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Alternating Current

155317 A resistance of $100 Ω$ , inductor of selfinductance $(\frac{4}{π^{2}}) H$ and a capacitor of unknown capacitance are connected in series to an a.c. source of $200 V$ and $50 Hz$ . When the current and voltage are in phase, the capacitance and power dissipated is respectively

1

2.5 \times 10^{- 5} F, 400 W

2

3.0 \times 10^{- 5} F, 50 W

3

2.0 \times 10^{- 5} F, 100 W

4

1.5 \times 10^{- 5} F, 200 W

Explanation:

A Given,
$R = 100 Ω$
$L = \frac{4}{π^{2}} H$
$V = 200 V$
$f = 50 Hz$

When voltage and current are in phase then
$X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
Capacitance $(C) = \frac{1}{ω^{2} L}$
$C = \frac{1}{L (2 π f)^{2}} = \frac{1 \times π^{2}}{4 (2 π \times 50)^{2}}$
$C = \frac{1}{40000}$
$C = 2.5 \times 10^{- 5} F$
We know that,
Power dissipated $(P) = \frac{V^{2}}{R}$
$P = \frac{40000}{100} = 400 W$

MHT-CET 2020

Alternating Current

155318 In an AC circuit, the current is given by $I = 5 \sin (100 t - π / 2)$ and the alternating potential is $V = 200 \sin (100) t$ volt. The power consumed in the circuit is

1

20 W

2

40 W

3

1000 W

4 zero

Explanation:

D Given, $I = 5 \sin (100 t - π / 2), V = 200$ $\sin (100) t$
We know that,
$I_{rms} = \frac{I_{p}}{\sqrt{2}} = \frac{5}{\sqrt{2}} A$
$V_{rms} = \frac{V_{p}}{\sqrt{2}} = \frac{200}{\sqrt{2}} V$
Phase difference $(ϕ) = \frac{π}{2}$
$∵ P = V_{rms} \times I_{rms} \times \cos ϕ$
$P = \frac{200}{\sqrt{2}} \times \frac{5}{\sqrt{2}} \times \cos \frac{π}{2}$
$P = 0$

AP EAMCET-23.09.2020

Alternating Current

155319 A lamp delivers a luminous flux of $100 W$ to an absorber of area $1 {cm}^{2}$ . The force due to radiation pressure is

1

3.3 \times 10^{- 4} N

2

16.5 \times 10^{- 7} N

3

3.3 \times 10^{- 6} N

4

3.3 \times 10^{- 7} N

Explanation:

D Given,
$A = 1 {cm}^{2} = 10^{- 4} m^{2}$
$P = 100 W$
$c = 3 \times 10^{8} m / \sec$
We know that,
Force $=$ Rate of change of linear momentum
$F = \frac{dp}{dt} = \frac{d}{dt} (\frac{v}{c})$
$F = \frac{1}{c} \frac{dv}{dt}$
Where, $v =$ Electromagnetic wave energy
$F = \frac{P}{c}$
$F = \frac{100}{3 \times 10^{8}}$
$F = 3.3 \times 10^{- 7} N$

AP EAMCET (20.04.2019) Shift-II

Alternating Current

155320 An electric bulb is rated as $200 V - 100 W$ . The power consumed by the bulb when operated at $100 V$ is

1

25 W

2

50 W

3

75 W

4

100 W

Explanation:

A Given that,
$P = 100 W$
$V = 200 V$
We know that, $P = \frac{V^{2}}{R}$
$R = \frac{(200)^{2}}{100} = \frac{200 \times 200}{100} = 400 Ω$
When bulb is operated at $100 V$
Then, power consumed
$P = \frac{V^{2}}{R} = \frac{(100)^{2}}{400} = \frac{100 \times 100}{400} = 25 W$
$P = 25 W$

UPSEE 2019

Alternating Current

155317 A resistance of $100 Ω$ , inductor of selfinductance $(\frac{4}{π^{2}}) H$ and a capacitor of unknown capacitance are connected in series to an a.c. source of $200 V$ and $50 Hz$ . When the current and voltage are in phase, the capacitance and power dissipated is respectively

1

2.5 \times 10^{- 5} F, 400 W

2

3.0 \times 10^{- 5} F, 50 W

3

2.0 \times 10^{- 5} F, 100 W

4

1.5 \times 10^{- 5} F, 200 W

Explanation:

A Given,
$R = 100 Ω$
$L = \frac{4}{π^{2}} H$
$V = 200 V$
$f = 50 Hz$

When voltage and current are in phase then
$X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
Capacitance $(C) = \frac{1}{ω^{2} L}$
$C = \frac{1}{L (2 π f)^{2}} = \frac{1 \times π^{2}}{4 (2 π \times 50)^{2}}$
$C = \frac{1}{40000}$
$C = 2.5 \times 10^{- 5} F$
We know that,
Power dissipated $(P) = \frac{V^{2}}{R}$
$P = \frac{40000}{100} = 400 W$

MHT-CET 2020

Alternating Current

155318 In an AC circuit, the current is given by $I = 5 \sin (100 t - π / 2)$ and the alternating potential is $V = 200 \sin (100) t$ volt. The power consumed in the circuit is

1

20 W

2

40 W

3

1000 W

4 zero

Explanation:

D Given, $I = 5 \sin (100 t - π / 2), V = 200$ $\sin (100) t$
We know that,
$I_{rms} = \frac{I_{p}}{\sqrt{2}} = \frac{5}{\sqrt{2}} A$
$V_{rms} = \frac{V_{p}}{\sqrt{2}} = \frac{200}{\sqrt{2}} V$
Phase difference $(ϕ) = \frac{π}{2}$
$∵ P = V_{rms} \times I_{rms} \times \cos ϕ$
$P = \frac{200}{\sqrt{2}} \times \frac{5}{\sqrt{2}} \times \cos \frac{π}{2}$
$P = 0$

