QBManager

PHXII07:ALTERNATING CURRENT

356166 The average power dissipated in pure inductor is

1

V I^{2}

2

z e r o

3

\frac{1}{2} V I

4

\frac{V I^{2}}{4}

Explanation:

The average power dissipated in a pure inductor is zero, as the phase angle between voltage and current is always $π / 2$

PHXII07:ALTERNATING CURRENT

356167 An $a c$ supply gives $(30 V)_{r, m . s}$ which passed through a $100 Ω$ resistance. The power dissipated in it is:

1

30 W

2

90 W

3

40 W

4

9 W

Explanation:

$P = (\frac{V_{r m s}^{2}}{R}) = \frac{{(30)}^{2}}{100} = \frac{900}{100} = 9 W$
So, correct option is (4).

PHXII07:ALTERNATING CURRENT

356168 Power factor value is one for

1 Pure inductor

2 Pure resistor

3 Either an inductor or a capacitor

4 Pure capacitor

PHXII07:ALTERNATING CURRENT

356169 A group of electric lamps having a total power rating of 1000 watt is supplied by an $A C$ voltage $E = 200 \sin (3 10 t + 60^{\circ}) .$ Then the rms value of the circuit current is[ Here $i = i_{o} \sin 310 t$ ]

1

20 \sqrt{2} A

2

10 A

3

20 A

4

10 \sqrt{2} A

Explanation:

$P = \frac{1}{2} V_{0} i_{0} \cos ϕ \Rightarrow 1000 = \frac{1}{2} \times 200 \times i_{0} \cos 60^{\circ}$
$\Rightarrow i_{0} = 20 A \Rightarrow i_{r m s} = \frac{i_{0}}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10 \sqrt{2} A .$

PHXII07:ALTERNATING CURRENT

356166 The average power dissipated in pure inductor is

1

V I^{2}

2

z e r o

3

\frac{1}{2} V I

4

\frac{V I^{2}}{4}

Explanation:

The average power dissipated in a pure inductor is zero, as the phase angle between voltage and current is always $π / 2$

PHXII07:ALTERNATING CURRENT

356167 An $a c$ supply gives $(30 V)_{r, m . s}$ which passed through a $100 Ω$ resistance. The power dissipated in it is:

1

30 W

2

90 W

3

40 W

4

9 W

Explanation:

$P = (\frac{V_{r m s}^{2}}{R}) = \frac{{(30)}^{2}}{100} = \frac{900}{100} = 9 W$
So, correct option is (4).

PHXII07:ALTERNATING CURRENT

356168 Power factor value is one for

1 Pure inductor

2 Pure resistor

3 Either an inductor or a capacitor

4 Pure capacitor

PHXII07:ALTERNATING CURRENT

356169 A group of electric lamps having a total power rating of 1000 watt is supplied by an $A C$ voltage $E = 200 \sin (3 10 t + 60^{\circ}) .$ Then the rms value of the circuit current is[ Here $i = i_{o} \sin 310 t$ ]

1

20 \sqrt{2} A

2

10 A

3

20 A

4

10 \sqrt{2} A

Explanation:

$P = \frac{1}{2} V_{0} i_{0} \cos ϕ \Rightarrow 1000 = \frac{1}{2} \times 200 \times i_{0} \cos 60^{\circ}$
$\Rightarrow i_{0} = 20 A \Rightarrow i_{r m s} = \frac{i_{0}}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10 \sqrt{2} A .$

PHXII07:ALTERNATING CURRENT

356166 The average power dissipated in pure inductor is

1

V I^{2}

2

z e r o

3

\frac{1}{2} V I

4

\frac{V I^{2}}{4}

Explanation:

The average power dissipated in a pure inductor is zero, as the phase angle between voltage and current is always $π / 2$

PHXII07:ALTERNATING CURRENT

356167 An $a c$ supply gives $(30 V)_{r, m . s}$ which passed through a $100 Ω$ resistance. The power dissipated in it is:

1

30 W

2

90 W

3

40 W

4

9 W

Explanation:

$P = (\frac{V_{r m s}^{2}}{R}) = \frac{{(30)}^{2}}{100} = \frac{900}{100} = 9 W$
So, correct option is (4).

PHXII07:ALTERNATING CURRENT

356168 Power factor value is one for

1 Pure inductor

2 Pure resistor

3 Either an inductor or a capacitor

4 Pure capacitor

PHXII07:ALTERNATING CURRENT

356169 A group of electric lamps having a total power rating of 1000 watt is supplied by an $A C$ voltage $E = 200 \sin (3 10 t + 60^{\circ}) .$ Then the rms value of the circuit current is[ Here $i = i_{o} \sin 310 t$ ]

1

20 \sqrt{2} A

2

10 A

3

20 A

4

10 \sqrt{2} A

Explanation:

$P = \frac{1}{2} V_{0} i_{0} \cos ϕ \Rightarrow 1000 = \frac{1}{2} \times 200 \times i_{0} \cos 60^{\circ}$
$\Rightarrow i_{0} = 20 A \Rightarrow i_{r m s} = \frac{i_{0}}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10 \sqrt{2} A .$

PHXII07:ALTERNATING CURRENT

356166 The average power dissipated in pure inductor is

1

V I^{2}

2

z e r o

3

\frac{1}{2} V I

4

\frac{V I^{2}}{4}

Explanation:

The average power dissipated in a pure inductor is zero, as the phase angle between voltage and current is always $π / 2$

PHXII07:ALTERNATING CURRENT

356167 An $a c$ supply gives $(30 V)_{r, m . s}$ which passed through a $100 Ω$ resistance. The power dissipated in it is:

1

30 W

2

90 W

3

40 W

4

9 W

Explanation:

$P = (\frac{V_{r m s}^{2}}{R}) = \frac{{(30)}^{2}}{100} = \frac{900}{100} = 9 W$
So, correct option is (4).

PHXII07:ALTERNATING CURRENT

356168 Power factor value is one for

1 Pure inductor

2 Pure resistor

3 Either an inductor or a capacitor

4 Pure capacitor

PHXII07:ALTERNATING CURRENT

356169 A group of electric lamps having a total power rating of 1000 watt is supplied by an $A C$ voltage $E = 200 \sin (3 10 t + 60^{\circ}) .$ Then the rms value of the circuit current is[ Here $i = i_{o} \sin 310 t$ ]

1

20 \sqrt{2} A

2

10 A

3

20 A

4

10 \sqrt{2} A

Explanation:

$P = \frac{1}{2} V_{0} i_{0} \cos ϕ \Rightarrow 1000 = \frac{1}{2} \times 200 \times i_{0} \cos 60^{\circ}$
$\Rightarrow i_{0} = 20 A \Rightarrow i_{r m s} = \frac{i_{0}}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10 \sqrt{2} A .$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series