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PHXII07:ALTERNATING CURRENT

356114 A capacitor of capacitance $100 μ F$ is charged to a potential of $12 V$ and connected to a $6.4 m H$ inductor to produce oscillations. The maximum current in the circuit would be

1

3.2 A

2

1.5 A

3

1.2 A

4

2.0 A

Explanation:

$C = 100 μ F = 100 \times 10^{- 6} F = 10^{- 4} F$
$V = 12 V$
$L = 6.4 m H = 6.4 \times 10^{- 3} H$
supporting img

Total energy is conserved.
max energy in $C =$ max energy in $L$
$\Rightarrow \frac{1}{2} C V^{2} = \frac{1}{2} L i^{2}$
$\Rightarrow i = \sqrt{\frac{C}{L}} V$
$\Rightarrow i = \sqrt{\frac{10^{- 4}}{6.4 \times 10^{- 3}}} \times 12 = \frac{12}{8} = 1.5 A$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356115 In the given circuit, the resonant frequency is
supporting img

1

1592 H z

2

15910 H z

3

15.92 H z

4

159.2 H z

Explanation:

Resonant frequency,
$v = \frac{1}{2 π \sqrt{L C}}$
$= \frac{1}{2 \times 3.14 \times \sqrt{0.5 \times 10^{- 3} \times 20 \times 10^{- 6}}} \approx 1592 H z$

PHXII07:ALTERNATING CURRENT

356116 The frequency of the output signal becomes $x$ times by doubling of the value of the capacitance in the $L C$ oscillator circuit. The value of $x$ is

1

\frac{1}{\sqrt{2}}

2

2

3

\sqrt{2}

4

\frac{1}{2}

Explanation:

$f = \frac{1}{2 π} \sqrt{\frac{1}{L C}}$
$f^{'} = \frac{1}{2 π} \sqrt{\frac{1}{2 (L C)}} = \frac{f}{\sqrt{2}}$

PHXII07:ALTERNATING CURRENT

356117 An $L - C$ circuit is in the state of resonance. If $C = 0.1 μ F$ and $L = 0.25 H$ , neglecting ohmic resistance of circuit, then what is the frequency of oscillations?

1

1007 H z

2

100 H z

3

109 H z

4

500 H z

Explanation:

From the formula, the frequency of
oscillation is
$f = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 π \sqrt{0.25 \times 0.1 \times 10^{- 6}}}$
$\approx 1007 H z$

PHXII07:ALTERNATING CURRENT

356114 A capacitor of capacitance $100 μ F$ is charged to a potential of $12 V$ and connected to a $6.4 m H$ inductor to produce oscillations. The maximum current in the circuit would be

1

3.2 A

2

1.5 A

3

1.2 A

4

2.0 A

Explanation:

$C = 100 μ F = 100 \times 10^{- 6} F = 10^{- 4} F$
$V = 12 V$
$L = 6.4 m H = 6.4 \times 10^{- 3} H$
supporting img

Total energy is conserved.
max energy in $C =$ max energy in $L$
$\Rightarrow \frac{1}{2} C V^{2} = \frac{1}{2} L i^{2}$
$\Rightarrow i = \sqrt{\frac{C}{L}} V$
$\Rightarrow i = \sqrt{\frac{10^{- 4}}{6.4 \times 10^{- 3}}} \times 12 = \frac{12}{8} = 1.5 A$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356115 In the given circuit, the resonant frequency is
supporting img

1

1592 H z

2

15910 H z

3

15.92 H z

4

159.2 H z

Explanation:

Resonant frequency,
$v = \frac{1}{2 π \sqrt{L C}}$
$= \frac{1}{2 \times 3.14 \times \sqrt{0.5 \times 10^{- 3} \times 20 \times 10^{- 6}}} \approx 1592 H z$

PHXII07:ALTERNATING CURRENT

356116 The frequency of the output signal becomes $x$ times by doubling of the value of the capacitance in the $L C$ oscillator circuit. The value of $x$ is

1

\frac{1}{\sqrt{2}}

2

2

3

\sqrt{2}

4

\frac{1}{2}

Explanation:

$f = \frac{1}{2 π} \sqrt{\frac{1}{L C}}$
$f^{'} = \frac{1}{2 π} \sqrt{\frac{1}{2 (L C)}} = \frac{f}{\sqrt{2}}$

PHXII07:ALTERNATING CURRENT

356117 An $L - C$ circuit is in the state of resonance. If $C = 0.1 μ F$ and $L = 0.25 H$ , neglecting ohmic resistance of circuit, then what is the frequency of oscillations?

