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PHXII07:ALTERNATING CURRENT

356200 The rms current in an ac circuit is $2 A$ . If the wattless current be $\sqrt{3} A$ . What is power factor?

1

\frac{1}{\sqrt{3}}

2

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

3

\frac{1}{2}

4

\frac{1}{3}

Explanation:

$I_{W L} = I_{r m s} \sin ϕ \Rightarrow \sqrt{3} = 2 \sin ϕ$
$\Rightarrow \sin ϕ = \frac{\sqrt{3}}{2} \Rightarrow ϕ = 60^{\circ}$
Power factor, $\cos ϕ = \cos 60^{\circ} = \frac{1}{2}$

PHXII07:ALTERNATING CURRENT

356201 $E m f$ and current in an ac circuit are given by $E = 20 V \sin (314 t)$ and $I = 2 A \sin (314 t - 60^{\circ})$ . Power consumed in the circuit per cycle is

1

40 W

2

20 W

3

10 W

4

5 W

Explanation:

$P_{a v} = \frac{E_{0} I_{0}}{2} \cos ϕ$
$= \frac{20 \times 2}{2} \times \cos 60^{\circ} = 10 W$

PHXII07:ALTERNATING CURRENT

356202 Assertion :
When an $A C$ circuit contains resistor only, its power is minimum.
Reason :
Power of an $A C$ circuit depends on phase angle.

1 Both assertion and reason are correct and reason isthe correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The power of an $A C$ circuit is given by $p = E I \cos ϕ$ where $\cos ϕ$ is power factor and $ϕ$ is phase angle. In case of circuit containing resistance only, phase angle is zero and power factor is equal to one. Therefore power is maximum in case of circuit containing only resistor. So option (4) is correct.

PHXII07:ALTERNATING CURRENT

356203 The potential difference across an instrument in an $A C$ circuit of frequency $f$ is $V$ and the current flowing through it is $i$ such that $V = 5 \cos (2 π f t)$ Volt and $i = 2 \sin (2 π f t) a m p .$ The power disspated in the instrument is

1

10 W

2

5 W

3

2.5 W

4

z e r o

Explanation:

As $V = 5 \cos (2 π f t) = 5 \sin (2 π f t + π / 2)$ and $i = 2 \sin (2 π f t)$
$∴$ Phase difference between $V$ and $i$ is $θ = \frac{π}{2}$
Average power, $P = V_{r m s} i_{r m s} \times \cos θ = 0$

PHXII07:ALTERNATING CURRENT

356204 In a $A C$ circuit the voltage and current are described by $V = 100 (319 t - \frac{π}{6}) v o l t s$ and $i = 200 \sin (319 t + \frac{π}{6}) m A$ respectively. The average power dissipated in the circuit is:

1

2.5 watts

2

5.0 watts

3

10.0 watts

4

50.0 watts

Explanation:

$P = \frac{V_{0} i_{0} \cos ϕ}{2} = \frac{100 \times 200 \times 10^{- 3}}{2} \cos \frac{π}{3} = 5 W$

PHXII07:ALTERNATING CURRENT

356200 The rms current in an ac circuit is $2 A$ . If the wattless current be $\sqrt{3} A$ . What is power factor?

1

\frac{1}{\sqrt{3}}

2

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

3

\frac{1}{2}

4

\frac{1}{3}

Explanation:

$I_{W L} = I_{r m s} \sin ϕ \Rightarrow \sqrt{3} = 2 \sin ϕ$
$\Rightarrow \sin ϕ = \frac{\sqrt{3}}{2} \Rightarrow ϕ = 60^{\circ}$
Power factor, $\cos ϕ = \cos 60^{\circ} = \frac{1}{2}$

PHXII07:ALTERNATING CURRENT

356201 $E m f$ and current in an ac circuit are given by $E = 20 V \sin (314 t)$ and $I = 2 A \sin (314 t - 60^{\circ})$ . Power consumed in the circuit per cycle is

1

40 W

2

20 W

3

10 W

4

5 W

Explanation:

$P_{a v} = \frac{E_{0} I_{0}}{2} \cos ϕ$
$= \frac{20 \times 2}{2} \times \cos 60^{\circ} = 10 W$

PHXII07:ALTERNATING CURRENT

356202 Assertion :
When an $A C$ circuit contains resistor only, its power is minimum.
Reason :
Power of an $A C$ circuit depends on phase angle.

1 Both assertion and reason are correct and reason isthe correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The power of an $A C$ circuit is given by $p = E I \cos ϕ$ where $\cos ϕ$ is power factor and $ϕ$ is phase angle. In case of circuit containing resistance only, phase angle is zero and power factor is equal to one. Therefore power is maximum in case of circuit containing only resistor. So option (4) is correct.

