155298
Assertion: A capacitor blocks direct current in the steady state. Reason: The capacitive reactance of the capacitor is inversely proportional to frequency $f$ of the source of emf.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
A Given that, $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ Frequency of direct current is zero that is i.e., $\mathrm{f}=0$ $\therefore \quad \mathrm{X}_{\mathrm{C}}=\infty$ So, it blocks the DC current, $\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ $\mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$ Hence, option (a) is correct.
AIIMS-2011
Alternating Current
155300
In L-C-R circuit power factor at resonance is
1 less than one
2 greater than one
3 unity
4 Can't predicted
Explanation:
C We know that, $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\cos \phi=\frac{\mathrm{R}}{\mathrm{R}}=1$ Current and voltage across the resistance are all in phase. So, phase angle $\phi=0^{\circ}$ Therefore, $\cos \phi=1$
MHT-CET 2011
Alternating Current
155301
The equation of $\mathrm{AC}$ voltage is $\mathrm{E}=\mathbf{2 2 0}$ sin $(\omega t+\pi / 6)$ and the $\mathrm{AC}$ current is $\mathrm{I}=10 \sin$ $(\omega t-\pi / 6)$. The average power dissipated is
155302
The natural frequency of the circuit shown in adjoining figure is
1 $\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
2 $\frac{1}{2 \pi \sqrt{2 \mathrm{LC}}}$
3 $\frac{2}{2 \pi \sqrt{\mathrm{LC}}}$
4 zero
Explanation:
A In the given circuit inductor and capacitor are in series- $\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}} \text { and } \mathrm{L}_{\mathrm{S}}=\mathrm{L}+\mathrm{L}$ $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C}}{2} \mathrm{~L}_{\mathrm{S}}=2 \mathrm{~L}$ Therefore natural frequency of the circuit $\mathrm{f}_{0}=\frac{1}{2 \pi \sqrt{\mathrm{L}_{\mathrm{S}} \mathrm{C}_{\mathrm{S}}}}$ $\mathrm{f}_{\mathrm{o}}=\frac{1}{2 \pi \sqrt{2 \mathrm{~L} \times \frac{\mathrm{C}}{2}}}$ $\mathrm{f}_{\mathrm{o}}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
VITEEE-2013
Alternating Current
155307
A current $I=I_{0} \sin \left(\omega t-\frac{\pi}{2}\right)$ flows in an A.C. potential of $E=E_{0} \sin \omega t$ has been applied, then the power consumption $P$ in the circuit will be
D Given that, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin \omega \mathrm{t}$ And $\quad \mathrm{I}=\mathrm{I}_{\mathrm{o}} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)$ Here, voltage and current are in out of phase $\left[\phi=\frac{\pi}{2}\right]$ $\therefore$ Power factor, $\cos \phi=\cos 90^{\circ}=0$ Hence, power consumption become zero.
155298
Assertion: A capacitor blocks direct current in the steady state. Reason: The capacitive reactance of the capacitor is inversely proportional to frequency $f$ of the source of emf.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
A Given that, $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ Frequency of direct current is zero that is i.e., $\mathrm{f}=0$ $\therefore \quad \mathrm{X}_{\mathrm{C}}=\infty$ So, it blocks the DC current, $\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ $\mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$ Hence, option (a) is correct.
AIIMS-2011
Alternating Current
155300
In L-C-R circuit power factor at resonance is
1 less than one
2 greater than one
3 unity
4 Can't predicted
Explanation:
C We know that, $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\cos \phi=\frac{\mathrm{R}}{\mathrm{R}}=1$ Current and voltage across the resistance are all in phase. So, phase angle $\phi=0^{\circ}$ Therefore, $\cos \phi=1$
MHT-CET 2011
Alternating Current
155301
The equation of $\mathrm{AC}$ voltage is $\mathrm{E}=\mathbf{2 2 0}$ sin $(\omega t+\pi / 6)$ and the $\mathrm{AC}$ current is $\mathrm{I}=10 \sin$ $(\omega t-\pi / 6)$. The average power dissipated is
155302
The natural frequency of the circuit shown in adjoining figure is
1 $\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
2 $\frac{1}{2 \pi \sqrt{2 \mathrm{LC}}}$
3 $\frac{2}{2 \pi \sqrt{\mathrm{LC}}}$
4 zero
Explanation:
A In the given circuit inductor and capacitor are in series- $\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}} \text { and } \mathrm{L}_{\mathrm{S}}=\mathrm{L}+\mathrm{L}$ $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C}}{2} \mathrm{~L}_{\mathrm{S}}=2 \mathrm{~L}$ Therefore natural frequency of the circuit $\mathrm{f}_{0}=\frac{1}{2 \pi \sqrt{\mathrm{L}_{\mathrm{S}} \mathrm{C}_{\mathrm{S}}}}$ $\mathrm{f}_{\mathrm{o}}=\frac{1}{2 \pi \sqrt{2 \mathrm{~L} \times \frac{\mathrm{C}}{2}}}$ $\mathrm{f}_{\mathrm{o}}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
VITEEE-2013
Alternating Current
155307
A current $I=I_{0} \sin \left(\omega t-\frac{\pi}{2}\right)$ flows in an A.C. potential of $E=E_{0} \sin \omega t$ has been applied, then the power consumption $P$ in the circuit will be
D Given that, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin \omega \mathrm{t}$ And $\quad \mathrm{I}=\mathrm{I}_{\mathrm{o}} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)$ Here, voltage and current are in out of phase $\left[\phi=\frac{\pi}{2}\right]$ $\therefore$ Power factor, $\cos \phi=\cos 90^{\circ}=0$ Hence, power consumption become zero.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
155298
Assertion: A capacitor blocks direct current in the steady state. Reason: The capacitive reactance of the capacitor is inversely proportional to frequency $f$ of the source of emf.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
A Given that, $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ Frequency of direct current is zero that is i.e., $\mathrm{f}=0$ $\therefore \quad \mathrm{X}_{\mathrm{C}}=\infty$ So, it blocks the DC current, $\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ $\mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$ Hence, option (a) is correct.
