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Alternating Current

155032 In an oscillating $L C$ circuit, the value of inductance is $1.6 mH$ and the value of capacitance is $4 μ F$ . If the maximum charge on the capacitor is $4 \times 10^{- 6} C$ , then the maximum current is

1

75 mA

2

12.5 mA

3

125 mA

4

50 mA

Explanation:

D Given,
$L = 1.6 mH = 1.6 \times 10^{- 3} H, C = 4 μ F = 4 \times 10^{- 6} F$ $Q = 4 \times 10^{- 6}$
Energy stored in capacitor $(U_{C}) = \frac{1}{2} \frac{Q^{2}}{C}$
Energy stored in $Inductor (U_{L}) = \frac{1}{2} {LI}^{2}$
$U_{C} = U_{L}$
$\frac{1}{2} \frac{Q^{2}}{C} = \frac{1}{2} {LI}^{2}$
$I^{2} = \frac{Q^{2}}{L C}$
$I = \frac{Q}{\sqrt{LC}} = \frac{4 \times 10^{- 6}}{\sqrt{1.6 \times 10^{- 3} \times 4 \times 10^{- 6}}}$
$= \frac{4 \times 10^{- 6}}{8 \times 10^{- 5}}$
$= 5 \times 10^{- 2} = 50 \times 10^{- 3} = 50 mA$

AP EAMCET-11.07.2022

Alternating Current

155035
Consider a pure inductive A.C. circuit as shown in the figure. If the average power consumed is $P$ , then

1

P > 0

2

P < 0

3

P = 0

4

P

is infinite

Explanation:

C In pure inductor current is tog by $90^{\circ}$ with voltage.
$∴$ average power dissipation $P$
$P = \frac{V_{0} I_{0}}{\sqrt{2} \times \sqrt{2}} \cos ϕ$
$P = \frac{V_{0} I_{0}}{2} \cos 90^{\circ}$
$P = 0$

WB JEE 2021

Alternating Current

155036 An ac source of angular frequency $ω$ is fed across a resistor $R$ and a capacitor $C$ in series. The current flowing in the circuit found to be ' $I$ ' now the frequency of the source is changed to $\frac{ω}{3}$ . (Maintaining the same voltage) the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?

1

\sqrt{\frac{5}{7}}

2

\sqrt{\frac{3}{4}}

3

\sqrt{\frac{3}{5}}

4

\sqrt{\frac{7}{5}}

Explanation:

C Given, For RC circuit,
$I = \frac{V_{rms}}{\sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}} = \frac{V_{rms}}{\sqrt{R^{2} + \frac{1}{ω^{2} C^{2}}}}$
$I = \frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}}$
When angular frequency of the source is $\frac{ω}{3}$ Then current becomes $\frac{I}{2}$ . Then,
$\frac{I}{2} = \frac{V_{rms} \frac{ω}{3} C}{\sqrt{R^{2} {(\frac{ω}{3})}^{2} C^{2} + 1}}$
$\frac{I}{2} = \frac{V_{rms} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 9}}$
Dividing equation (i) by (ii) -
$\frac{I}{I / 2} = \frac{\frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}}}{\frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 9}}}$
$2 = \frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}} \times \frac{\sqrt{R^{2} ω^{2} C^{2} + 9}}{V_{r m s} ω C}$
$2 = \frac{\sqrt{R^{2} ω^{2} C^{2} + 9}}{\sqrt{R^{2} ω^{2} C^{2} + 1}}$
Squaring both sides,
$4 = \frac{R^{2} ω^{2} C^{2} + 9}{R^{2} ω^{2} C^{2} + 1}$
4 $R^{2} ω^{2} C^{2} + 4 = R^{2} ω^{2} C^{2} + 9$
3 $R^{2} ω^{2} C^{2} = 5$
$R^{2} ω^{2} C^{2} = \frac{5}{3}$
$R = \sqrt{\frac{5}{3} \frac{1}{ω^{2} C^{2}}} = \sqrt{\frac{5}{3}} \times \frac{1}{ω C} = \sqrt{\frac{5}{3}} \times X_{C}$
$\frac{X_{C}}{R} = \sqrt{\frac{3}{5}}$

AP EAMCET-25.08.2021

Alternating Current

155038 If $120 V, 60 Hz$ Ac voltage is applied to an $LR$ having $R = 100 Ω$ and $L = 2 H$ , then current in the circuit is