PHXII07:ALTERNATING CURRENT

356203 The potential difference across an instrument in an $A C$ circuit of frequency $f$ is $V$ and the current flowing through it is $i$ such that $V = 5 \cos (2 π f t)$ Volt and $i = 2 \sin (2 π f t) a m p .$ The power disspated in the instrument is

1

10 W

2

5 W

3

2.5 W

4

z e r o

Explanation:

As $V = 5 \cos (2 π f t) = 5 \sin (2 π f t + π / 2)$ and $i = 2 \sin (2 π f t)$
$∴$ Phase difference between $V$ and $i$ is $θ = \frac{π}{2}$
Average power, $P = V_{r m s} i_{r m s} \times \cos θ = 0$

PHXII07:ALTERNATING CURRENT

356204 In a $A C$ circuit the voltage and current are described by $V = 100 (319 t - \frac{π}{6}) v o l t s$ and $i = 200 \sin (319 t + \frac{π}{6}) m A$ respectively. The average power dissipated in the circuit is:

1

2.5 watts

2

5.0 watts

3

10.0 watts

4

50.0 watts

Explanation:

$P = \frac{V_{0} i_{0} \cos ϕ}{2} = \frac{100 \times 200 \times 10^{- 3}}{2} \cos \frac{π}{3} = 5 W$

PHXII07:ALTERNATING CURRENT

356200 The rms current in an ac circuit is $2 A$ . If the wattless current be $\sqrt{3} A$ . What is power factor?

1

\frac{1}{\sqrt{3}}

2

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

3

\frac{1}{2}

4

\frac{1}{3}

Explanation:

$I_{W L} = I_{r m s} \sin ϕ \Rightarrow \sqrt{3} = 2 \sin ϕ$
$\Rightarrow \sin ϕ = \frac{\sqrt{3}}{2} \Rightarrow ϕ = 60^{\circ}$
Power factor, $\cos ϕ = \cos 60^{\circ} = \frac{1}{2}$

PHXII07:ALTERNATING CURRENT

356201 $E m f$ and current in an ac circuit are given by $E = 20 V \sin (314 t)$ and $I = 2 A \sin (314 t - 60^{\circ})$ . Power consumed in the circuit per cycle is

1

40 W

2

20 W

3

10 W

4

5 W

Explanation:

$P_{a v} = \frac{E_{0} I_{0}}{2} \cos ϕ$
$= \frac{20 \times 2}{2} \times \cos 60^{\circ} = 10 W$

PHXII07:ALTERNATING CURRENT

356202 Assertion :
When an $A C$ circuit contains resistor only, its power is minimum.
Reason :
Power of an $A C$ circuit depends on phase angle.

1 Both assertion and reason are correct and reason isthe correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The power of an $A C$ circuit is given by $p = E I \cos ϕ$ where $\cos ϕ$ is power factor and $ϕ$ is phase angle. In case of circuit containing resistance only, phase angle is zero and power factor is equal to one. Therefore power is maximum in case of circuit containing only resistor. So option (4) is correct.

PHXII07:ALTERNATING CURRENT

356203 The potential difference across an instrument in an $A C$ circuit of frequency $f$ is $V$ and the current flowing through it is $i$ such that $V = 5 \cos (2 π f t)$ Volt and $i = 2 \sin (2 π f t) a m p .$ The power disspated in the instrument is

1

10 W

2

5 W

3

2.5 W

4

z e r o

Explanation:

As $V = 5 \cos (2 π f t) = 5 \sin (2 π f t + π / 2)$ and $i = 2 \sin (2 π f t)$
$∴$ Phase difference between $V$ and $i$ is $θ = \frac{π}{2}$
Average power, $P = V_{r m s} i_{r m s} \times \cos θ = 0$

PHXII07:ALTERNATING CURRENT

356204 In a $A C$ circuit the voltage and current are described by $V = 100 (319 t - \frac{π}{6}) v o l t s$ and $i = 200 \sin (319 t + \frac{π}{6}) m A$ respectively. The average power dissipated in the circuit is:

1

2.5 watts

2

5.0 watts

3

10.0 watts

4

50.0 watts

Explanation:

$P = \frac{V_{0} i_{0} \cos ϕ}{2} = \frac{100 \times 200 \times 10^{- 3}}{2} \cos \frac{π}{3} = 5 W$

PHXII07:ALTERNATING CURRENT

356200 The rms current in an ac circuit is $2 A$ . If the wattless current be $\sqrt{3} A$ . What is power factor?

1

\frac{1}{\sqrt{3}}

2

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

3

\frac{1}{2}

4

\frac{1}{3}

Explanation:

$I_{W L} = I_{r m s} \sin ϕ \Rightarrow \sqrt{3} = 2 \sin ϕ$
$\Rightarrow \sin ϕ = \frac{\sqrt{3}}{2} \Rightarrow ϕ = 60^{\circ}$
Power factor, $\cos ϕ = \cos 60^{\circ} = \frac{1}{2}$

PHXII07:ALTERNATING CURRENT

356201 $E m f$ and current in an ac circuit are given by $E = 20 V \sin (314 t)$ and $I = 2 A \sin (314 t - 60^{\circ})$ . Power consumed in the circuit per cycle is

1

40 W

2

20 W

3

10 W

4

5 W

Explanation:

$P_{a v} = \frac{E_{0} I_{0}}{2} \cos ϕ$
$= \frac{20 \times 2}{2} \times \cos 60^{\circ} = 10 W$

PHXII07:ALTERNATING CURRENT

356202 Assertion :
When an $A C$ circuit contains resistor only, its power is minimum.
Reason :
Power of an $A C$ circuit depends on phase angle.