AIIMS-2011
Alternating Current
155300
In L-C-R circuit power factor at resonance is
1 less than one
2 greater than one
3 unity
4 Can't predicted
Explanation:
C We know that, $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\cos \phi=\frac{\mathrm{R}}{\mathrm{R}}=1$ Current and voltage across the resistance are all in phase. So, phase angle $\phi=0^{\circ}$ Therefore, $\cos \phi=1$
MHT-CET 2011
Alternating Current
155301
The equation of $\mathrm{AC}$ voltage is $\mathrm{E}=\mathbf{2 2 0}$ sin $(\omega t+\pi / 6)$ and the $\mathrm{AC}$ current is $\mathrm{I}=10 \sin$ $(\omega t-\pi / 6)$. The average power dissipated is
155302
The natural frequency of the circuit shown in adjoining figure is
1 $\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
2 $\frac{1}{2 \pi \sqrt{2 \mathrm{LC}}}$
3 $\frac{2}{2 \pi \sqrt{\mathrm{LC}}}$
4 zero
Explanation:
A In the given circuit inductor and capacitor are in series- $\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}} \text { and } \mathrm{L}_{\mathrm{S}}=\mathrm{L}+\mathrm{L}$ $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C}}{2} \mathrm{~L}_{\mathrm{S}}=2 \mathrm{~L}$ Therefore natural frequency of the circuit $\mathrm{f}_{0}=\frac{1}{2 \pi \sqrt{\mathrm{L}_{\mathrm{S}} \mathrm{C}_{\mathrm{S}}}}$ $\mathrm{f}_{\mathrm{o}}=\frac{1}{2 \pi \sqrt{2 \mathrm{~L} \times \frac{\mathrm{C}}{2}}}$ $\mathrm{f}_{\mathrm{o}}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
VITEEE-2013
Alternating Current
155307
A current $I=I_{0} \sin \left(\omega t-\frac{\pi}{2}\right)$ flows in an A.C. potential of $E=E_{0} \sin \omega t$ has been applied, then the power consumption $P$ in the circuit will be
D Given that, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin \omega \mathrm{t}$ And $\quad \mathrm{I}=\mathrm{I}_{\mathrm{o}} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)$ Here, voltage and current are in out of phase $\left[\phi=\frac{\pi}{2}\right]$ $\therefore$ Power factor, $\cos \phi=\cos 90^{\circ}=0$ Hence, power consumption become zero.
155298
Assertion: A capacitor blocks direct current in the steady state. Reason: The capacitive reactance of the capacitor is inversely proportional to frequency $f$ of the source of emf.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
A Given that, $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ Frequency of direct current is zero that is i.e., $\mathrm{f}=0$ $\therefore \quad \mathrm{X}_{\mathrm{C}}=\infty$ So, it blocks the DC current, $\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ $\mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$ Hence, option (a) is correct.
