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PHXII07:ALTERNATING CURRENT

356218 Assertion :
In series $L C R$ circuit resonance can take place.
Reason :
Resonance takes place if inductance and capacitive reactances are equal and opposite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Resonance occurs when the inductive reactance $(X_{L})$ and capacitive reactance $(X_{C})$ are equal in magnitude and opposite in phase, resulting in the impedance being purely resistive. At this point, the circuit allows for maximum current flow and is in a state of resonance. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So, correct option is (1).

PHXII07:ALTERNATING CURRENT

356219 For the series $L C R$ circuit shown in the figure, the resonance frequency is
supporting img

1

2500 r a d / s

2

250 r a d / s

3

25 r a d / s

4

2.5 r a d / s

Explanation:

$V = \frac{1}{2 π \sqrt{L C}}$
$ω = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{8 \times 10^{- 3} \times 20 \times 10^{- 6}}}$
$= \frac{1}{\sqrt{16 \times 10^{- 8}}} = \frac{1}{4 \times 10^{- 4}} = 2500 r a d / s$

PHXII07:ALTERNATING CURRENT

356220 To reduce the resonant frequency in an $L - C - R$ series circuit with a generator

1 The iron core of the inductor should be removed

2 Dielectric in the capacitor should be removed

3 The generator frequency should be reduced

4 Another capacitor should be added in parallel to the first

Explanation:

Reasonant frequency in an $L - C - R$ series circuit is
$υ_{r} = \frac{1}{2 π \sqrt{L C}}$
To reduce $υ_{r}; C$ can be increased, by adding another capacitor in parallel to the first capacitor

NCERT Exemplar

PHXII07:ALTERNATING CURRENT

356221 Which of the following combinations should be selected for better tuning of an $L - C - R$ circuit used for communication ?

1

R = 25 Ω, L = 2.5 H, C = 45 μ F

2

R = 20 Ω, L = 1.5 H, C = 35 μ F

3

R = 25 Ω, L = 1.5 H, C = 45 μ F

4

R = 15 Ω, L = 3.5 H, C = 30 μ F

Explanation:

For better tuning the resonance curve should be sharp. When $Q$ -factor is large, the sharpness of resonance curve is more and vice-verse.
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
For large value of $Q$ , $R$ should be small, $L$ should be high and $C$ should be low. These conditions are best satisfied by the values in option (4)

NCERT Exemplar

PHXII07:ALTERNATING CURRENT

356222 An $L C R$ circuit contains $R = 50 Ω, L = 1 m H$ and $C = 0.1 μ F$ . The impedance of the circuit will be minimum for a frequency of

1

\frac{10^{5}}{2 π} H z

2

\frac{10^{6}}{2 π} H z

3

2 π \times 10^{5} H z

4

2 π \times 10^{6} H z

Explanation:

The impedance of the $L C R$ circuit will be minimum for a resonant frequency $v_{r}$ .
$v_{r} = \frac{1}{2 π \sqrt{L C}}$
Here, $L = 1 m H = 1 \times 10^{- 3} H$
$C = 0.1 μ F = 0.1 \times 10^{- 6} F$
$∴ v_{r} = \frac{1}{2 π \sqrt{(1 \times 10^{- 3} H) (0.1 \times 10^{- 6} F)}}$
$= \frac{10^{5}}{2 π} H z$

PHXII07:ALTERNATING CURRENT

356218 Assertion :
In series $L C R$ circuit resonance can take place.
Reason :
Resonance takes place if inductance and capacitive reactances are equal and opposite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Resonance occurs when the inductive reactance $(X_{L})$ and capacitive reactance $(X_{C})$ are equal in magnitude and opposite in phase, resulting in the impedance being purely resistive. At this point, the circuit allows for maximum current flow and is in a state of resonance. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So, correct option is (1).