1 Both assertion and reason are correct and reason isthe correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The power of an $A C$ circuit is given by $p = E I \cos ϕ$ where $\cos ϕ$ is power factor and $ϕ$ is phase angle. In case of circuit containing resistance only, phase angle is zero and power factor is equal to one. Therefore power is maximum in case of circuit containing only resistor. So option (4) is correct.

PHXII07:ALTERNATING CURRENT

356203 The potential difference across an instrument in an $A C$ circuit of frequency $f$ is $V$ and the current flowing through it is $i$ such that $V = 5 \cos (2 π f t)$ Volt and $i = 2 \sin (2 π f t) a m p .$ The power disspated in the instrument is

1

10 W

2

5 W

3

2.5 W

4

z e r o

Explanation:

As $V = 5 \cos (2 π f t) = 5 \sin (2 π f t + π / 2)$ and $i = 2 \sin (2 π f t)$
$∴$ Phase difference between $V$ and $i$ is $θ = \frac{π}{2}$
Average power, $P = V_{r m s} i_{r m s} \times \cos θ = 0$

PHXII07:ALTERNATING CURRENT

356204 In a $A C$ circuit the voltage and current are described by $V = 100 (319 t - \frac{π}{6}) v o l t s$ and $i = 200 \sin (319 t + \frac{π}{6}) m A$ respectively. The average power dissipated in the circuit is:

1

2.5 watts

2

5.0 watts

3

10.0 watts

4

50.0 watts

Explanation:

$P = \frac{V_{0} i_{0} \cos ϕ}{2} = \frac{100 \times 200 \times 10^{- 3}}{2} \cos \frac{π}{3} = 5 W$

PHXII07:ALTERNATING CURRENT

356200 The rms current in an ac circuit is $2 A$ . If the wattless current be $\sqrt{3} A$ . What is power factor?

1

\frac{1}{\sqrt{3}}

2

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

3

\frac{1}{2}

4

\frac{1}{3}

Explanation:

$I_{W L} = I_{r m s} \sin ϕ \Rightarrow \sqrt{3} = 2 \sin ϕ$
$\Rightarrow \sin ϕ = \frac{\sqrt{3}}{2} \Rightarrow ϕ = 60^{\circ}$
Power factor, $\cos ϕ = \cos 60^{\circ} = \frac{1}{2}$

PHXII07:ALTERNATING CURRENT

356201 $E m f$ and current in an ac circuit are given by $E = 20 V \sin (314 t)$ and $I = 2 A \sin (314 t - 60^{\circ})$ . Power consumed in the circuit per cycle is

1

40 W

2

20 W

3

10 W

4

5 W

Explanation:

$P_{a v} = \frac{E_{0} I_{0}}{2} \cos ϕ$
$= \frac{20 \times 2}{2} \times \cos 60^{\circ} = 10 W$

PHXII07:ALTERNATING CURRENT

356202 Assertion :
When an $A C$ circuit contains resistor only, its power is minimum.
Reason :
Power of an $A C$ circuit depends on phase angle.

1 Both assertion and reason are correct and reason isthe correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The power of an $A C$ circuit is given by $p = E I \cos ϕ$ where $\cos ϕ$ is power factor and $ϕ$ is phase angle. In case of circuit containing resistance only, phase angle is zero and power factor is equal to one. Therefore power is maximum in case of circuit containing only resistor. So option (4) is correct.

PHXII07:ALTERNATING CURRENT

356203 The potential difference across an instrument in an $A C$ circuit of frequency $f$ is $V$ and the current flowing through it is $i$ such that $V = 5 \cos (2 π f t)$ Volt and $i = 2 \sin (2 π f t) a m p .$ The power disspated in the instrument is

1

10 W

2

5 W

3

2.5 W

4

z e r o

Explanation:

As $V = 5 \cos (2 π f t) = 5 \sin (2 π f t + π / 2)$ and $i = 2 \sin (2 π f t)$
$∴$ Phase difference between $V$ and $i$ is $θ = \frac{π}{2}$
Average power, $P = V_{r m s} i_{r m s} \times \cos θ = 0$

PHXII07:ALTERNATING CURRENT

356204 In a $A C$ circuit the voltage and current are described by $V = 100 (319 t - \frac{π}{6}) v o l t s$ and $i = 200 \sin (319 t + \frac{π}{6}) m A$ respectively. The average power dissipated in the circuit is:

1

2.5 watts

2

5.0 watts

3

10.0 watts

4

50.0 watts

Explanation:

$P = \frac{V_{0} i_{0} \cos ϕ}{2} = \frac{100 \times 200 \times 10^{- 3}}{2} \cos \frac{π}{3} = 5 W$