1

0.32 A

2

0.16 A

3

0.48 A

4

0.80 A

Explanation:

B Given that, $V = 120 V, f = 60 Hz, R = 100 Ω$ , $L = 2 H$
Impedance of LR circuit,
$Z = \sqrt{R^{2} + {(X_{L})}^{2}}$
$Z = \sqrt{(100)^{2} + (2 π \times 60 \times 2)^{2}}$
$Z = \sqrt{(100)^{2} + (240 π)^{2}}$
$Z = 760.58$
$I = \frac{V}{Z} = \frac{120}{760.58}$
$I = 0.157 \approx 0.16 A$

AP EAMCET-05.10.2021

Alternating Current

155039 A coil of resistance $50 Ω$ is connected across a $5.0 V$ battery. If the current in the coil is found to be $50 mA$ after time $t = 0.1$ battery is connected, then the inductance of the coil is

1

\frac{5}{\ln (2)}

2

10 \ln (2)

3

5 e^{4}

4

\frac{10}{e^{4}}

Explanation:

A Given, $R = 50 Ω, E = 5.0 V, i = 50 mA = 50 \times$ $10^{- 3} A, t = 0.1$
$i = i_{0} (1 - e^{- t R / L})$
$50 \times 10^{- 3} = \frac{5}{50} (1 - e^{- 0.1 \times \frac{50}{L}})$
$10 \times 50 \times 10^{- 3} = 1 - e^{- \frac{5}{L}}$
$e^{- 5 / L} = 1 - 50 \times 10^{- 2}$
$e^{- 5 / L} = \frac{1}{2}$
$\frac{- 5}{L} = \log (\frac{1}{2})$
$L = \frac{5}{\log (2)}$

TS EAMCET 05.08.2021

Alternating Current

155032 In an oscillating $L C$ circuit, the value of inductance is $1.6 mH$ and the value of capacitance is $4 μ F$ . If the maximum charge on the capacitor is $4 \times 10^{- 6} C$ , then the maximum current is

1

75 mA

2

12.5 mA

3

125 mA

4

50 mA

Explanation:

D Given,
$L = 1.6 mH = 1.6 \times 10^{- 3} H, C = 4 μ F = 4 \times 10^{- 6} F$ $Q = 4 \times 10^{- 6}$
Energy stored in capacitor $(U_{C}) = \frac{1}{2} \frac{Q^{2}}{C}$
Energy stored in $Inductor (U_{L}) = \frac{1}{2} {LI}^{2}$
$U_{C} = U_{L}$
$\frac{1}{2} \frac{Q^{2}}{C} = \frac{1}{2} {LI}^{2}$
$I^{2} = \frac{Q^{2}}{L C}$
$I = \frac{Q}{\sqrt{LC}} = \frac{4 \times 10^{- 6}}{\sqrt{1.6 \times 10^{- 3} \times 4 \times 10^{- 6}}}$
$= \frac{4 \times 10^{- 6}}{8 \times 10^{- 5}}$
$= 5 \times 10^{- 2} = 50 \times 10^{- 3} = 50 mA$

AP EAMCET-11.07.2022

Alternating Current

155035
Consider a pure inductive A.C. circuit as shown in the figure. If the average power consumed is $P$ , then

1

P > 0

2

P < 0

3

P = 0

4

P

is infinite

Explanation:

C In pure inductor current is tog by $90^{\circ}$ with voltage.
$∴$ average power dissipation $P$
$P = \frac{V_{0} I_{0}}{\sqrt{2} \times \sqrt{2}} \cos ϕ$
$P = \frac{V_{0} I_{0}}{2} \cos 90^{\circ}$
$P = 0$

WB JEE 2021

Alternating Current

155036 An ac source of angular frequency $ω$ is fed across a resistor $R$ and a capacitor $C$ in series. The current flowing in the circuit found to be ' $I$ ' now the frequency of the source is changed to $\frac{ω}{3}$ . (Maintaining the same voltage) the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?

