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PHXII07:ALTERNATING CURRENT

356089 In India, the normal domestic power supply $A C$ is 220 $V$ , 50 $H z$ , Here 220 $V$ and 50 $H z$ refer to

1 Peak value of voltage and frequency

2 Rms value of voltage and frequency

3 Mean value of voltage and frequency

4 Peak, value of voltage and angular frequency.

Explanation:

220 $V$ and 50 $H z$ refer to rms value of voltage and frequency.

PHXII07:ALTERNATING CURRENT

356090 Virtual value of a current in $A C$ circuit is equal to

1 Zero

2 rms value

3 Peak value

4 None of these

PHXII07:ALTERNATING CURRENT

356091 The current vs time is shown in the graph. The given graph is half wave rectified. Find the rms and average currents.
supporting img

1

\frac{i_{0}}{2}, \frac{i_{0}}{π}

2

i_{0}, i_{0}

3

\frac{i_{0}}{π}, \frac{i_{0}}{4}

4

\frac{i_{0}}{π}, \frac{i_{0}}{π}

Explanation:

$i_{a v g} = \frac{\int_{0}^{2 π / ω} i d t}{(\frac{2 π}{ω})}$
$= \frac{ω}{2 π} [\int_{0}^{π / ω} i_{0} \sin (ω t) d t + \int_{π / ω}^{2 π / ω} (0) d t]$
$i_{a v g} = \frac{i_{0}}{2 π} [- \cos ω t]_{0}^{π} = - \frac{i_{0}}{2 π} [\cos (ω \frac{π}{ω}) - \cos 0]$
$i_{a v g} = \frac{2 i_{0}}{2 π} = \frac{i_{0}}{π}$
$⟨ i^{2} ⟩ = \frac{ω}{2 π} \int_{0}^{2 π / ω} i^{2} d t$
$= \frac{ω}{2 π} [\int_{0}^{π / ω} i_{0}^{2} \sin^{2} (ω t) d t + \int_{π / ω}^{2 π / ω} (0) d t]$
$⟨ i^{2} ⟩ = \frac{ω i_{0}^{2}}{2 π} [\int_{0}^{π / ω} \sin^{2} (ω t) d t]$
$\int_{0}^{π / ω} \sin^{2} (ω t) d t = \frac{1}{2} \int_{0}^{π / ω} (1 - \cos (2 ω t)) d t = \frac{π}{2 ω}$
$\Rightarrow ⟨ i^{2} ⟩ = \frac{ω i_{0}^{2}}{2 π} \cdot \frac{π}{2 ω} = \frac{i_{o}^{2}}{4} \Rightarrow i_{r m s} = \frac{i_{0}}{2}$

PHXII07:ALTERNATING CURRENT

356092 The average and effective values for the wave shape shown in the figure are:
supporting img

1

\frac{2}{π} V_{m}

and

\frac{V_{m}}{2}

2

\frac{V_{m}}{π}

and

\frac{V_{m}}{\sqrt{2}}

3

\frac{2}{π} V_{m}

and

\frac{V_{m}}{\sqrt{2}}

4

\frac{V_{m}}{π \sqrt{2}}

and

\frac{V_{m}}{\sqrt{2}}

Explanation:

The equation of the graph from 0 to $π$ is $V = V_{m} \sin θ$
The average value is $V_{a v g} = \frac{\int_{0}^{π} V d θ}{π}$
$\begin{aligned} V_{avg} = \frac{V_{m}}{π} \int_{0}^{π} \sin θ d θ = \frac{2 V_{m}}{π} \\ V_{eff}^{2} = \frac{V_{m}^{2}}{π} \int_{0}^{π} \sin^{2} θ d θ = \frac{V_{m}^{2}}{2} \\ V_{eff} = V_{m} / \sqrt{2} \end{aligned}$

PHXII07:ALTERNATING CURRENT

356089 In India, the normal domestic power supply $A C$ is 220 $V$ , 50 $H z$ , Here 220 $V$ and 50 $H z$ refer to

1 Peak value of voltage and frequency

2 Rms value of voltage and frequency

3 Mean value of voltage and frequency

4 Peak, value of voltage and angular frequency.

Explanation:

220 $V$ and 50 $H z$ refer to rms value of voltage and frequency.

PHXII07:ALTERNATING CURRENT

356090 Virtual value of a current in $A C$ circuit is equal to

1 Zero

2 rms value

3 Peak value

4 None of these

PHXII07:ALTERNATING CURRENT

356091 The current vs time is shown in the graph. The given graph is half wave rectified. Find the rms and average currents.
supporting img

1

\frac{i_{0}}{2}, \frac{i_{0}}{π}

2

i_{0}, i_{0}

3

\frac{i_{0}}{π}, \frac{i_{0}}{4}

4

\frac{i_{0}}{π}, \frac{i_{0}}{π}

Explanation:

