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PHXII07:ALTERNATING CURRENT

356248 In an $L C R$ circuit, at resonance

1 The impedance is maximum

2 The current leads the voltage by

π / 2

3 The current and voltage are in phase

4 The current is minimum

Explanation:

In an $L C R$ circuit at resonance, the phase angle $(ϕ)$ between current and voltage is zero as the total reactance is zero.
Thus the current and voltage are in phase. The current is maximum and the impedance is minimum at resonance in $L C R$ circuit.

PHXII07:ALTERNATING CURRENT

356249 In the joining ac circuit the voltmeter whose reading will be zero at resonance is
supporting img

1

V_{4}

2

V_{3}

3

V_{2}

4

V_{1}

Explanation:

At resonance net voltage across $L$ and $C$ is zero

PHXII07:ALTERNATING CURRENT

356250 In the figure shown, three $A C$ voltmeters are connected. At resonance,
supporting img

1

V_{3} = 0

2

V_{1} = V_{2} \neq 0

3

V_{2} = 0

4

V_{1} = 0

Explanation:

At resonance, $V_{L}$ and $V_{C}$ are equal in magnitude but have phase difference of relative to each other.
$∴ V_{L C} = V_{L} - V_{C} = 0$
Hence, voltmeter $V_{2}$ read 0 volt.
So correct option is (4).

PHXII07:ALTERNATING CURRENT

356251 In a series $L C R$ circuit, the inductance $L$ is 10 $m H$ , capacitance $C$ is $1 μ F$ and resistance $R i s 100 Ω$ . The frequency at which resonance occurs is

1

15.9 k H z

2

1.59 r a d / s

3

1.59 k H z

4

15.9 r a d / s

Explanation:

At resonance the two reactances are equal
$X_{L} = X_{C}$ .
$\Rightarrow ω L = \frac{1}{ω C}$
$\Rightarrow ω = \frac{1}{\sqrt{L C}}$
$\Rightarrow f = \frac{1}{2 π \sqrt{L C}}$
$\Rightarrow f = \frac{1}{2 π \sqrt{10 \times 10^{- 3} \times 10^{- 6}}} = 1.59 k H z$
Correct option is (3).

NEET - 2023

PHXII07:ALTERNATING CURRENT

356252 Read the Statement A and Statement B carefully to mark the correct options given below.
Statement A :
In an $L C R$ series circuit, current is maximum at resonance .
Statement B :
Current in a purely resistive circuit can never be less than that in a series $L C R$ circuit when connected to same voltage source.

1 Statement A is correct but Statement B is incorrect

2 Statement A is incorrect but Statement B is correct

3 Both Statements are true

4 Both Statements are incorrect

Explanation:

In a series $L C R$ circuit, current is maximum at resonance.
At resonance, $X_{L} = X_{C}$ ,
$∴$ Impedance,
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{R^{2}}$
or $Z = R$
$\Rightarrow Z$ is minimum at resonance
$∴ I_{m} = \frac{V_{m}}{Z} = \frac{V_{m}}{R}$
In a purely resistive circuit, $I = \frac{V}{R}$ and in a series LCR circuit, $I^{'} = \frac{V}{Z}$
$\Rightarrow I > I^{'}$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356248 In an $L C R$ circuit, at resonance

1 The impedance is maximum

2 The current leads the voltage by

π / 2

3 The current and voltage are in phase

4 The current is minimum

Explanation:

In an $L C R$ circuit at resonance, the phase angle $(ϕ)$ between current and voltage is zero as the total reactance is zero.
Thus the current and voltage are in phase. The current is maximum and the impedance is minimum at resonance in $L C R$ circuit.

