NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155252
The power loss in an A.C circuit will be minimum, when
1 resistance is high, inductance is high
2 resistance is high, inductance is low
3 resistance is low, inductance is high
4 none of these
Explanation:
C We know that, Power loss in AC circuit - $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}$ will be minimum when $\cos \phi$ will be minimum The power loss in an A.C circuit will be minimum, Then, resistance is low, inductance is high.
J and K CET- 2004
Alternating Current
155254
A series LCR circuit is tuned to resonance. The impedance of the circuit at resonance is
D We know that, Impedance $(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $Z=\sqrt{R^{2}+\left(X_{L}-X_{L}\right)^{2}} Z=\sqrt{R^{2}}$ $\mathrm{Z}=\mathrm{R}$
J and K CET- 2000
Alternating Current
155256
What is the resonance frequency of a driven LCR oscillator?
1 $\frac{1}{\mathrm{LC}}$
2 $\frac{1}{2 \pi \mathrm{LC}}$
3 $(\mathrm{LC})^{-1 / 2}$
4 $(2 \pi \mathrm{LC})^{-1 / 2}$
Explanation:
C The resonance frequency of a driven LCR oscillator, $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\omega=(\mathrm{LC})^{-1 / 2}$
J and K CET-2012
Alternating Current
155276
For a series LCR circuit, the power loss at resonance is
C We know at resonance, the impedance of the circuit is equal to the resistance - $\text { So, } \mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}} \left(\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\right)$ $\mathrm{Z}=\mathrm{R}$ $\mathrm{P}_{\text {loss }}=\mathrm{VI} \cos \phi=\mathrm{VI} {[\cos \phi=1]}$ $\mathrm{P}_{\text {loss }}=\mathrm{I}^{2} \mathrm{R} {[\mathrm{V}=\mathrm{IR}]}$
155252
The power loss in an A.C circuit will be minimum, when
1 resistance is high, inductance is high
2 resistance is high, inductance is low
3 resistance is low, inductance is high
4 none of these
Explanation:
C We know that, Power loss in AC circuit - $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}$ will be minimum when $\cos \phi$ will be minimum The power loss in an A.C circuit will be minimum, Then, resistance is low, inductance is high.
J and K CET- 2004
Alternating Current
155254
A series LCR circuit is tuned to resonance. The impedance of the circuit at resonance is
D We know that, Impedance $(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $Z=\sqrt{R^{2}+\left(X_{L}-X_{L}\right)^{2}} Z=\sqrt{R^{2}}$ $\mathrm{Z}=\mathrm{R}$
J and K CET- 2000
Alternating Current
155256
What is the resonance frequency of a driven LCR oscillator?
1 $\frac{1}{\mathrm{LC}}$
2 $\frac{1}{2 \pi \mathrm{LC}}$
3 $(\mathrm{LC})^{-1 / 2}$
4 $(2 \pi \mathrm{LC})^{-1 / 2}$
Explanation:
C The resonance frequency of a driven LCR oscillator, $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\omega=(\mathrm{LC})^{-1 / 2}$
J and K CET-2012
Alternating Current
155276
For a series LCR circuit, the power loss at resonance is
C We know at resonance, the impedance of the circuit is equal to the resistance - $\text { So, } \mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}} \left(\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\right)$ $\mathrm{Z}=\mathrm{R}$ $\mathrm{P}_{\text {loss }}=\mathrm{VI} \cos \phi=\mathrm{VI} {[\cos \phi=1]}$ $\mathrm{P}_{\text {loss }}=\mathrm{I}^{2} \mathrm{R} {[\mathrm{V}=\mathrm{IR}]}$
155252
The power loss in an A.C circuit will be minimum, when
1 resistance is high, inductance is high
2 resistance is high, inductance is low
3 resistance is low, inductance is high
4 none of these
Explanation:
C We know that, Power loss in AC circuit - $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}$ will be minimum when $\cos \phi$ will be minimum The power loss in an A.C circuit will be minimum, Then, resistance is low, inductance is high.
J and K CET- 2004
Alternating Current
155254
A series LCR circuit is tuned to resonance. The impedance of the circuit at resonance is
D We know that, Impedance $(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $Z=\sqrt{R^{2}+\left(X_{L}-X_{L}\right)^{2}} Z=\sqrt{R^{2}}$ $\mathrm{Z}=\mathrm{R}$
J and K CET- 2000
Alternating Current
155256
What is the resonance frequency of a driven LCR oscillator?
1 $\frac{1}{\mathrm{LC}}$
2 $\frac{1}{2 \pi \mathrm{LC}}$
3 $(\mathrm{LC})^{-1 / 2}$
4 $(2 \pi \mathrm{LC})^{-1 / 2}$
Explanation:
C The resonance frequency of a driven LCR oscillator, $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\omega=(\mathrm{LC})^{-1 / 2}$
J and K CET-2012
Alternating Current
155276
For a series LCR circuit, the power loss at resonance is
C We know at resonance, the impedance of the circuit is equal to the resistance - $\text { So, } \mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}} \left(\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\right)$ $\mathrm{Z}=\mathrm{R}$ $\mathrm{P}_{\text {loss }}=\mathrm{VI} \cos \phi=\mathrm{VI} {[\cos \phi=1]}$ $\mathrm{P}_{\text {loss }}=\mathrm{I}^{2} \mathrm{R} {[\mathrm{V}=\mathrm{IR}]}$
155252
The power loss in an A.C circuit will be minimum, when
1 resistance is high, inductance is high
2 resistance is high, inductance is low
3 resistance is low, inductance is high
4 none of these
Explanation:
C We know that, Power loss in AC circuit - $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}$ will be minimum when $\cos \phi$ will be minimum The power loss in an A.C circuit will be minimum, Then, resistance is low, inductance is high.
J and K CET- 2004
Alternating Current
155254
A series LCR circuit is tuned to resonance. The impedance of the circuit at resonance is
D We know that, Impedance $(Z)=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $Z=\sqrt{R^{2}+\left(X_{L}-X_{L}\right)^{2}} Z=\sqrt{R^{2}}$ $\mathrm{Z}=\mathrm{R}$
J and K CET- 2000
Alternating Current
155256
What is the resonance frequency of a driven LCR oscillator?
1 $\frac{1}{\mathrm{LC}}$
2 $\frac{1}{2 \pi \mathrm{LC}}$
3 $(\mathrm{LC})^{-1 / 2}$
4 $(2 \pi \mathrm{LC})^{-1 / 2}$
Explanation:
C The resonance frequency of a driven LCR oscillator, $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $\omega=(\mathrm{LC})^{-1 / 2}$
J and K CET-2012
Alternating Current
155276
For a series LCR circuit, the power loss at resonance is
C We know at resonance, the impedance of the circuit is equal to the resistance - $\text { So, } \mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}} \left(\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\right)$ $\mathrm{Z}=\mathrm{R}$ $\mathrm{P}_{\text {loss }}=\mathrm{VI} \cos \phi=\mathrm{VI} {[\cos \phi=1]}$ $\mathrm{P}_{\text {loss }}=\mathrm{I}^{2} \mathrm{R} {[\mathrm{V}=\mathrm{IR}]}$