1

1007 H z

2

100 H z

3

109 H z

4

500 H z

Explanation:

From the formula, the frequency of
oscillation is
$f = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 π \sqrt{0.25 \times 0.1 \times 10^{- 6}}}$
$\approx 1007 H z$

PHXII07:ALTERNATING CURRENT

356114 A capacitor of capacitance $100 μ F$ is charged to a potential of $12 V$ and connected to a $6.4 m H$ inductor to produce oscillations. The maximum current in the circuit would be

1

3.2 A

2

1.5 A

3

1.2 A

4

2.0 A

Explanation:

$C = 100 μ F = 100 \times 10^{- 6} F = 10^{- 4} F$
$V = 12 V$
$L = 6.4 m H = 6.4 \times 10^{- 3} H$
supporting img

Total energy is conserved.
max energy in $C =$ max energy in $L$
$\Rightarrow \frac{1}{2} C V^{2} = \frac{1}{2} L i^{2}$
$\Rightarrow i = \sqrt{\frac{C}{L}} V$
$\Rightarrow i = \sqrt{\frac{10^{- 4}}{6.4 \times 10^{- 3}}} \times 12 = \frac{12}{8} = 1.5 A$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356115 In the given circuit, the resonant frequency is
supporting img

1

1592 H z

2

15910 H z

3

15.92 H z

4

159.2 H z

Explanation:

Resonant frequency,
$v = \frac{1}{2 π \sqrt{L C}}$
$= \frac{1}{2 \times 3.14 \times \sqrt{0.5 \times 10^{- 3} \times 20 \times 10^{- 6}}} \approx 1592 H z$

PHXII07:ALTERNATING CURRENT

356116 The frequency of the output signal becomes $x$ times by doubling of the value of the capacitance in the $L C$ oscillator circuit. The value of $x$ is

1

\frac{1}{\sqrt{2}}

2

2

3

\sqrt{2}

4

\frac{1}{2}

Explanation:

$f = \frac{1}{2 π} \sqrt{\frac{1}{L C}}$
$f^{'} = \frac{1}{2 π} \sqrt{\frac{1}{2 (L C)}} = \frac{f}{\sqrt{2}}$

PHXII07:ALTERNATING CURRENT

356117 An $L - C$ circuit is in the state of resonance. If $C = 0.1 μ F$ and $L = 0.25 H$ , neglecting ohmic resistance of circuit, then what is the frequency of oscillations?

1

1007 H z

2

100 H z

3

109 H z

4

500 H z

Explanation:

From the formula, the frequency of
oscillation is
$f = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 π \sqrt{0.25 \times 0.1 \times 10^{- 6}}}$
$\approx 1007 H z$

PHXII07:ALTERNATING CURRENT

356114 A capacitor of capacitance $100 μ F$ is charged to a potential of $12 V$ and connected to a $6.4 m H$ inductor to produce oscillations. The maximum current in the circuit would be

1

3.2 A

2

1.5 A

3

1.2 A

4

2.0 A

Explanation:

$C = 100 μ F = 100 \times 10^{- 6} F = 10^{- 4} F$
$V = 12 V$
$L = 6.4 m H = 6.4 \times 10^{- 3} H$
supporting img

Total energy is conserved.
max energy in $C =$ max energy in $L$
$\Rightarrow \frac{1}{2} C V^{2} = \frac{1}{2} L i^{2}$
$\Rightarrow i = \sqrt{\frac{C}{L}} V$
$\Rightarrow i = \sqrt{\frac{10^{- 4}}{6.4 \times 10^{- 3}}} \times 12 = \frac{12}{8} = 1.5 A$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356115 In the given circuit, the resonant frequency is
supporting img

1

1592 H z

2

15910 H z

3

15.92 H z

4

159.2 H z

Explanation:

Resonant frequency,
$v = \frac{1}{2 π \sqrt{L C}}$
$= \frac{1}{2 \times 3.14 \times \sqrt{0.5 \times 10^{- 3} \times 20 \times 10^{- 6}}} \approx 1592 H z$

PHXII07:ALTERNATING CURRENT

356116 The frequency of the output signal becomes $x$ times by doubling of the value of the capacitance in the $L C$ oscillator circuit. The value of $x$ is

1

\frac{1}{\sqrt{2}}

2

2

3

\sqrt{2}

4

\frac{1}{2}

Explanation:

$f = \frac{1}{2 π} \sqrt{\frac{1}{L C}}$
$f^{'} = \frac{1}{2 π} \sqrt{\frac{1}{2 (L C)}} = \frac{f}{\sqrt{2}}$

PHXII07:ALTERNATING CURRENT

356117 An $L - C$ circuit is in the state of resonance. If $C = 0.1 μ F$ and $L = 0.25 H$ , neglecting ohmic resistance of circuit, then what is the frequency of oscillations?

1

1007 H z

2

100 H z

3

109 H z

4

500 H z

Explanation:

From the formula, the frequency of
oscillation is
$f = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 π \sqrt{0.25 \times 0.1 \times 10^{- 6}}}$
$\approx 1007 H z$