PHXII07:ALTERNATING CURRENT

356249 In the joining ac circuit the voltmeter whose reading will be zero at resonance is
supporting img

1

V_{4}

2

V_{3}

3

V_{2}

4

V_{1}

Explanation:

At resonance net voltage across $L$ and $C$ is zero

PHXII07:ALTERNATING CURRENT

356250 In the figure shown, three $A C$ voltmeters are connected. At resonance,
supporting img

1

V_{3} = 0

2

V_{1} = V_{2} \neq 0

3

V_{2} = 0

4

V_{1} = 0

Explanation:

At resonance, $V_{L}$ and $V_{C}$ are equal in magnitude but have phase difference of relative to each other.
$∴ V_{L C} = V_{L} - V_{C} = 0$
Hence, voltmeter $V_{2}$ read 0 volt.
So correct option is (4).

PHXII07:ALTERNATING CURRENT

356251 In a series $L C R$ circuit, the inductance $L$ is 10 $m H$ , capacitance $C$ is $1 μ F$ and resistance $R i s 100 Ω$ . The frequency at which resonance occurs is

1

15.9 k H z

2

1.59 r a d / s

3

1.59 k H z

4

15.9 r a d / s

Explanation:

At resonance the two reactances are equal
$X_{L} = X_{C}$ .
$\Rightarrow ω L = \frac{1}{ω C}$
$\Rightarrow ω = \frac{1}{\sqrt{L C}}$
$\Rightarrow f = \frac{1}{2 π \sqrt{L C}}$
$\Rightarrow f = \frac{1}{2 π \sqrt{10 \times 10^{- 3} \times 10^{- 6}}} = 1.59 k H z$
Correct option is (3).

NEET - 2023

PHXII07:ALTERNATING CURRENT

356252 Read the Statement A and Statement B carefully to mark the correct options given below.
Statement A :
In an $L C R$ series circuit, current is maximum at resonance .
Statement B :
Current in a purely resistive circuit can never be less than that in a series $L C R$ circuit when connected to same voltage source.

1 Statement A is correct but Statement B is incorrect

2 Statement A is incorrect but Statement B is correct

3 Both Statements are true

4 Both Statements are incorrect

Explanation:

In a series $L C R$ circuit, current is maximum at resonance.
At resonance, $X_{L} = X_{C}$ ,
$∴$ Impedance,
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{R^{2}}$
or $Z = R$
$\Rightarrow Z$ is minimum at resonance
$∴ I_{m} = \frac{V_{m}}{Z} = \frac{V_{m}}{R}$
In a purely resistive circuit, $I = \frac{V}{R}$ and in a series LCR circuit, $I^{'} = \frac{V}{Z}$
$\Rightarrow I > I^{'}$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356248 In an $L C R$ circuit, at resonance

1 The impedance is maximum

2 The current leads the voltage by

π / 2

3 The current and voltage are in phase

4 The current is minimum

Explanation:

In an $L C R$ circuit at resonance, the phase angle $(ϕ)$ between current and voltage is zero as the total reactance is zero.
Thus the current and voltage are in phase. The current is maximum and the impedance is minimum at resonance in $L C R$ circuit.

PHXII07:ALTERNATING CURRENT

356249 In the joining ac circuit the voltmeter whose reading will be zero at resonance is
supporting img

1

V_{4}

2

V_{3}

3

V_{2}

4

V_{1}

Explanation:

At resonance net voltage across $L$ and $C$ is zero

PHXII07:ALTERNATING CURRENT

356250 In the figure shown, three $A C$ voltmeters are connected. At resonance,
supporting img

1

V_{3} = 0

2

V_{1} = V_{2} \neq 0

3

V_{2} = 0

4

V_{1} = 0

Explanation:

At resonance, $V_{L}$ and $V_{C}$ are equal in magnitude but have phase difference of relative to each other.
$∴ V_{L C} = V_{L} - V_{C} = 0$
Hence, voltmeter $V_{2}$ read 0 volt.
So correct option is (4).