PHXII07:ALTERNATING CURRENT

356219 For the series $L C R$ circuit shown in the figure, the resonance frequency is
supporting img

1

2500 r a d / s

2

250 r a d / s

3

25 r a d / s

4

2.5 r a d / s

Explanation:

$V = \frac{1}{2 π \sqrt{L C}}$
$ω = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{8 \times 10^{- 3} \times 20 \times 10^{- 6}}}$
$= \frac{1}{\sqrt{16 \times 10^{- 8}}} = \frac{1}{4 \times 10^{- 4}} = 2500 r a d / s$

PHXII07:ALTERNATING CURRENT

356220 To reduce the resonant frequency in an $L - C - R$ series circuit with a generator

1 The iron core of the inductor should be removed

2 Dielectric in the capacitor should be removed

3 The generator frequency should be reduced

4 Another capacitor should be added in parallel to the first

Explanation:

Reasonant frequency in an $L - C - R$ series circuit is
$υ_{r} = \frac{1}{2 π \sqrt{L C}}$
To reduce $υ_{r}; C$ can be increased, by adding another capacitor in parallel to the first capacitor

NCERT Exemplar

PHXII07:ALTERNATING CURRENT

356221 Which of the following combinations should be selected for better tuning of an $L - C - R$ circuit used for communication ?

1

R = 25 Ω, L = 2.5 H, C = 45 μ F

2

R = 20 Ω, L = 1.5 H, C = 35 μ F

3

R = 25 Ω, L = 1.5 H, C = 45 μ F

4

R = 15 Ω, L = 3.5 H, C = 30 μ F

Explanation:

For better tuning the resonance curve should be sharp. When $Q$ -factor is large, the sharpness of resonance curve is more and vice-verse.
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
For large value of $Q$ , $R$ should be small, $L$ should be high and $C$ should be low. These conditions are best satisfied by the values in option (4)

NCERT Exemplar

PHXII07:ALTERNATING CURRENT

356222 An $L C R$ circuit contains $R = 50 Ω, L = 1 m H$ and $C = 0.1 μ F$ . The impedance of the circuit will be minimum for a frequency of

1

\frac{10^{5}}{2 π} H z

2

\frac{10^{6}}{2 π} H z

3

2 π \times 10^{5} H z

4

2 π \times 10^{6} H z

Explanation:

The impedance of the $L C R$ circuit will be minimum for a resonant frequency $v_{r}$ .
$v_{r} = \frac{1}{2 π \sqrt{L C}}$
Here, $L = 1 m H = 1 \times 10^{- 3} H$
$C = 0.1 μ F = 0.1 \times 10^{- 6} F$
$∴ v_{r} = \frac{1}{2 π \sqrt{(1 \times 10^{- 3} H) (0.1 \times 10^{- 6} F)}}$
$= \frac{10^{5}}{2 π} H z$

PHXII07:ALTERNATING CURRENT

356218 Assertion :
In series $L C R$ circuit resonance can take place.
Reason :
Resonance takes place if inductance and capacitive reactances are equal and opposite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Resonance occurs when the inductive reactance $(X_{L})$ and capacitive reactance $(X_{C})$ are equal in magnitude and opposite in phase, resulting in the impedance being purely resistive. At this point, the circuit allows for maximum current flow and is in a state of resonance. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So, correct option is (1).

PHXII07:ALTERNATING CURRENT

356219 For the series $L C R$ circuit shown in the figure, the resonance frequency is
supporting img

1

2500 r a d / s

2

250 r a d / s

3

25 r a d / s

4

2.5 r a d / s

Explanation:

$V = \frac{1}{2 π \sqrt{L C}}$
$ω = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{8 \times 10^{- 3} \times 20 \times 10^{- 6}}}$
$= \frac{1}{\sqrt{16 \times 10^{- 8}}} = \frac{1}{4 \times 10^{- 4}} = 2500 r a d / s$

PHXII07:ALTERNATING CURRENT

356220 To reduce the resonant frequency in an $L - C - R$ series circuit with a generator

1 The iron core of the inductor should be removed

2 Dielectric in the capacitor should be removed

3 The generator frequency should be reduced

4 Another capacitor should be added in parallel to the first

Explanation:

Reasonant frequency in an $L - C - R$ series circuit is
$υ_{r} = \frac{1}{2 π \sqrt{L C}}$
To reduce $υ_{r}; C$ can be increased, by adding another capacitor in parallel to the first capacitor

NCERT Exemplar

PHXII07:ALTERNATING CURRENT

356221 Which of the following combinations should be selected for better tuning of an $L - C - R$ circuit used for communication ?