$i_{a v g} = \frac{\int_{0}^{2 π / ω} i d t}{(\frac{2 π}{ω})}$
$= \frac{ω}{2 π} [\int_{0}^{π / ω} i_{0} \sin (ω t) d t + \int_{π / ω}^{2 π / ω} (0) d t]$
$i_{a v g} = \frac{i_{0}}{2 π} [- \cos ω t]_{0}^{π} = - \frac{i_{0}}{2 π} [\cos (ω \frac{π}{ω}) - \cos 0]$
$i_{a v g} = \frac{2 i_{0}}{2 π} = \frac{i_{0}}{π}$
$⟨ i^{2} ⟩ = \frac{ω}{2 π} \int_{0}^{2 π / ω} i^{2} d t$
$= \frac{ω}{2 π} [\int_{0}^{π / ω} i_{0}^{2} \sin^{2} (ω t) d t + \int_{π / ω}^{2 π / ω} (0) d t]$
$⟨ i^{2} ⟩ = \frac{ω i_{0}^{2}}{2 π} [\int_{0}^{π / ω} \sin^{2} (ω t) d t]$
$\int_{0}^{π / ω} \sin^{2} (ω t) d t = \frac{1}{2} \int_{0}^{π / ω} (1 - \cos (2 ω t)) d t = \frac{π}{2 ω}$
$\Rightarrow ⟨ i^{2} ⟩ = \frac{ω i_{0}^{2}}{2 π} \cdot \frac{π}{2 ω} = \frac{i_{o}^{2}}{4} \Rightarrow i_{r m s} = \frac{i_{0}}{2}$

PHXII07:ALTERNATING CURRENT

356092 The average and effective values for the wave shape shown in the figure are:
supporting img

1

\frac{2}{π} V_{m}

and

\frac{V_{m}}{2}

2

\frac{V_{m}}{π}

and

\frac{V_{m}}{\sqrt{2}}

3

\frac{2}{π} V_{m}

and

\frac{V_{m}}{\sqrt{2}}

4

\frac{V_{m}}{π \sqrt{2}}

and

\frac{V_{m}}{\sqrt{2}}

Explanation:

The equation of the graph from 0 to $π$ is $V = V_{m} \sin θ$
The average value is $V_{a v g} = \frac{\int_{0}^{π} V d θ}{π}$
$\begin{aligned} V_{avg} = \frac{V_{m}}{π} \int_{0}^{π} \sin θ d θ = \frac{2 V_{m}}{π} \\ V_{eff}^{2} = \frac{V_{m}^{2}}{π} \int_{0}^{π} \sin^{2} θ d θ = \frac{V_{m}^{2}}{2} \\ V_{eff} = V_{m} / \sqrt{2} \end{aligned}$

PHXII07:ALTERNATING CURRENT

356089 In India, the normal domestic power supply $A C$ is 220 $V$ , 50 $H z$ , Here 220 $V$ and 50 $H z$ refer to

1 Peak value of voltage and frequency

2 Rms value of voltage and frequency

3 Mean value of voltage and frequency

4 Peak, value of voltage and angular frequency.

Explanation:

220 $V$ and 50 $H z$ refer to rms value of voltage and frequency.

PHXII07:ALTERNATING CURRENT

356090 Virtual value of a current in $A C$ circuit is equal to

1 Zero

2 rms value

3 Peak value

4 None of these

PHXII07:ALTERNATING CURRENT

356091 The current vs time is shown in the graph. The given graph is half wave rectified. Find the rms and average currents.
supporting img

1

\frac{i_{0}}{2}, \frac{i_{0}}{π}

2

i_{0}, i_{0}

3

\frac{i_{0}}{π}, \frac{i_{0}}{4}

4

\frac{i_{0}}{π}, \frac{i_{0}}{π}

Explanation:

$i_{a v g} = \frac{\int_{0}^{2 π / ω} i d t}{(\frac{2 π}{ω})}$
$= \frac{ω}{2 π} [\int_{0}^{π / ω} i_{0} \sin (ω t) d t + \int_{π / ω}^{2 π / ω} (0) d t]$
$i_{a v g} = \frac{i_{0}}{2 π} [- \cos ω t]_{0}^{π} = - \frac{i_{0}}{2 π} [\cos (ω \frac{π}{ω}) - \cos 0]$
$i_{a v g} = \frac{2 i_{0}}{2 π} = \frac{i_{0}}{π}$
$⟨ i^{2} ⟩ = \frac{ω}{2 π} \int_{0}^{2 π / ω} i^{2} d t$
$= \frac{ω}{2 π} [\int_{0}^{π / ω} i_{0}^{2} \sin^{2} (ω t) d t + \int_{π / ω}^{2 π / ω} (0) d t]$
$⟨ i^{2} ⟩ = \frac{ω i_{0}^{2}}{2 π} [\int_{0}^{π / ω} \sin^{2} (ω t) d t]$
$\int_{0}^{π / ω} \sin^{2} (ω t) d t = \frac{1}{2} \int_{0}^{π / ω} (1 - \cos (2 ω t)) d t = \frac{π}{2 ω}$
$\Rightarrow ⟨ i^{2} ⟩ = \frac{ω i_{0}^{2}}{2 π} \cdot \frac{π}{2 ω} = \frac{i_{o}^{2}}{4} \Rightarrow i_{r m s} = \frac{i_{0}}{2}$