1

\sqrt{\frac{5}{7}}

2

\sqrt{\frac{3}{4}}

3

\sqrt{\frac{3}{5}}

4

\sqrt{\frac{7}{5}}

Explanation:

C Given, For RC circuit,
$I = \frac{V_{rms}}{\sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}} = \frac{V_{rms}}{\sqrt{R^{2} + \frac{1}{ω^{2} C^{2}}}}$
$I = \frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}}$
When angular frequency of the source is $\frac{ω}{3}$ Then current becomes $\frac{I}{2}$ . Then,
$\frac{I}{2} = \frac{V_{rms} \frac{ω}{3} C}{\sqrt{R^{2} {(\frac{ω}{3})}^{2} C^{2} + 1}}$
$\frac{I}{2} = \frac{V_{rms} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 9}}$
Dividing equation (i) by (ii) -
$\frac{I}{I / 2} = \frac{\frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}}}{\frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 9}}}$
$2 = \frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}} \times \frac{\sqrt{R^{2} ω^{2} C^{2} + 9}}{V_{r m s} ω C}$
$2 = \frac{\sqrt{R^{2} ω^{2} C^{2} + 9}}{\sqrt{R^{2} ω^{2} C^{2} + 1}}$
Squaring both sides,
$4 = \frac{R^{2} ω^{2} C^{2} + 9}{R^{2} ω^{2} C^{2} + 1}$
4 $R^{2} ω^{2} C^{2} + 4 = R^{2} ω^{2} C^{2} + 9$
3 $R^{2} ω^{2} C^{2} = 5$
$R^{2} ω^{2} C^{2} = \frac{5}{3}$
$R = \sqrt{\frac{5}{3} \frac{1}{ω^{2} C^{2}}} = \sqrt{\frac{5}{3}} \times \frac{1}{ω C} = \sqrt{\frac{5}{3}} \times X_{C}$
$\frac{X_{C}}{R} = \sqrt{\frac{3}{5}}$

AP EAMCET-25.08.2021

Alternating Current

155038 If $120 V, 60 Hz$ Ac voltage is applied to an $LR$ having $R = 100 Ω$ and $L = 2 H$ , then current in the circuit is

1

0.32 A

2

0.16 A

3

0.48 A

4

0.80 A

Explanation:

B Given that, $V = 120 V, f = 60 Hz, R = 100 Ω$ , $L = 2 H$
Impedance of LR circuit,
$Z = \sqrt{R^{2} + {(X_{L})}^{2}}$
$Z = \sqrt{(100)^{2} + (2 π \times 60 \times 2)^{2}}$
$Z = \sqrt{(100)^{2} + (240 π)^{2}}$
$Z = 760.58$
$I = \frac{V}{Z} = \frac{120}{760.58}$
$I = 0.157 \approx 0.16 A$

AP EAMCET-05.10.2021

Alternating Current

155039 A coil of resistance $50 Ω$ is connected across a $5.0 V$ battery. If the current in the coil is found to be $50 mA$ after time $t = 0.1$ battery is connected, then the inductance of the coil is

1

\frac{5}{\ln (2)}

2

10 \ln (2)

3

5 e^{4}

4

\frac{10}{e^{4}}

Explanation:

A Given, $R = 50 Ω, E = 5.0 V, i = 50 mA = 50 \times$ $10^{- 3} A, t = 0.1$
$i = i_{0} (1 - e^{- t R / L})$
$50 \times 10^{- 3} = \frac{5}{50} (1 - e^{- 0.1 \times \frac{50}{L}})$
$10 \times 50 \times 10^{- 3} = 1 - e^{- \frac{5}{L}}$
$e^{- 5 / L} = 1 - 50 \times 10^{- 2}$
$e^{- 5 / L} = \frac{1}{2}$
$\frac{- 5}{L} = \log (\frac{1}{2})$
$L = \frac{5}{\log (2)}$

TS EAMCET 05.08.2021

Alternating Current

155032 In an oscillating $L C$ circuit, the value of inductance is $1.6 mH$ and the value of capacitance is $4 μ F$ . If the maximum charge on the capacitor is $4 \times 10^{- 6} C$ , then the maximum current is

1

75 mA

2

12.5 mA

3

125 mA

4

50 mA

Explanation:

D Given,
$L = 1.6 mH = 1.6 \times 10^{- 3} H, C = 4 μ F = 4 \times 10^{- 6} F$ $Q = 4 \times 10^{- 6}$
Energy stored in capacitor $(U_{C}) = \frac{1}{2} \frac{Q^{2}}{C}$
Energy stored in $Inductor (U_{L}) = \frac{1}{2} {LI}^{2}$
$U_{C} = U_{L}$
$\frac{1}{2} \frac{Q^{2}}{C} = \frac{1}{2} {LI}^{2}$
$I^{2} = \frac{Q^{2}}{L C}$
$I = \frac{Q}{\sqrt{LC}} = \frac{4 \times 10^{- 6}}{\sqrt{1.6 \times 10^{- 3} \times 4 \times 10^{- 6}}}$
$= \frac{4 \times 10^{- 6}}{8 \times 10^{- 5}}$
$= 5 \times 10^{- 2} = 50 \times 10^{- 3} = 50 mA$

