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Alternating Current

155335 A lamp consumes only $50 %$ of maximum power applied in an A.C. circuit. What will be the phase difference between applied voltage and circuit current?

1

π / 6 rad

2

π / 3 rad

3

π / 4 rad

4

π / 2 rad

Explanation:

B Given,
$P = \frac{P_{P}}{2}$
We know that,
$P = V_{rms} I_{rms} \cos ϕ$
Where, $V_{rms} = \frac{V_{p}}{\sqrt{2}}$ and $R_{rms} = \frac{I_{p}}{\sqrt{2}}$
So, $P = \frac{V_{P} I_{P}}{2} \cos ϕ$
$P = P_{P} \cos ϕ$
$\frac{P_{P}}{2} = P_{P} \cos ϕ$
Therefore, we get-
$ϕ = \cos^{- 1} \frac{1}{2}$
$ϕ = \frac{π}{3}$
Thus, the phase difference between the applied voltage and circuit current when a lamp consumer only $50 %$ of peak power in an AC circuit is $π / 3$ .

GUJCET 2014

Alternating Current

155336 Amount of heat produced per second in calories when a bulb of $50 W, 200 V$ glows (assuming that only $20 %$ of the electric energy is converted into light)
( $J = 4.2 J / cal$ )

1

40 cal / s

2

28 cal / s

3

18.22 cal / s

4

9.52 cal / s

Explanation:

D Given, power of bulb $= 50 W$ ,
Time taken $= 1 s$
Electric energy consumed per second $= 50 W$
Heat produced $= 100 % - 20 % = 80 %$
$= \frac{80}{100} \times 50 = 40 J$
We know that,
$1 cal = 4.2 J$
$1 J = \frac{1}{4.2} cal$
$= \frac{40}{4.2} cal / \sec = 9.52 cal / \sec$

CG PET -2016

Alternating Current

155338 A coil has resistance $30 ohm$ and inductive reactance $20 ohm$ at $50 Hz$ frequency. If an ac source, of 200 volt, $100 Hz$ , is connected across the coil, the current in the coil will be

1

4.0 A

2

8.0 A

3

7.2 A

4

2.0 A

Explanation:

A Given that,
$R = 30 Ω$
$X_{L} = 20 Ω$
$f = 50 Hz$
$V = 200 V$
We know that, $ω = 50 \times 2 π$ then, $ω L = 20 Ω$
$ω^{'} = 100 \times 2 π then, ω^{'} L = 40 Ω$
$∴ I = \frac{200}{Z} \times \frac{200}{\sqrt{R^{2} + {(ω^{'} L)}^{2}}}$
$I = \frac{200}{\sqrt{30^{2} + 40^{2}}}$
$I = 4 A$

BITSAT-2009

Alternating Current

155339 The impedance of a circuit consists of $3 Ω$ resistance and $4 Ω$ reactance. The power factor of the circuit is

1 0.4

2 0.6

3 0.8

4 1.0

Explanation:

B Given,
$R = 3 Ω$
$X_{L} = 4 Ω$
We know that, impedance of the circuit -
$Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{3^{2} + 4^{2}}$
$Z = 5 Ω$
The power factor is,
$\cos ϕ = \frac{R}{Z} = \frac{3}{5}$
$\cos ϕ = 0.6$

BITSAT-2012

Alternating Current

155335 A lamp consumes only $50 %$ of maximum power applied in an A.C. circuit. What will be the phase difference between applied voltage and circuit current?

1

π / 6 rad

2

π / 3 rad

3

π / 4 rad

4

π / 2 rad

Explanation:

B Given,
$P = \frac{P_{P}}{2}$
We know that,
$P = V_{rms} I_{rms} \cos ϕ$
Where, $V_{rms} = \frac{V_{p}}{\sqrt{2}}$ and $R_{rms} = \frac{I_{p}}{\sqrt{2}}$
So, $P = \frac{V_{P} I_{P}}{2} \cos ϕ$
$P = P_{P} \cos ϕ$
$\frac{P_{P}}{2} = P_{P} \cos ϕ$
Therefore, we get-
$ϕ = \cos^{- 1} \frac{1}{2}$
$ϕ = \frac{π}{3}$
Thus, the phase difference between the applied voltage and circuit current when a lamp consumer only $50 %$ of peak power in an AC circuit is $π / 3$ .

