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Alternating Current

155053 An L-R circuit has a cell of emf. $E$ , which is switched $O N$ a time $t = 0$ . The current in the circuit after a long time will be

1 zero

2

\frac{E}{R}

3

\frac{E}{L}

4

\frac{E}{\sqrt{L^{2} + R^{2}}}

Explanation:

B
In L-R circuit has a cell of emf $E$ . The growth of current in L-R circuit, the current in the circuit grows exponentially with the time $t = 0$ . Then maximum value of current,
$I_{0} = E / R$

CG PET 2019

Alternating Current

155089 In a $R - L$ circuit reactance offered by the coil is

1

ω L

2

\frac{1}{ω L}

3

ω RL

4

ω^{2} L

Explanation:

A In RL circuit,
Inductive reactance $X_{L} = 2 π fL$
$X_{L} = ω L$

J and K CET- 1998

Alternating Current

155054 An ac voltage of $100 V, \frac{500}{π} Hz$ is connected across a $20 Ω$ resistor and $15 mH$ inductor in series. Then the impedance of the circuit is

1

25 Ω

2

40 Ω

3

5 Ω

4

35 Ω

Explanation:

A Given, $V = 100 V, f = \frac{500}{π} Hz, R = 20 Ω, L =$ $15 mH = 15 \times 10^{- 3} H$
Impedance of the circuit $(Z) = \sqrt{R^{2} + ω^{2} L^{2}}$
$= \sqrt{(20)^{2} + (2 π f)^{2} \times {(15 \times 10^{- 3})}^{2}}$
$= \sqrt{400 + {(2 π \times \frac{500}{π})}^{2} \times {(15 \times 10^{- 3})}^{2}}$
$= \sqrt{400 + (1000)^{2} \times {(15 \times 10^{- 3})}^{2}} = 25 Ω$

AP EAMCET-25.04.2018

Alternating Current

155057 A charged capacitor $C = 30 μ F$ is connected to an inductor $L = 27 mH$ . The angular frequency of their oscillations is-

1

9.1 \times 10^{3} rad / s

2

3 \times 10^{3} rad / s

3

1.1 \times 10^{3} rad / s

4

0.3 \times 10^{3} rad / s

Explanation:

C Given that, $C = 30 μ F = 30 \times 10^{- 6} F$ $L = 27 mH = 27 \times 10^{- 3} H$
Angular frequency of the oscillation is,
$ω = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{27 \times 10^{- 3} \times 30 \times 10^{- 6}}}$
$= \frac{1}{9 \times 10^{- 4}} = 1.11 \times 10^{3} rad / \sec$

BCECE-2018

Alternating Current

155058 A capacitor ' $C$ ' connected across a D.C. source, the reactance of capacitor will be

1 Zero

2 High

3 Low

4 Infinite

Explanation:

D Reactance of capacitance $X_{C} = \frac{1}{2 π f C}$
Where, $f = 0$ for $DC$ source
$X_{C} = \infty$

GUJCET 2018

Alternating Current

155053 An L-R circuit has a cell of emf. $E$ , which is switched $O N$ a time $t = 0$ . The current in the circuit after a long time will be

1 zero

2

\frac{E}{R}

3

\frac{E}{L}

4

\frac{E}{\sqrt{L^{2} + R^{2}}}

Explanation:

B
In L-R circuit has a cell of emf $E$ . The growth of current in L-R circuit, the current in the circuit grows exponentially with the time $t = 0$ . Then maximum value of current,
$I_{0} = E / R$

CG PET 2019

Alternating Current

155089 In a $R - L$ circuit reactance offered by the coil is

1

ω L

2

\frac{1}{ω L}

3

ω RL

4

ω^{2} L

Explanation:

A In RL circuit,
Inductive reactance $X_{L} = 2 π fL$
$X_{L} = ω L$

J and K CET- 1998

Alternating Current

155054 An ac voltage of $100 V, \frac{500}{π} Hz$ is connected across a $20 Ω$ resistor and $15 mH$ inductor in series. Then the impedance of the circuit is

1

25 Ω

2

40 Ω

3

5 Ω

4

35 Ω

Explanation:

A Given, $V = 100 V, f = \frac{500}{π} Hz, R = 20 Ω, L =$ $15 mH = 15 \times 10^{- 3} H$
Impedance of the circuit $(Z) = \sqrt{R^{2} + ω^{2} L^{2}}$
$= \sqrt{(20)^{2} + (2 π f)^{2} \times {(15 \times 10^{- 3})}^{2}}$
$= \sqrt{400 + {(2 π \times \frac{500}{π})}^{2} \times {(15 \times 10^{- 3})}^{2}}$
$= \sqrt{400 + (1000)^{2} \times {(15 \times 10^{- 3})}^{2}} = 25 Ω$

AP EAMCET-25.04.2018

Alternating Current

155057 A charged capacitor $C = 30 μ F$ is connected to an inductor $L = 27 mH$ . The angular frequency of their oscillations is-

