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Alternating Current

155096 In a series $L C R$ circuit, the inductance $L$ is 10 $m H$ , capacitance $C$ is $1 μ F$ and resistance $R$ is $100 Ω$ . The frequency at which resonance occurs is :-

1

15.9 kHz

2

1.59 rad / s

3

1.59 kHz

4

15.9 rad / s

Explanation:

C Given,
Inductance $(L) = 10 \times 10^{- 3} H$
Capacitance $(C) = 1 \times 10^{- 6} F$
Resistance $(R) = 100 Ω$
At resonance $X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
$f = \frac{1}{2 π \sqrt{LC}} = \frac{1}{2 π \sqrt{10 \times 10^{- 3} \times 10^{- 6}}}$
$= 1.59 KHz$

NEET (UG)-07.05.2023

Alternating Current

155098 As per the given graph choose the correct representation for curve $A$ and curve $B$ . \{Where $X_{C} =$ reactance of pure capacitive circuit connected with A.C. source
$X_{L} =$ reactance of pure inductive circuit connected with A.C. source
$R =$ impedance of pure resistive circuit connected with A.C. source
$Z =$ Impedance of the LCR series circuit $}$

1

A = X_{C}, B = R

2

A = X_{L}, B = Z

3

A = X_{C}, B = X_{L}

4

A = X_{L}, B = R

Explanation:

C For curve A,
$X_{C} = \frac{1}{ω C} = \frac{1}{(2 π f) C}$
$∴ X_{C} \propto \frac{1}{f}$

For curve $B$ ,
$X_{L} = ω L = (2 π f) L$
$∴ X_{L} \propto f$

JEE Main-11.04.2023

Alternating Current

155099 The current in resistance $R$ at resonance is

1 zero

2 minimum but finite

3 maximum but finite

4 infinite

Explanation:

C The current in resistance $R$ at resonance is maximum but finite because $X_{L} = X_{C}$ and $I_{max} = \frac{E}{R}$ Where, $E =$ Applied voltage

$V = V_{0} \sin ω t$

AIIMS-2012

Alternating Current

155100 In the given circuit, rms value of current $(I_{rms})$ through the resistor $R$ is

1

\frac{1}{2} A

2

20 A

3

2 A

4

2 \sqrt{2 A}

Explanation:

C From the above circuit we have
$R = 100 Ω$
$X_{L} = 200 Ω$
$X_{C} = 100 Ω$
$V_{rms} = 200 \sqrt{2} V$
Since we have,
$I_{rms} = \frac{V_{rms}}{Z}$
Where, $Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$= \sqrt{100^{2} + (200 - 100)^{2}} = 100 \sqrt{2}$
$∴ I_{rms} = \frac{200 \sqrt{2}}{100 \sqrt{2}}$
$I_{rms} = 2 A$

JEE Main-30.01.2023

Alternating Current

155101 In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $x$ times its initial resonant frequency $ω_{0}$ . The value of $x$ is.

1

1 / 4

2 4

3 16

4

1 / 16

Explanation:

A Since, we know that -
$ω_{0} = \frac{1}{\sqrt{LC}}$
If the values of inductance becomes $2 L$ and capacitance becomes $8 C$ then -
$ω^{'} = \frac{1}{\sqrt{2 L \times 8 C}} = \frac{1}{\sqrt{16 LC}} = \frac{1}{4 \sqrt{LC}}$
$ω^{'} = \frac{1}{4} \times \frac{1}{\sqrt{LC}}$
$ω^{'} = \frac{1}{4} \times ω_{0}, \frac{ω^{'}}{ω_{0}} = \frac{1}{4}$

JEE Main-25.01.2023

Alternating Current

155096 In a series $L C R$ circuit, the inductance $L$ is 10 $m H$ , capacitance $C$ is $1 μ F$ and resistance $R$ is $100 Ω$ . The frequency at which resonance occurs is :-

1

15.9 kHz

2

1.59 rad / s

3

1.59 kHz

4

15.9 rad / s

Explanation:

C Given,
Inductance $(L) = 10 \times 10^{- 3} H$
Capacitance $(C) = 1 \times 10^{- 6} F$
Resistance $(R) = 100 Ω$
At resonance $X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
$f = \frac{1}{2 π \sqrt{LC}} = \frac{1}{2 π \sqrt{10 \times 10^{- 3} \times 10^{- 6}}}$
$= 1.59 KHz$