GUJCET 2014

Alternating Current

155336 Amount of heat produced per second in calories when a bulb of $50 W, 200 V$ glows (assuming that only $20 %$ of the electric energy is converted into light)
( $J = 4.2 J / cal$ )

1

40 cal / s

2

28 cal / s

3

18.22 cal / s

4

9.52 cal / s

Explanation:

D Given, power of bulb $= 50 W$ ,
Time taken $= 1 s$
Electric energy consumed per second $= 50 W$
Heat produced $= 100 % - 20 % = 80 %$
$= \frac{80}{100} \times 50 = 40 J$
We know that,
$1 cal = 4.2 J$
$1 J = \frac{1}{4.2} cal$
$= \frac{40}{4.2} cal / \sec = 9.52 cal / \sec$

CG PET -2016

Alternating Current

155338 A coil has resistance $30 ohm$ and inductive reactance $20 ohm$ at $50 Hz$ frequency. If an ac source, of 200 volt, $100 Hz$ , is connected across the coil, the current in the coil will be

1

4.0 A

2

8.0 A

3

7.2 A

4

2.0 A

Explanation:

A Given that,
$R = 30 Ω$
$X_{L} = 20 Ω$
$f = 50 Hz$
$V = 200 V$
We know that, $ω = 50 \times 2 π$ then, $ω L = 20 Ω$
$ω^{'} = 100 \times 2 π then, ω^{'} L = 40 Ω$
$∴ I = \frac{200}{Z} \times \frac{200}{\sqrt{R^{2} + {(ω^{'} L)}^{2}}}$
$I = \frac{200}{\sqrt{30^{2} + 40^{2}}}$
$I = 4 A$

BITSAT-2009

Alternating Current

155339 The impedance of a circuit consists of $3 Ω$ resistance and $4 Ω$ reactance. The power factor of the circuit is

1 0.4

2 0.6

3 0.8

4 1.0

Explanation:

B Given,
$R = 3 Ω$
$X_{L} = 4 Ω$
We know that, impedance of the circuit -
$Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{3^{2} + 4^{2}}$
$Z = 5 Ω$
The power factor is,
$\cos ϕ = \frac{R}{Z} = \frac{3}{5}$
$\cos ϕ = 0.6$

BITSAT-2012

Alternating Current

155335 A lamp consumes only $50 %$ of maximum power applied in an A.C. circuit. What will be the phase difference between applied voltage and circuit current?

1

π / 6 rad

2

π / 3 rad

3

π / 4 rad

4

π / 2 rad

Explanation:

B Given,
$P = \frac{P_{P}}{2}$
We know that,
$P = V_{rms} I_{rms} \cos ϕ$
Where, $V_{rms} = \frac{V_{p}}{\sqrt{2}}$ and $R_{rms} = \frac{I_{p}}{\sqrt{2}}$
So, $P = \frac{V_{P} I_{P}}{2} \cos ϕ$
$P = P_{P} \cos ϕ$
$\frac{P_{P}}{2} = P_{P} \cos ϕ$
Therefore, we get-
$ϕ = \cos^{- 1} \frac{1}{2}$
$ϕ = \frac{π}{3}$
Thus, the phase difference between the applied voltage and circuit current when a lamp consumer only $50 %$ of peak power in an AC circuit is $π / 3$ .

GUJCET 2014

Alternating Current

155336 Amount of heat produced per second in calories when a bulb of $50 W, 200 V$ glows (assuming that only $20 %$ of the electric energy is converted into light)
( $J = 4.2 J / cal$ )

1

40 cal / s

2

28 cal / s

3

18.22 cal / s

4

9.52 cal / s

Explanation:

D Given, power of bulb $= 50 W$ ,
Time taken $= 1 s$
Electric energy consumed per second $= 50 W$
Heat produced $= 100 % - 20 % = 80 %$
$= \frac{80}{100} \times 50 = 40 J$
We know that,
$1 cal = 4.2 J$
$1 J = \frac{1}{4.2} cal$
$= \frac{40}{4.2} cal / \sec = 9.52 cal / \sec$

CG PET -2016

Alternating Current

155338 A coil has resistance $30 ohm$ and inductive reactance $20 ohm$ at $50 Hz$ frequency. If an ac source, of 200 volt, $100 Hz$ , is connected across the coil, the current in the coil will be

1

4.0 A

2

8.0 A

3

7.2 A

4

2.0 A

Explanation:

A Given that,
$R = 30 Ω$
$X_{L} = 20 Ω$
$f = 50 Hz$
$V = 200 V$
We know that, $ω = 50 \times 2 π$ then, $ω L = 20 Ω$
$ω^{'} = 100 \times 2 π then, ω^{'} L = 40 Ω$
$∴ I = \frac{200}{Z} \times \frac{200}{\sqrt{R^{2} + {(ω^{'} L)}^{2}}}$
$I = \frac{200}{\sqrt{30^{2} + 40^{2}}}$
$I = 4 A$

BITSAT-2009

Alternating Current

155339 The impedance of a circuit consists of $3 Ω$ resistance and $4 Ω$ reactance. The power factor of the circuit is

1 0.4

2 0.6

3 0.8

4 1.0

Explanation:

B Given,
$R = 3 Ω$
$X_{L} = 4 Ω$
We know that, impedance of the circuit -
$Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{3^{2} + 4^{2}}$
$Z = 5 Ω$
The power factor is,
$\cos ϕ = \frac{R}{Z} = \frac{3}{5}$
$\cos ϕ = 0.6$

BITSAT-2012

Alternating Current

155335 A lamp consumes only $50 %$ of maximum power applied in an A.C. circuit. What will be the phase difference between applied voltage and circuit current?

1

π / 6 rad

2

π / 3 rad

3

π / 4 rad

4

π / 2 rad

Explanation:

B Given,
$P = \frac{P_{P}}{2}$
We know that,
$P = V_{rms} I_{rms} \cos ϕ$
Where, $V_{rms} = \frac{V_{p}}{\sqrt{2}}$ and $R_{rms} = \frac{I_{p}}{\sqrt{2}}$
So, $P = \frac{V_{P} I_{P}}{2} \cos ϕ$
$P = P_{P} \cos ϕ$
$\frac{P_{P}}{2} = P_{P} \cos ϕ$
Therefore, we get-
$ϕ = \cos^{- 1} \frac{1}{2}$
$ϕ = \frac{π}{3}$
Thus, the phase difference between the applied voltage and circuit current when a lamp consumer only $50 %$ of peak power in an AC circuit is $π / 3$ .

GUJCET 2014

Alternating Current

155336 Amount of heat produced per second in calories when a bulb of $50 W, 200 V$ glows (assuming that only $20 %$ of the electric energy is converted into light)
( $J = 4.2 J / cal$ )

1

40 cal / s

2

28 cal / s

3

18.22 cal / s

4

9.52 cal / s

Explanation:

D Given, power of bulb $= 50 W$ ,
Time taken $= 1 s$
Electric energy consumed per second $= 50 W$
Heat produced $= 100 % - 20 % = 80 %$
$= \frac{80}{100} \times 50 = 40 J$
We know that,
$1 cal = 4.2 J$
$1 J = \frac{1}{4.2} cal$
$= \frac{40}{4.2} cal / \sec = 9.52 cal / \sec$

CG PET -2016

Alternating Current

155338 A coil has resistance $30 ohm$ and inductive reactance $20 ohm$ at $50 Hz$ frequency. If an ac source, of 200 volt, $100 Hz$ , is connected across the coil, the current in the coil will be

1

4.0 A

2

8.0 A

3

7.2 A

4

2.0 A

Explanation:

A Given that,
$R = 30 Ω$
$X_{L} = 20 Ω$
$f = 50 Hz$
$V = 200 V$
We know that, $ω = 50 \times 2 π$ then, $ω L = 20 Ω$
$ω^{'} = 100 \times 2 π then, ω^{'} L = 40 Ω$
$∴ I = \frac{200}{Z} \times \frac{200}{\sqrt{R^{2} + {(ω^{'} L)}^{2}}}$
$I = \frac{200}{\sqrt{30^{2} + 40^{2}}}$
$I = 4 A$

BITSAT-2009

Alternating Current

155339 The impedance of a circuit consists of $3 Ω$ resistance and $4 Ω$ reactance. The power factor of the circuit is

1 0.4

2 0.6

3 0.8

4 1.0

Explanation:

B Given,
$R = 3 Ω$
$X_{L} = 4 Ω$
We know that, impedance of the circuit -
$Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{3^{2} + 4^{2}}$
$Z = 5 Ω$
The power factor is,
$\cos ϕ = \frac{R}{Z} = \frac{3}{5}$
$\cos ϕ = 0.6$

BITSAT-2012