AP EAMCET-11.07.2022

Alternating Current

155035
Consider a pure inductive A.C. circuit as shown in the figure. If the average power consumed is $P$ , then

1

P > 0

2

P < 0

3

P = 0

4

P

is infinite

Explanation:

C In pure inductor current is tog by $90^{\circ}$ with voltage.
$∴$ average power dissipation $P$
$P = \frac{V_{0} I_{0}}{\sqrt{2} \times \sqrt{2}} \cos ϕ$
$P = \frac{V_{0} I_{0}}{2} \cos 90^{\circ}$
$P = 0$

WB JEE 2021

Alternating Current

155036 An ac source of angular frequency $ω$ is fed across a resistor $R$ and a capacitor $C$ in series. The current flowing in the circuit found to be ' $I$ ' now the frequency of the source is changed to $\frac{ω}{3}$ . (Maintaining the same voltage) the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?

1

\sqrt{\frac{5}{7}}

2

\sqrt{\frac{3}{4}}

3

\sqrt{\frac{3}{5}}

4

\sqrt{\frac{7}{5}}

Explanation:

C Given, For RC circuit,
$I = \frac{V_{rms}}{\sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}} = \frac{V_{rms}}{\sqrt{R^{2} + \frac{1}{ω^{2} C^{2}}}}$
$I = \frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}}$
When angular frequency of the source is $\frac{ω}{3}$ Then current becomes $\frac{I}{2}$ . Then,
$\frac{I}{2} = \frac{V_{rms} \frac{ω}{3} C}{\sqrt{R^{2} {(\frac{ω}{3})}^{2} C^{2} + 1}}$
$\frac{I}{2} = \frac{V_{rms} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 9}}$
Dividing equation (i) by (ii) -
$\frac{I}{I / 2} = \frac{\frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}}}{\frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 9}}}$
$2 = \frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}} \times \frac{\sqrt{R^{2} ω^{2} C^{2} + 9}}{V_{r m s} ω C}$
$2 = \frac{\sqrt{R^{2} ω^{2} C^{2} + 9}}{\sqrt{R^{2} ω^{2} C^{2} + 1}}$
Squaring both sides,
$4 = \frac{R^{2} ω^{2} C^{2} + 9}{R^{2} ω^{2} C^{2} + 1}$
4 $R^{2} ω^{2} C^{2} + 4 = R^{2} ω^{2} C^{2} + 9$
3 $R^{2} ω^{2} C^{2} = 5$
$R^{2} ω^{2} C^{2} = \frac{5}{3}$
$R = \sqrt{\frac{5}{3} \frac{1}{ω^{2} C^{2}}} = \sqrt{\frac{5}{3}} \times \frac{1}{ω C} = \sqrt{\frac{5}{3}} \times X_{C}$
$\frac{X_{C}}{R} = \sqrt{\frac{3}{5}}$

AP EAMCET-25.08.2021

Alternating Current

155038 If $120 V, 60 Hz$ Ac voltage is applied to an $LR$ having $R = 100 Ω$ and $L = 2 H$ , then current in the circuit is

1

0.32 A

2

0.16 A

3

0.48 A

4

0.80 A

Explanation:

B Given that, $V = 120 V, f = 60 Hz, R = 100 Ω$ , $L = 2 H$
Impedance of LR circuit,
$Z = \sqrt{R^{2} + {(X_{L})}^{2}}$
$Z = \sqrt{(100)^{2} + (2 π \times 60 \times 2)^{2}}$
$Z = \sqrt{(100)^{2} + (240 π)^{2}}$
$Z = 760.58$
$I = \frac{V}{Z} = \frac{120}{760.58}$
$I = 0.157 \approx 0.16 A$

AP EAMCET-05.10.2021

Alternating Current

155039 A coil of resistance $50 Ω$ is connected across a $5.0 V$ battery. If the current in the coil is found to be $50 mA$ after time $t = 0.1$ battery is connected, then the inductance of the coil is

1

\frac{5}{\ln (2)}

2

10 \ln (2)