NEET (UG)-07.05.2023

Alternating Current

155098 As per the given graph choose the correct representation for curve $A$ and curve $B$ . \{Where $X_{C} =$ reactance of pure capacitive circuit connected with A.C. source
$X_{L} =$ reactance of pure inductive circuit connected with A.C. source
$R =$ impedance of pure resistive circuit connected with A.C. source
$Z =$ Impedance of the LCR series circuit $}$

1

A = X_{C}, B = R

2

A = X_{L}, B = Z

3

A = X_{C}, B = X_{L}

4

A = X_{L}, B = R

Explanation:

C For curve A,
$X_{C} = \frac{1}{ω C} = \frac{1}{(2 π f) C}$
$∴ X_{C} \propto \frac{1}{f}$

For curve $B$ ,
$X_{L} = ω L = (2 π f) L$
$∴ X_{L} \propto f$

JEE Main-11.04.2023

Alternating Current

155099 The current in resistance $R$ at resonance is

1 zero

2 minimum but finite

3 maximum but finite

4 infinite

Explanation:

C The current in resistance $R$ at resonance is maximum but finite because $X_{L} = X_{C}$ and $I_{max} = \frac{E}{R}$ Where, $E =$ Applied voltage

$V = V_{0} \sin ω t$

AIIMS-2012

Alternating Current

155100 In the given circuit, rms value of current $(I_{rms})$ through the resistor $R$ is

1

\frac{1}{2} A

2

20 A

3

2 A

4

2 \sqrt{2 A}

Explanation:

C From the above circuit we have
$R = 100 Ω$
$X_{L} = 200 Ω$
$X_{C} = 100 Ω$
$V_{rms} = 200 \sqrt{2} V$
Since we have,
$I_{rms} = \frac{V_{rms}}{Z}$
Where, $Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$= \sqrt{100^{2} + (200 - 100)^{2}} = 100 \sqrt{2}$
$∴ I_{rms} = \frac{200 \sqrt{2}}{100 \sqrt{2}}$
$I_{rms} = 2 A$

JEE Main-30.01.2023

Alternating Current

155101 In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $x$ times its initial resonant frequency $ω_{0}$ . The value of $x$ is.

1

1 / 4

2 4

3 16

4

1 / 16

Explanation:

A Since, we know that -
$ω_{0} = \frac{1}{\sqrt{LC}}$
If the values of inductance becomes $2 L$ and capacitance becomes $8 C$ then -
$ω^{'} = \frac{1}{\sqrt{2 L \times 8 C}} = \frac{1}{\sqrt{16 LC}} = \frac{1}{4 \sqrt{LC}}$
$ω^{'} = \frac{1}{4} \times \frac{1}{\sqrt{LC}}$
$ω^{'} = \frac{1}{4} \times ω_{0}, \frac{ω^{'}}{ω_{0}} = \frac{1}{4}$

JEE Main-25.01.2023

Alternating Current

155096 In a series $L C R$ circuit, the inductance $L$ is 10 $m H$ , capacitance $C$ is $1 μ F$ and resistance $R$ is $100 Ω$ . The frequency at which resonance occurs is :-

1

15.9 kHz

2

1.59 rad / s

3

1.59 kHz

4

15.9 rad / s

Explanation:

C Given,
Inductance $(L) = 10 \times 10^{- 3} H$
Capacitance $(C) = 1 \times 10^{- 6} F$
Resistance $(R) = 100 Ω$
At resonance $X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
$f = \frac{1}{2 π \sqrt{LC}} = \frac{1}{2 π \sqrt{10 \times 10^{- 3} \times 10^{- 6}}}$
$= 1.59 KHz$

NEET (UG)-07.05.2023

Alternating Current

155098 As per the given graph choose the correct representation for curve $A$ and curve $B$ . \{Where $X_{C} =$ reactance of pure capacitive circuit connected with A.C. source
$X_{L} =$ reactance of pure inductive circuit connected with A.C. source
$R =$ impedance of pure resistive circuit connected with A.C. source
$Z =$ Impedance of the LCR series circuit $}$

1

A = X_{C}, B = R

2

A = X_{L}, B = Z

3

A = X_{C}, B = X_{L}

4

A = X_{L}, B = R

Explanation:

C For curve A,
$X_{C} = \frac{1}{ω C} = \frac{1}{(2 π f) C}$
$∴ X_{C} \propto \frac{1}{f}$

For curve $B$ ,
$X_{L} = ω L = (2 π f) L$
$∴ X_{L} \propto f$

JEE Main-11.04.2023

Alternating Current

155099 The current in resistance $R$ at resonance is

1 zero

2 minimum but finite

3 maximum but finite

4 infinite

Explanation:

C The current in resistance $R$ at resonance is maximum but finite because $X_{L} = X_{C}$ and $I_{max} = \frac{E}{R}$ Where, $E =$ Applied voltage

$V = V_{0} \sin ω t$

AIIMS-2012

Alternating Current

155100 In the given circuit, rms value of current $(I_{rms})$ through the resistor $R$ is

1

\frac{1}{2} A

2

20 A

3

2 A

4

2 \sqrt{2 A}

Explanation:

C From the above circuit we have
$R = 100 Ω$
$X_{L} = 200 Ω$
$X_{C} = 100 Ω$
$V_{rms} = 200 \sqrt{2} V$
Since we have,
$I_{rms} = \frac{V_{rms}}{Z}$
Where, $Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$= \sqrt{100^{2} + (200 - 100)^{2}} = 100 \sqrt{2}$
$∴ I_{rms} = \frac{200 \sqrt{2}}{100 \sqrt{2}}$
$I_{rms} = 2 A$

JEE Main-30.01.2023

Alternating Current

155101 In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $x$ times its initial resonant frequency $ω_{0}$ . The value of $x$ is.

1

1 / 4

2 4

3 16

4

1 / 16

Explanation:

A Since, we know that -
$ω_{0} = \frac{1}{\sqrt{LC}}$
If the values of inductance becomes $2 L$ and capacitance becomes $8 C$ then -
$ω^{'} = \frac{1}{\sqrt{2 L \times 8 C}} = \frac{1}{\sqrt{16 LC}} = \frac{1}{4 \sqrt{LC}}$
$ω^{'} = \frac{1}{4} \times \frac{1}{\sqrt{LC}}$
$ω^{'} = \frac{1}{4} \times ω_{0}, \frac{ω^{'}}{ω_{0}} = \frac{1}{4}$

JEE Main-25.01.2023

Alternating Current

155096 In a series $L C R$ circuit, the inductance $L$ is 10 $m H$ , capacitance $C$ is $1 μ F$ and resistance $R$ is $100 Ω$ . The frequency at which resonance occurs is :-

1

15.9 kHz

2

1.59 rad / s

3

1.59 kHz

4

15.9 rad / s

Explanation:

C Given,
Inductance $(L) = 10 \times 10^{- 3} H$
Capacitance $(C) = 1 \times 10^{- 6} F$
Resistance $(R) = 100 Ω$
At resonance $X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
$f = \frac{1}{2 π \sqrt{LC}} = \frac{1}{2 π \sqrt{10 \times 10^{- 3} \times 10^{- 6}}}$
$= 1.59 KHz$

NEET (UG)-07.05.2023

Alternating Current

155098 As per the given graph choose the correct representation for curve $A$ and curve $B$ . \{Where $X_{C} =$ reactance of pure capacitive circuit connected with A.C. source
$X_{L} =$ reactance of pure inductive circuit connected with A.C. source
$R =$ impedance of pure resistive circuit connected with A.C. source
$Z =$ Impedance of the LCR series circuit $}$

1

A = X_{C}, B = R

2

A = X_{L}, B = Z

3

A = X_{C}, B = X_{L}

4

A = X_{L}, B = R

Explanation:

C For curve A,
$X_{C} = \frac{1}{ω C} = \frac{1}{(2 π f) C}$
$∴ X_{C} \propto \frac{1}{f}$

For curve $B$ ,
$X_{L} = ω L = (2 π f) L$
$∴ X_{L} \propto f$

JEE Main-11.04.2023

Alternating Current

155099 The current in resistance $R$ at resonance is

1 zero

2 minimum but finite

3 maximum but finite

4 infinite

Explanation:

C The current in resistance $R$ at resonance is maximum but finite because $X_{L} = X_{C}$ and $I_{max} = \frac{E}{R}$ Where, $E =$ Applied voltage

$V = V_{0} \sin ω t$

AIIMS-2012

Alternating Current

155100 In the given circuit, rms value of current $(I_{rms})$ through the resistor $R$ is

1

\frac{1}{2} A

2

20 A

3

2 A

4

2 \sqrt{2 A}

Explanation:

C From the above circuit we have
$R = 100 Ω$
$X_{L} = 200 Ω$
$X_{C} = 100 Ω$
$V_{rms} = 200 \sqrt{2} V$
Since we have,
$I_{rms} = \frac{V_{rms}}{Z}$
Where, $Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$= \sqrt{100^{2} + (200 - 100)^{2}} = 100 \sqrt{2}$
$∴ I_{rms} = \frac{200 \sqrt{2}}{100 \sqrt{2}}$
$I_{rms} = 2 A$

JEE Main-30.01.2023

Alternating Current

155101 In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $x$ times its initial resonant frequency $ω_{0}$ . The value of $x$ is.