1

9.1 \times 10^{3} rad / s

2

3 \times 10^{3} rad / s

3

1.1 \times 10^{3} rad / s

4

0.3 \times 10^{3} rad / s

Explanation:

C Given that, $C = 30 μ F = 30 \times 10^{- 6} F$ $L = 27 mH = 27 \times 10^{- 3} H$
Angular frequency of the oscillation is,
$ω = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{27 \times 10^{- 3} \times 30 \times 10^{- 6}}}$
$= \frac{1}{9 \times 10^{- 4}} = 1.11 \times 10^{3} rad / \sec$

BCECE-2018

Alternating Current

155058 A capacitor ' $C$ ' connected across a D.C. source, the reactance of capacitor will be

1 Zero

2 High

3 Low

4 Infinite

Explanation:

D Reactance of capacitance $X_{C} = \frac{1}{2 π f C}$
Where, $f = 0$ for $DC$ source
$X_{C} = \infty$

GUJCET 2018

Alternating Current

155053 An L-R circuit has a cell of emf. $E$ , which is switched $O N$ a time $t = 0$ . The current in the circuit after a long time will be

1 zero

2

\frac{E}{R}

3

\frac{E}{L}

4

\frac{E}{\sqrt{L^{2} + R^{2}}}

Explanation:

B
In L-R circuit has a cell of emf $E$ . The growth of current in L-R circuit, the current in the circuit grows exponentially with the time $t = 0$ . Then maximum value of current,
$I_{0} = E / R$

CG PET 2019

Alternating Current

155089 In a $R - L$ circuit reactance offered by the coil is

1

ω L

2

\frac{1}{ω L}

3

ω RL

4

ω^{2} L

Explanation:

A In RL circuit,
Inductive reactance $X_{L} = 2 π fL$
$X_{L} = ω L$

J and K CET- 1998

Alternating Current

155054 An ac voltage of $100 V, \frac{500}{π} Hz$ is connected across a $20 Ω$ resistor and $15 mH$ inductor in series. Then the impedance of the circuit is

1

25 Ω

2

40 Ω

3

5 Ω

4

35 Ω

Explanation:

A Given, $V = 100 V, f = \frac{500}{π} Hz, R = 20 Ω, L =$ $15 mH = 15 \times 10^{- 3} H$
Impedance of the circuit $(Z) = \sqrt{R^{2} + ω^{2} L^{2}}$
$= \sqrt{(20)^{2} + (2 π f)^{2} \times {(15 \times 10^{- 3})}^{2}}$
$= \sqrt{400 + {(2 π \times \frac{500}{π})}^{2} \times {(15 \times 10^{- 3})}^{2}}$
$= \sqrt{400 + (1000)^{2} \times {(15 \times 10^{- 3})}^{2}} = 25 Ω$

AP EAMCET-25.04.2018

Alternating Current

155057 A charged capacitor $C = 30 μ F$ is connected to an inductor $L = 27 mH$ . The angular frequency of their oscillations is-

1

9.1 \times 10^{3} rad / s

2

3 \times 10^{3} rad / s

3

1.1 \times 10^{3} rad / s

4

0.3 \times 10^{3} rad / s

Explanation:

C Given that, $C = 30 μ F = 30 \times 10^{- 6} F$ $L = 27 mH = 27 \times 10^{- 3} H$
Angular frequency of the oscillation is,
$ω = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{27 \times 10^{- 3} \times 30 \times 10^{- 6}}}$
$= \frac{1}{9 \times 10^{- 4}} = 1.11 \times 10^{3} rad / \sec$

BCECE-2018

Alternating Current

155058 A capacitor ' $C$ ' connected across a D.C. source, the reactance of capacitor will be

1 Zero

2 High

3 Low

4 Infinite

Explanation:

D Reactance of capacitance $X_{C} = \frac{1}{2 π f C}$
Where, $f = 0$ for $DC$ source
$X_{C} = \infty$

GUJCET 2018

Alternating Current

155053 An L-R circuit has a cell of emf. $E$ , which is switched $O N$ a time $t = 0$ . The current in the circuit after a long time will be

1 zero

2

\frac{E}{R}

3

\frac{E}{L}

4

\frac{E}{\sqrt{L^{2} + R^{2}}}

Explanation:

B
In L-R circuit has a cell of emf $E$ . The growth of current in L-R circuit, the current in the circuit grows exponentially with the time $t = 0$ . Then maximum value of current,
$I_{0} = E / R$

CG PET 2019

Alternating Current

155089 In a $R - L$ circuit reactance offered by the coil is

1

ω L

2

\frac{1}{ω L}

3

ω RL

4

ω^{2} L

Explanation:

A In RL circuit,
Inductive reactance $X_{L} = 2 π fL$
$X_{L} = ω L$

J and K CET- 1998

Alternating Current

155054 An ac voltage of $100 V, \frac{500}{π} Hz$ is connected across a $20 Ω$ resistor and $15 mH$ inductor in series. Then the impedance of the circuit is

