155134
In series $L C R$ circuit ' $R$ ' represents an electric bulb. If the frequency of ac supply is doubled, the value of ' $L$ ' and ' $C$ ' should be
1 both doubled simultaneously.
2 four times the original value.
3 eight times the original value.
4 both halved simultaneously.
Explanation:
D According to question, In series LCR circuit for the same current value the total impedance, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ When frequency is double $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(2 \pi \mathrm{fL}-\frac{1}{2 \pi \mathrm{fC}}\right)^{2}}$ $\mathrm{Z}$ must be same when frequency become double. Then $\mathrm{L}$ and $\mathrm{C}$ both halved simultaneously.
MHT-CET 2020
Alternating Current
155135
For a series LCR circuit at resonance, the statement which is not true is
1 Wattless current is zero
2 Power factor is zero
3 Peak energy stored by a capacitor $=$ peak energy stored by an inductor
4 Average power= apparent power
Explanation:
B At resonance condition, $\mathrm{Z}=\mathrm{R}$ Now, Power factor, $\cos \phi=\frac{Z}{R}$ $\cos \phi=\frac{Z}{Z}$ $\cos \phi=1$ Therefore, power factor equals to 1 , not to zero.
COMEDK 2020
Alternating Current
155136
A sinusoidal voltage having maximum value of $283 \mathrm{~V} \&$ frequency of $50 \mathrm{~Hz}$ is applied to LCR series connection where $R=3 \Omega, L=25.48 \mathrm{mH}$ $\& \mathrm{C}=796 \mu \mathrm{F}$. Then impedance is at resonance condition.
155137
In a series $\mathrm{L}-\mathrm{C}-\mathrm{R}$ circuit, the inductive reactance is twice the resistance and the capacitive reactance is $1 / 3^{\text {rd }}$ of the inductive reactance. The power factor of the circuit is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155134
In series $L C R$ circuit ' $R$ ' represents an electric bulb. If the frequency of ac supply is doubled, the value of ' $L$ ' and ' $C$ ' should be
1 both doubled simultaneously.
2 four times the original value.
3 eight times the original value.
4 both halved simultaneously.
Explanation:
D According to question, In series LCR circuit for the same current value the total impedance, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ When frequency is double $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(2 \pi \mathrm{fL}-\frac{1}{2 \pi \mathrm{fC}}\right)^{2}}$ $\mathrm{Z}$ must be same when frequency become double. Then $\mathrm{L}$ and $\mathrm{C}$ both halved simultaneously.
MHT-CET 2020
Alternating Current
155135
For a series LCR circuit at resonance, the statement which is not true is
1 Wattless current is zero
2 Power factor is zero
3 Peak energy stored by a capacitor $=$ peak energy stored by an inductor
4 Average power= apparent power
Explanation:
B At resonance condition, $\mathrm{Z}=\mathrm{R}$ Now, Power factor, $\cos \phi=\frac{Z}{R}$ $\cos \phi=\frac{Z}{Z}$ $\cos \phi=1$ Therefore, power factor equals to 1 , not to zero.
COMEDK 2020
Alternating Current
155136
A sinusoidal voltage having maximum value of $283 \mathrm{~V} \&$ frequency of $50 \mathrm{~Hz}$ is applied to LCR series connection where $R=3 \Omega, L=25.48 \mathrm{mH}$ $\& \mathrm{C}=796 \mu \mathrm{F}$. Then impedance is at resonance condition.
155137
In a series $\mathrm{L}-\mathrm{C}-\mathrm{R}$ circuit, the inductive reactance is twice the resistance and the capacitive reactance is $1 / 3^{\text {rd }}$ of the inductive reactance. The power factor of the circuit is
155134
In series $L C R$ circuit ' $R$ ' represents an electric bulb. If the frequency of ac supply is doubled, the value of ' $L$ ' and ' $C$ ' should be
1 both doubled simultaneously.
2 four times the original value.
3 eight times the original value.
4 both halved simultaneously.
Explanation:
D According to question, In series LCR circuit for the same current value the total impedance, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ When frequency is double $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(2 \pi \mathrm{fL}-\frac{1}{2 \pi \mathrm{fC}}\right)^{2}}$ $\mathrm{Z}$ must be same when frequency become double. Then $\mathrm{L}$ and $\mathrm{C}$ both halved simultaneously.
MHT-CET 2020
Alternating Current
155135
For a series LCR circuit at resonance, the statement which is not true is
1 Wattless current is zero
2 Power factor is zero
3 Peak energy stored by a capacitor $=$ peak energy stored by an inductor
4 Average power= apparent power
Explanation:
B At resonance condition, $\mathrm{Z}=\mathrm{R}$ Now, Power factor, $\cos \phi=\frac{Z}{R}$ $\cos \phi=\frac{Z}{Z}$ $\cos \phi=1$ Therefore, power factor equals to 1 , not to zero.
COMEDK 2020
Alternating Current
155136
A sinusoidal voltage having maximum value of $283 \mathrm{~V} \&$ frequency of $50 \mathrm{~Hz}$ is applied to LCR series connection where $R=3 \Omega, L=25.48 \mathrm{mH}$ $\& \mathrm{C}=796 \mu \mathrm{F}$. Then impedance is at resonance condition.
155137
In a series $\mathrm{L}-\mathrm{C}-\mathrm{R}$ circuit, the inductive reactance is twice the resistance and the capacitive reactance is $1 / 3^{\text {rd }}$ of the inductive reactance. The power factor of the circuit is
155134
In series $L C R$ circuit ' $R$ ' represents an electric bulb. If the frequency of ac supply is doubled, the value of ' $L$ ' and ' $C$ ' should be
1 both doubled simultaneously.
2 four times the original value.
3 eight times the original value.
4 both halved simultaneously.
Explanation:
D According to question, In series LCR circuit for the same current value the total impedance, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ When frequency is double $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(2 \pi \mathrm{fL}-\frac{1}{2 \pi \mathrm{fC}}\right)^{2}}$ $\mathrm{Z}$ must be same when frequency become double. Then $\mathrm{L}$ and $\mathrm{C}$ both halved simultaneously.
MHT-CET 2020
Alternating Current
155135
For a series LCR circuit at resonance, the statement which is not true is
1 Wattless current is zero
2 Power factor is zero
3 Peak energy stored by a capacitor $=$ peak energy stored by an inductor
4 Average power= apparent power
Explanation:
B At resonance condition, $\mathrm{Z}=\mathrm{R}$ Now, Power factor, $\cos \phi=\frac{Z}{R}$ $\cos \phi=\frac{Z}{Z}$ $\cos \phi=1$ Therefore, power factor equals to 1 , not to zero.
COMEDK 2020
Alternating Current
155136
A sinusoidal voltage having maximum value of $283 \mathrm{~V} \&$ frequency of $50 \mathrm{~Hz}$ is applied to LCR series connection where $R=3 \Omega, L=25.48 \mathrm{mH}$ $\& \mathrm{C}=796 \mu \mathrm{F}$. Then impedance is at resonance condition.
155137
In a series $\mathrm{L}-\mathrm{C}-\mathrm{R}$ circuit, the inductive reactance is twice the resistance and the capacitive reactance is $1 / 3^{\text {rd }}$ of the inductive reactance. The power factor of the circuit is