PHXII07:ALTERNATING CURRENT

356251 In a series $L C R$ circuit, the inductance $L$ is 10 $m H$ , capacitance $C$ is $1 μ F$ and resistance $R i s 100 Ω$ . The frequency at which resonance occurs is

1

15.9 k H z

2

1.59 r a d / s

3

1.59 k H z

4

15.9 r a d / s

Explanation:

At resonance the two reactances are equal
$X_{L} = X_{C}$ .
$\Rightarrow ω L = \frac{1}{ω C}$
$\Rightarrow ω = \frac{1}{\sqrt{L C}}$
$\Rightarrow f = \frac{1}{2 π \sqrt{L C}}$
$\Rightarrow f = \frac{1}{2 π \sqrt{10 \times 10^{- 3} \times 10^{- 6}}} = 1.59 k H z$
Correct option is (3).

NEET - 2023

PHXII07:ALTERNATING CURRENT

356252 Read the Statement A and Statement B carefully to mark the correct options given below.
Statement A :
In an $L C R$ series circuit, current is maximum at resonance .
Statement B :
Current in a purely resistive circuit can never be less than that in a series $L C R$ circuit when connected to same voltage source.

1 Statement A is correct but Statement B is incorrect

2 Statement A is incorrect but Statement B is correct

3 Both Statements are true

4 Both Statements are incorrect

Explanation:

In a series $L C R$ circuit, current is maximum at resonance.
At resonance, $X_{L} = X_{C}$ ,
$∴$ Impedance,
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{R^{2}}$
or $Z = R$
$\Rightarrow Z$ is minimum at resonance
$∴ I_{m} = \frac{V_{m}}{Z} = \frac{V_{m}}{R}$
In a purely resistive circuit, $I = \frac{V}{R}$ and in a series LCR circuit, $I^{'} = \frac{V}{Z}$
$\Rightarrow I > I^{'}$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356248 In an $L C R$ circuit, at resonance

1 The impedance is maximum

2 The current leads the voltage by

π / 2

3 The current and voltage are in phase

4 The current is minimum

Explanation:

In an $L C R$ circuit at resonance, the phase angle $(ϕ)$ between current and voltage is zero as the total reactance is zero.
Thus the current and voltage are in phase. The current is maximum and the impedance is minimum at resonance in $L C R$ circuit.

PHXII07:ALTERNATING CURRENT

356249 In the joining ac circuit the voltmeter whose reading will be zero at resonance is
supporting img

1

V_{4}

2

V_{3}

3

V_{2}

4

V_{1}

Explanation:

At resonance net voltage across $L$ and $C$ is zero

PHXII07:ALTERNATING CURRENT

356250 In the figure shown, three $A C$ voltmeters are connected. At resonance,
supporting img

1

V_{3} = 0

2

V_{1} = V_{2} \neq 0

3

V_{2} = 0

4

V_{1} = 0

Explanation:

At resonance, $V_{L}$ and $V_{C}$ are equal in magnitude but have phase difference of relative to each other.
$∴ V_{L C} = V_{L} - V_{C} = 0$
Hence, voltmeter $V_{2}$ read 0 volt.
So correct option is (4).

PHXII07:ALTERNATING CURRENT

356251 In a series $L C R$ circuit, the inductance $L$ is 10 $m H$ , capacitance $C$ is $1 μ F$ and resistance $R i s 100 Ω$ . The frequency at which resonance occurs is

1

15.9 k H z

2

1.59 r a d / s

3

1.59 k H z

4

15.9 r a d / s

Explanation:

At resonance the two reactances are equal
$X_{L} = X_{C}$ .
$\Rightarrow ω L = \frac{1}{ω C}$
$\Rightarrow ω = \frac{1}{\sqrt{L C}}$
$\Rightarrow f = \frac{1}{2 π \sqrt{L C}}$
$\Rightarrow f = \frac{1}{2 π \sqrt{10 \times 10^{- 3} \times 10^{- 6}}} = 1.59 k H z$
Correct option is (3).

NEET - 2023

PHXII07:ALTERNATING CURRENT

356252 Read the Statement A and Statement B carefully to mark the correct options given below.
Statement A :
In an $L C R$ series circuit, current is maximum at resonance .
Statement B :
Current in a purely resistive circuit can never be less than that in a series $L C R$ circuit when connected to same voltage source.