1

R = 25 Ω, L = 2.5 H, C = 45 μ F

2

R = 20 Ω, L = 1.5 H, C = 35 μ F

3

R = 25 Ω, L = 1.5 H, C = 45 μ F

4

R = 15 Ω, L = 3.5 H, C = 30 μ F

Explanation:

For better tuning the resonance curve should be sharp. When $Q$ -factor is large, the sharpness of resonance curve is more and vice-verse.
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
For large value of $Q$ , $R$ should be small, $L$ should be high and $C$ should be low. These conditions are best satisfied by the values in option (4)

NCERT Exemplar

PHXII07:ALTERNATING CURRENT

356222 An $L C R$ circuit contains $R = 50 Ω, L = 1 m H$ and $C = 0.1 μ F$ . The impedance of the circuit will be minimum for a frequency of

1

\frac{10^{5}}{2 π} H z

2

\frac{10^{6}}{2 π} H z

3

2 π \times 10^{5} H z

4

2 π \times 10^{6} H z

Explanation:

The impedance of the $L C R$ circuit will be minimum for a resonant frequency $v_{r}$ .
$v_{r} = \frac{1}{2 π \sqrt{L C}}$
Here, $L = 1 m H = 1 \times 10^{- 3} H$
$C = 0.1 μ F = 0.1 \times 10^{- 6} F$
$∴ v_{r} = \frac{1}{2 π \sqrt{(1 \times 10^{- 3} H) (0.1 \times 10^{- 6} F)}}$
$= \frac{10^{5}}{2 π} H z$

PHXII07:ALTERNATING CURRENT

356218 Assertion :
In series $L C R$ circuit resonance can take place.
Reason :
Resonance takes place if inductance and capacitive reactances are equal and opposite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Resonance occurs when the inductive reactance $(X_{L})$ and capacitive reactance $(X_{C})$ are equal in magnitude and opposite in phase, resulting in the impedance being purely resistive. At this point, the circuit allows for maximum current flow and is in a state of resonance. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So, correct option is (1).

PHXII07:ALTERNATING CURRENT

356219 For the series $L C R$ circuit shown in the figure, the resonance frequency is
supporting img

1

2500 r a d / s

2

250 r a d / s

3

25 r a d / s

4

2.5 r a d / s

Explanation:

$V = \frac{1}{2 π \sqrt{L C}}$
$ω = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{8 \times 10^{- 3} \times 20 \times 10^{- 6}}}$
$= \frac{1}{\sqrt{16 \times 10^{- 8}}} = \frac{1}{4 \times 10^{- 4}} = 2500 r a d / s$

PHXII07:ALTERNATING CURRENT

356220 To reduce the resonant frequency in an $L - C - R$ series circuit with a generator

1 The iron core of the inductor should be removed

2 Dielectric in the capacitor should be removed

3 The generator frequency should be reduced

4 Another capacitor should be added in parallel to the first

Explanation:

Reasonant frequency in an $L - C - R$ series circuit is
$υ_{r} = \frac{1}{2 π \sqrt{L C}}$
To reduce $υ_{r}; C$ can be increased, by adding another capacitor in parallel to the first capacitor

NCERT Exemplar

PHXII07:ALTERNATING CURRENT

356221 Which of the following combinations should be selected for better tuning of an $L - C - R$ circuit used for communication ?

1

R = 25 Ω, L = 2.5 H, C = 45 μ F

2

R = 20 Ω, L = 1.5 H, C = 35 μ F

3

R = 25 Ω, L = 1.5 H, C = 45 μ F

4

R = 15 Ω, L = 3.5 H, C = 30 μ F

Explanation:

For better tuning the resonance curve should be sharp. When $Q$ -factor is large, the sharpness of resonance curve is more and vice-verse.
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
For large value of $Q$ , $R$ should be small, $L$ should be high and $C$ should be low. These conditions are best satisfied by the values in option (4)

NCERT Exemplar

PHXII07:ALTERNATING CURRENT

356222 An $L C R$ circuit contains $R = 50 Ω, L = 1 m H$ and $C = 0.1 μ F$ . The impedance of the circuit will be minimum for a frequency of

