\omega_0\) At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\omega_{0} \mathrm{~L}=\frac{1}{\omega_{0} \mathrm{C}}$ $\omega_{0}^{2} \mathrm{LC}=1$ $\omega_{0}^{2}=\frac{1}{\mathrm{LC}}$ $\omega_{0}=\frac{1}{\sqrt{\mathrm{LC}}}$
J and K CET- 2008
Alternating Current
155255
The resonance frequency of a series LCR circuit containing $L=12.5 \mathrm{mH}, C=500 \mu F$ and $R=160 \Omega$ is
1 $\frac{100}{2 \pi}$
2 $\frac{400}{2 \pi}$
3 $\frac{2 \pi}{300}$
4 $\frac{2 \pi}{600}$
Explanation:
B Given that, $\mathrm{L}=12.5 \mathrm{mH}=12.5 \times 10^{-3} \mathrm{H}$ $\mathrm{C}=500 \mu \mathrm{F}=500 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=160 \Omega$ We know that, resonance frequency - $2 \pi \mathrm{f}=\frac{1}{\sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{12.5 \times 10^{-3} \times 500 \times 10^{-6}}}$ $\mathrm{f}=\frac{400}{2 \pi}$
J and K CET- 1997
Alternating Current
155258
A $100 \mathrm{~V}$, AC source of frequency $500 \mathrm{~Hz}$ is connected to an $L C R$ circuit with $L=8.1 \mathrm{mH}$, $C=12.5 \mu F, R=10 \Omega$ all connected in series as shown in figure. What is the quality factor of circuit?
1 2.02
2 2.5434
3 20.54
4 200.54
Explanation:
B Given, $\mathrm{L}=8.1 \mathrm{mH}=8.1 \times 10^{-3} \mathrm{H}$, $\mathrm{C}=12.5 \mu \mathrm{F}=12.5 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=10 \Omega$ Quality factor $(\mathrm{Q})=\frac{\text { Voltage across } \mathrm{L} \text { or } \mathrm{C}}{\text { Applied voltage }}$ $\mathrm{Q}=\frac{(\omega \mathrm{L}) \mathrm{I}}{\mathrm{IR}}=\frac{\omega \mathrm{L}}{\mathrm{R}}$ Or $\quad \frac{\left(\frac{1}{\omega C}\right) I}{I R}=\frac{1}{R C \omega}$ at resonence, $\left.X_{L}=X_{C}\right]$ $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\because \quad \mathrm{Q}=\frac{\mathrm{L}}{\mathrm{R}} \times \frac{1}{\sqrt{\mathrm{LC}}}$ Or $\quad \mathrm{Q}=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$ $\therefore \quad \mathrm{Q}=\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{2 \pi \mathrm{fL}}{\mathrm{R}}$ $=\frac{2 \times \pi \times 500 \times 8.1 \times 10^{-3}}{10}$ $=\frac{8.1 \pi}{10}$ $\mathrm{Q}=2.5434$
155250
The average power dissipated in a pure capacitance $A . C$ circuit is
1 $\mathrm{CV}$
2 zero
3 $\frac{1}{2} \mathrm{CV}^{2}$
4 $\frac{1}{4} \mathrm{CV}^{2}$
Explanation:
B Average power in AC circuit is given by $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi$ For pure capacitive circuit $(\phi)=90^{\circ}$ $\mathrm{P}=\mathrm{V}_{\text {rms }} \cdot \mathrm{I}_{\text {rms }} \cdot \cos 90^{\circ}$ Since, $\cos 90^{\circ}=0$ If we put $\cos 90^{\circ}=0$, we get - Then, power will be zero.
J and K CET- 2009
Alternating Current
155251
The natural frequency $\left(\omega_{0}\right)$ of oscillations in LC-circuit is given by
1 $\frac{1}{2 \pi} \frac{1}{\sqrt{\mathrm{LC}}}$
2 $\frac{1}{2 \pi} \sqrt{\mathrm{LC}}$
3 $\frac{1}{\sqrt{\mathrm{LC}}}$
4 $\sqrt{\mathrm{LC}}$
Explanation:
C Let, natural frequency \(=\omega_0\) At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\omega_{0} \mathrm{~L}=\frac{1}{\omega_{0} \mathrm{C}}$ $\omega_{0}^{2} \mathrm{LC}=1$ $\omega_{0}^{2}=\frac{1}{\mathrm{LC}}$ $\omega_{0}=\frac{1}{\sqrt{\mathrm{LC}}}$
J and K CET- 2008
Alternating Current
155255
The resonance frequency of a series LCR circuit containing $L=12.5 \mathrm{mH}, C=500 \mu F$ and $R=160 \Omega$ is
1 $\frac{100}{2 \pi}$
2 $\frac{400}{2 \pi}$
3 $\frac{2 \pi}{300}$
4 $\frac{2 \pi}{600}$
Explanation:
B Given that, $\mathrm{L}=12.5 \mathrm{mH}=12.5 \times 10^{-3} \mathrm{H}$ $\mathrm{C}=500 \mu \mathrm{F}=500 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=160 \Omega$ We know that, resonance frequency - $2 \pi \mathrm{f}=\frac{1}{\sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{12.5 \times 10^{-3} \times 500 \times 10^{-6}}}$ $\mathrm{f}=\frac{400}{2 \pi}$
J and K CET- 1997
Alternating Current
155258
A $100 \mathrm{~V}$, AC source of frequency $500 \mathrm{~Hz}$ is connected to an $L C R$ circuit with $L=8.1 \mathrm{mH}$, $C=12.5 \mu F, R=10 \Omega$ all connected in series as shown in figure. What is the quality factor of circuit?
1 2.02
2 2.5434
3 20.54
4 200.54
Explanation:
B Given, $\mathrm{L}=8.1 \mathrm{mH}=8.1 \times 10^{-3} \mathrm{H}$, $\mathrm{C}=12.5 \mu \mathrm{F}=12.5 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=10 \Omega$ Quality factor $(\mathrm{Q})=\frac{\text { Voltage across } \mathrm{L} \text { or } \mathrm{C}}{\text { Applied voltage }}$ $\mathrm{Q}=\frac{(\omega \mathrm{L}) \mathrm{I}}{\mathrm{IR}}=\frac{\omega \mathrm{L}}{\mathrm{R}}$ Or $\quad \frac{\left(\frac{1}{\omega C}\right) I}{I R}=\frac{1}{R C \omega}$ at resonence, $\left.X_{L}=X_{C}\right]$ $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\because \quad \mathrm{Q}=\frac{\mathrm{L}}{\mathrm{R}} \times \frac{1}{\sqrt{\mathrm{LC}}}$ Or $\quad \mathrm{Q}=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$ $\therefore \quad \mathrm{Q}=\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{2 \pi \mathrm{fL}}{\mathrm{R}}$ $=\frac{2 \times \pi \times 500 \times 8.1 \times 10^{-3}}{10}$ $=\frac{8.1 \pi}{10}$ $\mathrm{Q}=2.5434$