AIIMS-2011
Alternating Current
155300
In L-C-R circuit power factor at resonance is
1 less than one
2 greater than one
3 unity
4 Can't predicted
Explanation:
C We know that, $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\cos \phi=\frac{\mathrm{R}}{\mathrm{R}}=1$ Current and voltage across the resistance are all in phase. So, phase angle $\phi=0^{\circ}$ Therefore, $\cos \phi=1$
MHT-CET 2011
Alternating Current
155301
The equation of $\mathrm{AC}$ voltage is $\mathrm{E}=\mathbf{2 2 0}$ sin $(\omega t+\pi / 6)$ and the $\mathrm{AC}$ current is $\mathrm{I}=10 \sin$ $(\omega t-\pi / 6)$. The average power dissipated is
155302
The natural frequency of the circuit shown in adjoining figure is
1 $\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
2 $\frac{1}{2 \pi \sqrt{2 \mathrm{LC}}}$
3 $\frac{2}{2 \pi \sqrt{\mathrm{LC}}}$
4 zero
Explanation:
A In the given circuit inductor and capacitor are in series- $\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}} \text { and } \mathrm{L}_{\mathrm{S}}=\mathrm{L}+\mathrm{L}$ $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C}}{2} \mathrm{~L}_{\mathrm{S}}=2 \mathrm{~L}$ Therefore natural frequency of the circuit $\mathrm{f}_{0}=\frac{1}{2 \pi \sqrt{\mathrm{L}_{\mathrm{S}} \mathrm{C}_{\mathrm{S}}}}$ $\mathrm{f}_{\mathrm{o}}=\frac{1}{2 \pi \sqrt{2 \mathrm{~L} \times \frac{\mathrm{C}}{2}}}$ $\mathrm{f}_{\mathrm{o}}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
VITEEE-2013
Alternating Current
155307
A current $I=I_{0} \sin \left(\omega t-\frac{\pi}{2}\right)$ flows in an A.C. potential of $E=E_{0} \sin \omega t$ has been applied, then the power consumption $P$ in the circuit will be
D Given that, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin \omega \mathrm{t}$ And $\quad \mathrm{I}=\mathrm{I}_{\mathrm{o}} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)$ Here, voltage and current are in out of phase $\left[\phi=\frac{\pi}{2}\right]$ $\therefore$ Power factor, $\cos \phi=\cos 90^{\circ}=0$ Hence, power consumption become zero.
155298
Assertion: A capacitor blocks direct current in the steady state. Reason: The capacitive reactance of the capacitor is inversely proportional to frequency $f$ of the source of emf.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
A Given that, $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ Frequency of direct current is zero that is i.e., $\mathrm{f}=0$ $\therefore \quad \mathrm{X}_{\mathrm{C}}=\infty$ So, it blocks the DC current, $\mathrm{X}_{\mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ $\mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$ Hence, option (a) is correct.
AIIMS-2011
Alternating Current
155300
In L-C-R circuit power factor at resonance is
1 less than one
2 greater than one
3 unity
4 Can't predicted
Explanation:
C We know that, $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ $\cos \phi=\frac{\mathrm{R}}{\mathrm{R}}=1$ Current and voltage across the resistance are all in phase. So, phase angle $\phi=0^{\circ}$ Therefore, $\cos \phi=1$
MHT-CET 2011
Alternating Current
155301
The equation of $\mathrm{AC}$ voltage is $\mathrm{E}=\mathbf{2 2 0}$ sin $(\omega t+\pi / 6)$ and the $\mathrm{AC}$ current is $\mathrm{I}=10 \sin$ $(\omega t-\pi / 6)$. The average power dissipated is
155302
The natural frequency of the circuit shown in adjoining figure is
1 $\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
2 $\frac{1}{2 \pi \sqrt{2 \mathrm{LC}}}$
3 $\frac{2}{2 \pi \sqrt{\mathrm{LC}}}$
4 zero
Explanation:
A In the given circuit inductor and capacitor are in series- $\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}} \text { and } \mathrm{L}_{\mathrm{S}}=\mathrm{L}+\mathrm{L}$ $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C}}{2} \mathrm{~L}_{\mathrm{S}}=2 \mathrm{~L}$ Therefore natural frequency of the circuit $\mathrm{f}_{0}=\frac{1}{2 \pi \sqrt{\mathrm{L}_{\mathrm{S}} \mathrm{C}_{\mathrm{S}}}}$ $\mathrm{f}_{\mathrm{o}}=\frac{1}{2 \pi \sqrt{2 \mathrm{~L} \times \frac{\mathrm{C}}{2}}}$ $\mathrm{f}_{\mathrm{o}}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
VITEEE-2013
Alternating Current
155307
A current $I=I_{0} \sin \left(\omega t-\frac{\pi}{2}\right)$ flows in an A.C. potential of $E=E_{0} \sin \omega t$ has been applied, then the power consumption $P$ in the circuit will be
D Given that, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin \omega \mathrm{t}$ And $\quad \mathrm{I}=\mathrm{I}_{\mathrm{o}} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)$ Here, voltage and current are in out of phase $\left[\phi=\frac{\pi}{2}\right]$ $\therefore$ Power factor, $\cos \phi=\cos 90^{\circ}=0$ Hence, power consumption become zero.