1

25 Ω

2

40 Ω

3

5 Ω

4

35 Ω

Explanation:

A Given, $V = 100 V, f = \frac{500}{π} Hz, R = 20 Ω, L =$ $15 mH = 15 \times 10^{- 3} H$
Impedance of the circuit $(Z) = \sqrt{R^{2} + ω^{2} L^{2}}$
$= \sqrt{(20)^{2} + (2 π f)^{2} \times {(15 \times 10^{- 3})}^{2}}$
$= \sqrt{400 + {(2 π \times \frac{500}{π})}^{2} \times {(15 \times 10^{- 3})}^{2}}$
$= \sqrt{400 + (1000)^{2} \times {(15 \times 10^{- 3})}^{2}} = 25 Ω$

AP EAMCET-25.04.2018

Alternating Current

155057 A charged capacitor $C = 30 μ F$ is connected to an inductor $L = 27 mH$ . The angular frequency of their oscillations is-

1

9.1 \times 10^{3} rad / s

2

3 \times 10^{3} rad / s

3

1.1 \times 10^{3} rad / s

4

0.3 \times 10^{3} rad / s

Explanation:

C Given that, $C = 30 μ F = 30 \times 10^{- 6} F$ $L = 27 mH = 27 \times 10^{- 3} H$
Angular frequency of the oscillation is,
$ω = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{27 \times 10^{- 3} \times 30 \times 10^{- 6}}}$
$= \frac{1}{9 \times 10^{- 4}} = 1.11 \times 10^{3} rad / \sec$

BCECE-2018

Alternating Current

155058 A capacitor ' $C$ ' connected across a D.C. source, the reactance of capacitor will be

1 Zero

2 High

3 Low

4 Infinite

Explanation:

D Reactance of capacitance $X_{C} = \frac{1}{2 π f C}$
Where, $f = 0$ for $DC$ source
$X_{C} = \infty$

GUJCET 2018

Alternating Current

155053 An L-R circuit has a cell of emf. $E$ , which is switched $O N$ a time $t = 0$ . The current in the circuit after a long time will be

1 zero

2

\frac{E}{R}

3

\frac{E}{L}

4

\frac{E}{\sqrt{L^{2} + R^{2}}}

Explanation:

B
In L-R circuit has a cell of emf $E$ . The growth of current in L-R circuit, the current in the circuit grows exponentially with the time $t = 0$ . Then maximum value of current,
$I_{0} = E / R$

CG PET 2019

Alternating Current

155089 In a $R - L$ circuit reactance offered by the coil is

1

ω L

2

\frac{1}{ω L}

3

ω RL

4

ω^{2} L

Explanation:

A In RL circuit,
Inductive reactance $X_{L} = 2 π fL$
$X_{L} = ω L$

J and K CET- 1998

Alternating Current

155054 An ac voltage of $100 V, \frac{500}{π} Hz$ is connected across a $20 Ω$ resistor and $15 mH$ inductor in series. Then the impedance of the circuit is

1

25 Ω

2

40 Ω

3

5 Ω

4

35 Ω

Explanation:

A Given, $V = 100 V, f = \frac{500}{π} Hz, R = 20 Ω, L =$ $15 mH = 15 \times 10^{- 3} H$
Impedance of the circuit $(Z) = \sqrt{R^{2} + ω^{2} L^{2}}$
$= \sqrt{(20)^{2} + (2 π f)^{2} \times {(15 \times 10^{- 3})}^{2}}$
$= \sqrt{400 + {(2 π \times \frac{500}{π})}^{2} \times {(15 \times 10^{- 3})}^{2}}$
$= \sqrt{400 + (1000)^{2} \times {(15 \times 10^{- 3})}^{2}} = 25 Ω$

AP EAMCET-25.04.2018

Alternating Current

155057 A charged capacitor $C = 30 μ F$ is connected to an inductor $L = 27 mH$ . The angular frequency of their oscillations is-

1

9.1 \times 10^{3} rad / s

2

3 \times 10^{3} rad / s

3

1.1 \times 10^{3} rad / s

4

0.3 \times 10^{3} rad / s

Explanation:

C Given that, $C = 30 μ F = 30 \times 10^{- 6} F$ $L = 27 mH = 27 \times 10^{- 3} H$
Angular frequency of the oscillation is,
$ω = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{27 \times 10^{- 3} \times 30 \times 10^{- 6}}}$
$= \frac{1}{9 \times 10^{- 4}} = 1.11 \times 10^{3} rad / \sec$

BCECE-2018

Alternating Current

155058 A capacitor ' $C$ ' connected across a D.C. source, the reactance of capacitor will be

1 Zero

2 High

3 Low

4 Infinite

Explanation:

D Reactance of capacitance $X_{C} = \frac{1}{2 π f C}$
Where, $f = 0$ for $DC$ source
$X_{C} = \infty$

GUJCET 2018