1

1 / 4

2 4

3 16

4

1 / 16

Explanation:

A Since, we know that -
$ω_{0} = \frac{1}{\sqrt{LC}}$
If the values of inductance becomes $2 L$ and capacitance becomes $8 C$ then -
$ω^{'} = \frac{1}{\sqrt{2 L \times 8 C}} = \frac{1}{\sqrt{16 LC}} = \frac{1}{4 \sqrt{LC}}$
$ω^{'} = \frac{1}{4} \times \frac{1}{\sqrt{LC}}$
$ω^{'} = \frac{1}{4} \times ω_{0}, \frac{ω^{'}}{ω_{0}} = \frac{1}{4}$

JEE Main-25.01.2023

Alternating Current

155096 In a series $L C R$ circuit, the inductance $L$ is 10 $m H$ , capacitance $C$ is $1 μ F$ and resistance $R$ is $100 Ω$ . The frequency at which resonance occurs is :-

1

15.9 kHz

2

1.59 rad / s

3

1.59 kHz

4

15.9 rad / s

Explanation:

C Given,
Inductance $(L) = 10 \times 10^{- 3} H$
Capacitance $(C) = 1 \times 10^{- 6} F$
Resistance $(R) = 100 Ω$
At resonance $X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
$f = \frac{1}{2 π \sqrt{LC}} = \frac{1}{2 π \sqrt{10 \times 10^{- 3} \times 10^{- 6}}}$
$= 1.59 KHz$

NEET (UG)-07.05.2023

Alternating Current

155098 As per the given graph choose the correct representation for curve $A$ and curve $B$ . \{Where $X_{C} =$ reactance of pure capacitive circuit connected with A.C. source
$X_{L} =$ reactance of pure inductive circuit connected with A.C. source
$R =$ impedance of pure resistive circuit connected with A.C. source
$Z =$ Impedance of the LCR series circuit $}$

1

A = X_{C}, B = R

2

A = X_{L}, B = Z

3

A = X_{C}, B = X_{L}

4

A = X_{L}, B = R

Explanation:

C For curve A,
$X_{C} = \frac{1}{ω C} = \frac{1}{(2 π f) C}$
$∴ X_{C} \propto \frac{1}{f}$

For curve $B$ ,
$X_{L} = ω L = (2 π f) L$
$∴ X_{L} \propto f$

JEE Main-11.04.2023

Alternating Current

155099 The current in resistance $R$ at resonance is

1 zero

2 minimum but finite

3 maximum but finite

4 infinite

Explanation:

C The current in resistance $R$ at resonance is maximum but finite because $X_{L} = X_{C}$ and $I_{max} = \frac{E}{R}$ Where, $E =$ Applied voltage

$V = V_{0} \sin ω t$

AIIMS-2012

Alternating Current

155100 In the given circuit, rms value of current $(I_{rms})$ through the resistor $R$ is

1

\frac{1}{2} A

2

20 A

3

2 A

4

2 \sqrt{2 A}

Explanation:

C From the above circuit we have
$R = 100 Ω$
$X_{L} = 200 Ω$
$X_{C} = 100 Ω$
$V_{rms} = 200 \sqrt{2} V$
Since we have,
$I_{rms} = \frac{V_{rms}}{Z}$
Where, $Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$= \sqrt{100^{2} + (200 - 100)^{2}} = 100 \sqrt{2}$
$∴ I_{rms} = \frac{200 \sqrt{2}}{100 \sqrt{2}}$
$I_{rms} = 2 A$

JEE Main-30.01.2023

Alternating Current

155101 In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $x$ times its initial resonant frequency $ω_{0}$ . The value of $x$ is.

1

1 / 4

2 4

3 16

4

1 / 16

Explanation:

A Since, we know that -
$ω_{0} = \frac{1}{\sqrt{LC}}$
If the values of inductance becomes $2 L$ and capacitance becomes $8 C$ then -
$ω^{'} = \frac{1}{\sqrt{2 L \times 8 C}} = \frac{1}{\sqrt{16 LC}} = \frac{1}{4 \sqrt{LC}}$
$ω^{'} = \frac{1}{4} \times \frac{1}{\sqrt{LC}}$
$ω^{'} = \frac{1}{4} \times ω_{0}, \frac{ω^{'}}{ω_{0}} = \frac{1}{4}$

JEE Main-25.01.2023