1 Statement A is correct but Statement B is incorrect

2 Statement A is incorrect but Statement B is correct

3 Both Statements are true

4 Both Statements are incorrect

Explanation:

In a series $L C R$ circuit, current is maximum at resonance.
At resonance, $X_{L} = X_{C}$ ,
$∴$ Impedance,
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{R^{2}}$
or $Z = R$
$\Rightarrow Z$ is minimum at resonance
$∴ I_{m} = \frac{V_{m}}{Z} = \frac{V_{m}}{R}$
In a purely resistive circuit, $I = \frac{V}{R}$ and in a series LCR circuit, $I^{'} = \frac{V}{Z}$
$\Rightarrow I > I^{'}$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356248 In an $L C R$ circuit, at resonance

1 The impedance is maximum

2 The current leads the voltage by

π / 2

3 The current and voltage are in phase

4 The current is minimum

Explanation:

In an $L C R$ circuit at resonance, the phase angle $(ϕ)$ between current and voltage is zero as the total reactance is zero.
Thus the current and voltage are in phase. The current is maximum and the impedance is minimum at resonance in $L C R$ circuit.

PHXII07:ALTERNATING CURRENT

356249 In the joining ac circuit the voltmeter whose reading will be zero at resonance is
supporting img

1

V_{4}

2

V_{3}

3

V_{2}

4

V_{1}

Explanation:

At resonance net voltage across $L$ and $C$ is zero

PHXII07:ALTERNATING CURRENT

356250 In the figure shown, three $A C$ voltmeters are connected. At resonance,
supporting img

1

V_{3} = 0

2

V_{1} = V_{2} \neq 0

3

V_{2} = 0

4

V_{1} = 0

Explanation:

At resonance, $V_{L}$ and $V_{C}$ are equal in magnitude but have phase difference of relative to each other.
$∴ V_{L C} = V_{L} - V_{C} = 0$
Hence, voltmeter $V_{2}$ read 0 volt.
So correct option is (4).

PHXII07:ALTERNATING CURRENT

356251 In a series $L C R$ circuit, the inductance $L$ is 10 $m H$ , capacitance $C$ is $1 μ F$ and resistance $R i s 100 Ω$ . The frequency at which resonance occurs is

1

15.9 k H z

2

1.59 r a d / s

3

1.59 k H z

4

15.9 r a d / s

Explanation:

At resonance the two reactances are equal
$X_{L} = X_{C}$ .
$\Rightarrow ω L = \frac{1}{ω C}$
$\Rightarrow ω = \frac{1}{\sqrt{L C}}$
$\Rightarrow f = \frac{1}{2 π \sqrt{L C}}$
$\Rightarrow f = \frac{1}{2 π \sqrt{10 \times 10^{- 3} \times 10^{- 6}}} = 1.59 k H z$
Correct option is (3).

NEET - 2023

PHXII07:ALTERNATING CURRENT

356252 Read the Statement A and Statement B carefully to mark the correct options given below.
Statement A :
In an $L C R$ series circuit, current is maximum at resonance .
Statement B :
Current in a purely resistive circuit can never be less than that in a series $L C R$ circuit when connected to same voltage source.

1 Statement A is correct but Statement B is incorrect

2 Statement A is incorrect but Statement B is correct

3 Both Statements are true

4 Both Statements are incorrect

Explanation:

In a series $L C R$ circuit, current is maximum at resonance.
At resonance, $X_{L} = X_{C}$ ,
$∴$ Impedance,
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{R^{2}}$
or $Z = R$
$\Rightarrow Z$ is minimum at resonance
$∴ I_{m} = \frac{V_{m}}{Z} = \frac{V_{m}}{R}$
In a purely resistive circuit, $I = \frac{V}{R}$ and in a series LCR circuit, $I^{'} = \frac{V}{Z}$
$\Rightarrow I > I^{'}$

JEE - 2024

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