1

\frac{10^{5}}{2 π} H z

2

\frac{10^{6}}{2 π} H z

3

2 π \times 10^{5} H z

4

2 π \times 10^{6} H z

Explanation:

The impedance of the $L C R$ circuit will be minimum for a resonant frequency $v_{r}$ .
$v_{r} = \frac{1}{2 π \sqrt{L C}}$
Here, $L = 1 m H = 1 \times 10^{- 3} H$
$C = 0.1 μ F = 0.1 \times 10^{- 6} F$
$∴ v_{r} = \frac{1}{2 π \sqrt{(1 \times 10^{- 3} H) (0.1 \times 10^{- 6} F)}}$
$= \frac{10^{5}}{2 π} H z$

PHXII07:ALTERNATING CURRENT

356218 Assertion :
In series $L C R$ circuit resonance can take place.
Reason :
Resonance takes place if inductance and capacitive reactances are equal and opposite.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Resonance occurs when the inductive reactance $(X_{L})$ and capacitive reactance $(X_{C})$ are equal in magnitude and opposite in phase, resulting in the impedance being purely resistive. At this point, the circuit allows for maximum current flow and is in a state of resonance. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So, correct option is (1).

PHXII07:ALTERNATING CURRENT

356219 For the series $L C R$ circuit shown in the figure, the resonance frequency is
supporting img

1

2500 r a d / s

2

250 r a d / s

3

25 r a d / s

4

2.5 r a d / s

Explanation:

$V = \frac{1}{2 π \sqrt{L C}}$
$ω = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{8 \times 10^{- 3} \times 20 \times 10^{- 6}}}$
$= \frac{1}{\sqrt{16 \times 10^{- 8}}} = \frac{1}{4 \times 10^{- 4}} = 2500 r a d / s$

PHXII07:ALTERNATING CURRENT

356220 To reduce the resonant frequency in an $L - C - R$ series circuit with a generator

1 The iron core of the inductor should be removed

2 Dielectric in the capacitor should be removed

3 The generator frequency should be reduced

4 Another capacitor should be added in parallel to the first

Explanation:

Reasonant frequency in an $L - C - R$ series circuit is
$υ_{r} = \frac{1}{2 π \sqrt{L C}}$
To reduce $υ_{r}; C$ can be increased, by adding another capacitor in parallel to the first capacitor

NCERT Exemplar

PHXII07:ALTERNATING CURRENT

356221 Which of the following combinations should be selected for better tuning of an $L - C - R$ circuit used for communication ?

1

R = 25 Ω, L = 2.5 H, C = 45 μ F

2

R = 20 Ω, L = 1.5 H, C = 35 μ F

3

R = 25 Ω, L = 1.5 H, C = 45 μ F

4

R = 15 Ω, L = 3.5 H, C = 30 μ F

Explanation:

For better tuning the resonance curve should be sharp. When $Q$ -factor is large, the sharpness of resonance curve is more and vice-verse.
$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
For large value of $Q$ , $R$ should be small, $L$ should be high and $C$ should be low. These conditions are best satisfied by the values in option (4)

NCERT Exemplar

PHXII07:ALTERNATING CURRENT

356222 An $L C R$ circuit contains $R = 50 Ω, L = 1 m H$ and $C = 0.1 μ F$ . The impedance of the circuit will be minimum for a frequency of

1

\frac{10^{5}}{2 π} H z

2

\frac{10^{6}}{2 π} H z

3

2 π \times 10^{5} H z

4

2 π \times 10^{6} H z

Explanation:

The impedance of the $L C R$ circuit will be minimum for a resonant frequency $v_{r}$ .
$v_{r} = \frac{1}{2 π \sqrt{L C}}$
Here, $L = 1 m H = 1 \times 10^{- 3} H$
$C = 0.1 μ F = 0.1 \times 10^{- 6} F$
$∴ v_{r} = \frac{1}{2 π \sqrt{(1 \times 10^{- 3} H) (0.1 \times 10^{- 6} F)}}$
$= \frac{10^{5}}{2 π} H z$

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