PHXII07:ALTERNATING CURRENT

356092 The average and effective values for the wave shape shown in the figure are:
supporting img

1

\frac{2}{π} V_{m}

and

\frac{V_{m}}{2}

2

\frac{V_{m}}{π}

and

\frac{V_{m}}{\sqrt{2}}

3

\frac{2}{π} V_{m}

and

\frac{V_{m}}{\sqrt{2}}

4

\frac{V_{m}}{π \sqrt{2}}

and

\frac{V_{m}}{\sqrt{2}}

Explanation:

The equation of the graph from 0 to $π$ is $V = V_{m} \sin θ$
The average value is $V_{a v g} = \frac{\int_{0}^{π} V d θ}{π}$
$\begin{aligned} V_{avg} = \frac{V_{m}}{π} \int_{0}^{π} \sin θ d θ = \frac{2 V_{m}}{π} \\ V_{eff}^{2} = \frac{V_{m}^{2}}{π} \int_{0}^{π} \sin^{2} θ d θ = \frac{V_{m}^{2}}{2} \\ V_{eff} = V_{m} / \sqrt{2} \end{aligned}$

PHXII07:ALTERNATING CURRENT

356089 In India, the normal domestic power supply $A C$ is 220 $V$ , 50 $H z$ , Here 220 $V$ and 50 $H z$ refer to

1 Peak value of voltage and frequency

2 Rms value of voltage and frequency

3 Mean value of voltage and frequency

4 Peak, value of voltage and angular frequency.

Explanation:

220 $V$ and 50 $H z$ refer to rms value of voltage and frequency.

PHXII07:ALTERNATING CURRENT

356090 Virtual value of a current in $A C$ circuit is equal to

1 Zero

2 rms value

3 Peak value

4 None of these

PHXII07:ALTERNATING CURRENT

356091 The current vs time is shown in the graph. The given graph is half wave rectified. Find the rms and average currents.
supporting img

1

\frac{i_{0}}{2}, \frac{i_{0}}{π}

2

i_{0}, i_{0}

3

\frac{i_{0}}{π}, \frac{i_{0}}{4}

4

\frac{i_{0}}{π}, \frac{i_{0}}{π}

Explanation:

$i_{a v g} = \frac{\int_{0}^{2 π / ω} i d t}{(\frac{2 π}{ω})}$
$= \frac{ω}{2 π} [\int_{0}^{π / ω} i_{0} \sin (ω t) d t + \int_{π / ω}^{2 π / ω} (0) d t]$
$i_{a v g} = \frac{i_{0}}{2 π} [- \cos ω t]_{0}^{π} = - \frac{i_{0}}{2 π} [\cos (ω \frac{π}{ω}) - \cos 0]$
$i_{a v g} = \frac{2 i_{0}}{2 π} = \frac{i_{0}}{π}$
$⟨ i^{2} ⟩ = \frac{ω}{2 π} \int_{0}^{2 π / ω} i^{2} d t$
$= \frac{ω}{2 π} [\int_{0}^{π / ω} i_{0}^{2} \sin^{2} (ω t) d t + \int_{π / ω}^{2 π / ω} (0) d t]$
$⟨ i^{2} ⟩ = \frac{ω i_{0}^{2}}{2 π} [\int_{0}^{π / ω} \sin^{2} (ω t) d t]$
$\int_{0}^{π / ω} \sin^{2} (ω t) d t = \frac{1}{2} \int_{0}^{π / ω} (1 - \cos (2 ω t)) d t = \frac{π}{2 ω}$
$\Rightarrow ⟨ i^{2} ⟩ = \frac{ω i_{0}^{2}}{2 π} \cdot \frac{π}{2 ω} = \frac{i_{o}^{2}}{4} \Rightarrow i_{r m s} = \frac{i_{0}}{2}$

PHXII07:ALTERNATING CURRENT

356092 The average and effective values for the wave shape shown in the figure are:
supporting img

1

\frac{2}{π} V_{m}

and

\frac{V_{m}}{2}

2

\frac{V_{m}}{π}

and

\frac{V_{m}}{\sqrt{2}}

3

\frac{2}{π} V_{m}

and

\frac{V_{m}}{\sqrt{2}}

4

\frac{V_{m}}{π \sqrt{2}}

and

\frac{V_{m}}{\sqrt{2}}

Explanation:

The equation of the graph from 0 to $π$ is $V = V_{m} \sin θ$
The average value is $V_{a v g} = \frac{\int_{0}^{π} V d θ}{π}$
$\begin{aligned} V_{avg} = \frac{V_{m}}{π} \int_{0}^{π} \sin θ d θ = \frac{2 V_{m}}{π} \\ V_{eff}^{2} = \frac{V_{m}^{2}}{π} \int_{0}^{π} \sin^{2} θ d θ = \frac{V_{m}^{2}}{2} \\ V_{eff} = V_{m} / \sqrt{2} \end{aligned}$

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