3

5 e^{4}

4

\frac{10}{e^{4}}

Explanation:

A Given, $R = 50 Ω, E = 5.0 V, i = 50 mA = 50 \times$ $10^{- 3} A, t = 0.1$
$i = i_{0} (1 - e^{- t R / L})$
$50 \times 10^{- 3} = \frac{5}{50} (1 - e^{- 0.1 \times \frac{50}{L}})$
$10 \times 50 \times 10^{- 3} = 1 - e^{- \frac{5}{L}}$
$e^{- 5 / L} = 1 - 50 \times 10^{- 2}$
$e^{- 5 / L} = \frac{1}{2}$
$\frac{- 5}{L} = \log (\frac{1}{2})$
$L = \frac{5}{\log (2)}$

TS EAMCET 05.08.2021

Alternating Current

155032 In an oscillating $L C$ circuit, the value of inductance is $1.6 mH$ and the value of capacitance is $4 μ F$ . If the maximum charge on the capacitor is $4 \times 10^{- 6} C$ , then the maximum current is

1

75 mA

2

12.5 mA

3

125 mA

4

50 mA

Explanation:

D Given,
$L = 1.6 mH = 1.6 \times 10^{- 3} H, C = 4 μ F = 4 \times 10^{- 6} F$ $Q = 4 \times 10^{- 6}$
Energy stored in capacitor $(U_{C}) = \frac{1}{2} \frac{Q^{2}}{C}$
Energy stored in $Inductor (U_{L}) = \frac{1}{2} {LI}^{2}$
$U_{C} = U_{L}$
$\frac{1}{2} \frac{Q^{2}}{C} = \frac{1}{2} {LI}^{2}$
$I^{2} = \frac{Q^{2}}{L C}$
$I = \frac{Q}{\sqrt{LC}} = \frac{4 \times 10^{- 6}}{\sqrt{1.6 \times 10^{- 3} \times 4 \times 10^{- 6}}}$
$= \frac{4 \times 10^{- 6}}{8 \times 10^{- 5}}$
$= 5 \times 10^{- 2} = 50 \times 10^{- 3} = 50 mA$

AP EAMCET-11.07.2022

Alternating Current

155035
Consider a pure inductive A.C. circuit as shown in the figure. If the average power consumed is $P$ , then

1

P > 0

2

P < 0

3

P = 0

4

P

is infinite

Explanation:

C In pure inductor current is tog by $90^{\circ}$ with voltage.
$∴$ average power dissipation $P$
$P = \frac{V_{0} I_{0}}{\sqrt{2} \times \sqrt{2}} \cos ϕ$
$P = \frac{V_{0} I_{0}}{2} \cos 90^{\circ}$
$P = 0$

WB JEE 2021

Alternating Current

155036 An ac source of angular frequency $ω$ is fed across a resistor $R$ and a capacitor $C$ in series. The current flowing in the circuit found to be ' $I$ ' now the frequency of the source is changed to $\frac{ω}{3}$ . (Maintaining the same voltage) the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?

1

\sqrt{\frac{5}{7}}

2

\sqrt{\frac{3}{4}}

3

\sqrt{\frac{3}{5}}

4

\sqrt{\frac{7}{5}}

Explanation:

C Given, For RC circuit,
$I = \frac{V_{rms}}{\sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}} = \frac{V_{rms}}{\sqrt{R^{2} + \frac{1}{ω^{2} C^{2}}}}$
$I = \frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}}$
When angular frequency of the source is $\frac{ω}{3}$ Then current becomes $\frac{I}{2}$ . Then,
$\frac{I}{2} = \frac{V_{rms} \frac{ω}{3} C}{\sqrt{R^{2} {(\frac{ω}{3})}^{2} C^{2} + 1}}$
$\frac{I}{2} = \frac{V_{rms} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 9}}$
Dividing equation (i) by (ii) -
$\frac{I}{I / 2} = \frac{\frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}}}{\frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 9}}}$
$2 = \frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}} \times \frac{\sqrt{R^{2} ω^{2} C^{2} + 9}}{V_{r m s} ω C}$
$2 = \frac{\sqrt{R^{2} ω^{2} C^{2} + 9}}{\sqrt{R^{2} ω^{2} C^{2} + 1}}$
Squaring both sides,
$4 = \frac{R^{2} ω^{2} C^{2} + 9}{R^{2} ω^{2} C^{2} + 1}$
4 $R^{2} ω^{2} C^{2} + 4 = R^{2} ω^{2} C^{2} + 9$
3 $R^{2} ω^{2} C^{2} = 5$
$R^{2} ω^{2} C^{2} = \frac{5}{3}$
$R = \sqrt{\frac{5}{3} \frac{1}{ω^{2} C^{2}}} = \sqrt{\frac{5}{3}} \times \frac{1}{ω C} = \sqrt{\frac{5}{3}} \times X_{C}$
$\frac{X_{C}}{R} = \sqrt{\frac{3}{5}}$

AP EAMCET-25.08.2021

Alternating Current

155038 If $120 V, 60 Hz$ Ac voltage is applied to an $LR$ having $R = 100 Ω$ and $L = 2 H$ , then current in the circuit is

1

0.32 A

2

0.16 A

3

0.48 A

4

0.80 A

Explanation:

B Given that, $V = 120 V, f = 60 Hz, R = 100 Ω$ , $L = 2 H$
Impedance of LR circuit,
$Z = \sqrt{R^{2} + {(X_{L})}^{2}}$
$Z = \sqrt{(100)^{2} + (2 π \times 60 \times 2)^{2}}$
$Z = \sqrt{(100)^{2} + (240 π)^{2}}$
$Z = 760.58$
$I = \frac{V}{Z} = \frac{120}{760.58}$
$I = 0.157 \approx 0.16 A$

AP EAMCET-05.10.2021

Alternating Current

155039 A coil of resistance $50 Ω$ is connected across a $5.0 V$ battery. If the current in the coil is found to be $50 mA$ after time $t = 0.1$ battery is connected, then the inductance of the coil is

1

\frac{5}{\ln (2)}

2

10 \ln (2)

3

5 e^{4}

4

\frac{10}{e^{4}}

Explanation:

A Given, $R = 50 Ω, E = 5.0 V, i = 50 mA = 50 \times$ $10^{- 3} A, t = 0.1$
$i = i_{0} (1 - e^{- t R / L})$
$50 \times 10^{- 3} = \frac{5}{50} (1 - e^{- 0.1 \times \frac{50}{L}})$
$10 \times 50 \times 10^{- 3} = 1 - e^{- \frac{5}{L}}$
$e^{- 5 / L} = 1 - 50 \times 10^{- 2}$
$e^{- 5 / L} = \frac{1}{2}$
$\frac{- 5}{L} = \log (\frac{1}{2})$
$L = \frac{5}{\log (2)}$

TS EAMCET 05.08.2021

Alternating Current

155032 In an oscillating $L C$ circuit, the value of inductance is $1.6 mH$ and the value of capacitance is $4 μ F$ . If the maximum charge on the capacitor is $4 \times 10^{- 6} C$ , then the maximum current is

1

75 mA

2

12.5 mA

3

125 mA

4

50 mA

Explanation:

D Given,
$L = 1.6 mH = 1.6 \times 10^{- 3} H, C = 4 μ F = 4 \times 10^{- 6} F$ $Q = 4 \times 10^{- 6}$
Energy stored in capacitor $(U_{C}) = \frac{1}{2} \frac{Q^{2}}{C}$
Energy stored in $Inductor (U_{L}) = \frac{1}{2} {LI}^{2}$
$U_{C} = U_{L}$
$\frac{1}{2} \frac{Q^{2}}{C} = \frac{1}{2} {LI}^{2}$
$I^{2} = \frac{Q^{2}}{L C}$
$I = \frac{Q}{\sqrt{LC}} = \frac{4 \times 10^{- 6}}{\sqrt{1.6 \times 10^{- 3} \times 4 \times 10^{- 6}}}$
$= \frac{4 \times 10^{- 6}}{8 \times 10^{- 5}}$
$= 5 \times 10^{- 2} = 50 \times 10^{- 3} = 50 mA$

AP EAMCET-11.07.2022

Alternating Current

155035
Consider a pure inductive A.C. circuit as shown in the figure. If the average power consumed is $P$ , then

1

P > 0

2

P < 0

3

P = 0

4

P

is infinite

Explanation:

C In pure inductor current is tog by $90^{\circ}$ with voltage.
$∴$ average power dissipation $P$
$P = \frac{V_{0} I_{0}}{\sqrt{2} \times \sqrt{2}} \cos ϕ$
$P = \frac{V_{0} I_{0}}{2} \cos 90^{\circ}$
$P = 0$

WB JEE 2021

Alternating Current

155036 An ac source of angular frequency $ω$ is fed across a resistor $R$ and a capacitor $C$ in series. The current flowing in the circuit found to be ' $I$ ' now the frequency of the source is changed to $\frac{ω}{3}$ . (Maintaining the same voltage) the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?

1

\sqrt{\frac{5}{7}}

2

\sqrt{\frac{3}{4}}

3

\sqrt{\frac{3}{5}}

4

\sqrt{\frac{7}{5}}

Explanation:

C Given, For RC circuit,
$I = \frac{V_{rms}}{\sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}} = \frac{V_{rms}}{\sqrt{R^{2} + \frac{1}{ω^{2} C^{2}}}}$
$I = \frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}}$
When angular frequency of the source is $\frac{ω}{3}$ Then current becomes $\frac{I}{2}$ . Then,
$\frac{I}{2} = \frac{V_{rms} \frac{ω}{3} C}{\sqrt{R^{2} {(\frac{ω}{3})}^{2} C^{2} + 1}}$
$\frac{I}{2} = \frac{V_{rms} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 9}}$
Dividing equation (i) by (ii) -
$\frac{I}{I / 2} = \frac{\frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}}}{\frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 9}}}$
$2 = \frac{V_{r m s} ω C}{\sqrt{R^{2} ω^{2} C^{2} + 1}} \times \frac{\sqrt{R^{2} ω^{2} C^{2} + 9}}{V_{r m s} ω C}$
$2 = \frac{\sqrt{R^{2} ω^{2} C^{2} + 9}}{\sqrt{R^{2} ω^{2} C^{2} + 1}}$
Squaring both sides,
$4 = \frac{R^{2} ω^{2} C^{2} + 9}{R^{2} ω^{2} C^{2} + 1}$
4 $R^{2} ω^{2} C^{2} + 4 = R^{2} ω^{2} C^{2} + 9$
3 $R^{2} ω^{2} C^{2} = 5$
$R^{2} ω^{2} C^{2} = \frac{5}{3}$
$R = \sqrt{\frac{5}{3} \frac{1}{ω^{2} C^{2}}} = \sqrt{\frac{5}{3}} \times \frac{1}{ω C} = \sqrt{\frac{5}{3}} \times X_{C}$
$\frac{X_{C}}{R} = \sqrt{\frac{3}{5}}$

AP EAMCET-25.08.2021

Alternating Current

155038 If $120 V, 60 Hz$ Ac voltage is applied to an $LR$ having $R = 100 Ω$ and $L = 2 H$ , then current in the circuit is

1

0.32 A

2

0.16 A

3

0.48 A

4

0.80 A

Explanation:

B Given that, $V = 120 V, f = 60 Hz, R = 100 Ω$ , $L = 2 H$
Impedance of LR circuit,
$Z = \sqrt{R^{2} + {(X_{L})}^{2}}$
$Z = \sqrt{(100)^{2} + (2 π \times 60 \times 2)^{2}}$
$Z = \sqrt{(100)^{2} + (240 π)^{2}}$
$Z = 760.58$
$I = \frac{V}{Z} = \frac{120}{760.58}$
$I = 0.157 \approx 0.16 A$

AP EAMCET-05.10.2021

Alternating Current

155039 A coil of resistance $50 Ω$ is connected across a $5.0 V$ battery. If the current in the coil is found to be $50 mA$ after time $t = 0.1$ battery is connected, then the inductance of the coil is

1

\frac{5}{\ln (2)}

2

10 \ln (2)

3

5 e^{4}

4

\frac{10}{e^{4}}

Explanation:

A Given, $R = 50 Ω, E = 5.0 V, i = 50 mA = 50 \times$ $10^{- 3} A, t = 0.1$
$i = i_{0} (1 - e^{- t R / L})$
$50 \times 10^{- 3} = \frac{5}{50} (1 - e^{- 0.1 \times \frac{50}{L}})$
$10 \times 50 \times 10^{- 3} = 1 - e^{- \frac{5}{L}}$
$e^{- 5 / L} = 1 - 50 \times 10^{- 2}$
$e^{- 5 / L} = \frac{1}{2}$
$\frac{- 5}{L} = \log (\frac{1}{2})$
$L = \frac{5}{\log (2)}$

TS EAMCET 05.08.2021

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