AP EAMCET-23.09.2020

Alternating Current

155319 A lamp delivers a luminous flux of $100 W$ to an absorber of area $1 {cm}^{2}$ . The force due to radiation pressure is

1

3.3 \times 10^{- 4} N

2

16.5 \times 10^{- 7} N

3

3.3 \times 10^{- 6} N

4

3.3 \times 10^{- 7} N

Explanation:

D Given,
$A = 1 {cm}^{2} = 10^{- 4} m^{2}$
$P = 100 W$
$c = 3 \times 10^{8} m / \sec$
We know that,
Force $=$ Rate of change of linear momentum
$F = \frac{dp}{dt} = \frac{d}{dt} (\frac{v}{c})$
$F = \frac{1}{c} \frac{dv}{dt}$
Where, $v =$ Electromagnetic wave energy
$F = \frac{P}{c}$
$F = \frac{100}{3 \times 10^{8}}$
$F = 3.3 \times 10^{- 7} N$

AP EAMCET (20.04.2019) Shift-II

Alternating Current

155320 An electric bulb is rated as $200 V - 100 W$ . The power consumed by the bulb when operated at $100 V$ is

1

25 W

2

50 W

3

75 W

4

100 W

Explanation:

A Given that,
$P = 100 W$
$V = 200 V$
We know that, $P = \frac{V^{2}}{R}$
$R = \frac{(200)^{2}}{100} = \frac{200 \times 200}{100} = 400 Ω$
When bulb is operated at $100 V$
Then, power consumed
$P = \frac{V^{2}}{R} = \frac{(100)^{2}}{400} = \frac{100 \times 100}{400} = 25 W$
$P = 25 W$

UPSEE 2019

Alternating Current

155317 A resistance of $100 Ω$ , inductor of selfinductance $(\frac{4}{π^{2}}) H$ and a capacitor of unknown capacitance are connected in series to an a.c. source of $200 V$ and $50 Hz$ . When the current and voltage are in phase, the capacitance and power dissipated is respectively

1

2.5 \times 10^{- 5} F, 400 W

2

3.0 \times 10^{- 5} F, 50 W

3

2.0 \times 10^{- 5} F, 100 W

4

1.5 \times 10^{- 5} F, 200 W

Explanation:

A Given,
$R = 100 Ω$
$L = \frac{4}{π^{2}} H$
$V = 200 V$
$f = 50 Hz$

When voltage and current are in phase then
$X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
Capacitance $(C) = \frac{1}{ω^{2} L}$
$C = \frac{1}{L (2 π f)^{2}} = \frac{1 \times π^{2}}{4 (2 π \times 50)^{2}}$
$C = \frac{1}{40000}$
$C = 2.5 \times 10^{- 5} F$
We know that,
Power dissipated $(P) = \frac{V^{2}}{R}$
$P = \frac{40000}{100} = 400 W$

MHT-CET 2020

Alternating Current

155318 In an AC circuit, the current is given by $I = 5 \sin (100 t - π / 2)$ and the alternating potential is $V = 200 \sin (100) t$ volt. The power consumed in the circuit is

1

20 W

2

40 W

3

1000 W

4 zero

Explanation:

D Given, $I = 5 \sin (100 t - π / 2), V = 200$ $\sin (100) t$
We know that,
$I_{rms} = \frac{I_{p}}{\sqrt{2}} = \frac{5}{\sqrt{2}} A$
$V_{rms} = \frac{V_{p}}{\sqrt{2}} = \frac{200}{\sqrt{2}} V$
Phase difference $(ϕ) = \frac{π}{2}$
$∵ P = V_{rms} \times I_{rms} \times \cos ϕ$
$P = \frac{200}{\sqrt{2}} \times \frac{5}{\sqrt{2}} \times \cos \frac{π}{2}$
$P = 0$

AP EAMCET-23.09.2020

Alternating Current

155319 A lamp delivers a luminous flux of $100 W$ to an absorber of area $1 {cm}^{2}$ . The force due to radiation pressure is

1

3.3 \times 10^{- 4} N

2

16.5 \times 10^{- 7} N

3

3.3 \times 10^{- 6} N

4

3.3 \times 10^{- 7} N

Explanation:

D Given,
$A = 1 {cm}^{2} = 10^{- 4} m^{2}$
$P = 100 W$
$c = 3 \times 10^{8} m / \sec$
We know that,
Force $=$ Rate of change of linear momentum
$F = \frac{dp}{dt} = \frac{d}{dt} (\frac{v}{c})$
$F = \frac{1}{c} \frac{dv}{dt}$
Where, $v =$ Electromagnetic wave energy
$F = \frac{P}{c}$
$F = \frac{100}{3 \times 10^{8}}$
$F = 3.3 \times 10^{- 7} N$

AP EAMCET (20.04.2019) Shift-II

Alternating Current

155320 An electric bulb is rated as $200 V - 100 W$ . The power consumed by the bulb when operated at $100 V$ is

1

25 W

2

50 W

3

75 W

4

100 W

Explanation:

A Given that,
$P = 100 W$
$V = 200 V$
We know that, $P = \frac{V^{2}}{R}$
$R = \frac{(200)^{2}}{100} = \frac{200 \times 200}{100} = 400 Ω$
When bulb is operated at $100 V$
Then, power consumed
$P = \frac{V^{2}}{R} = \frac{(100)^{2}}{400} = \frac{100 \times 100}{400} = 25 W$
$P = 25 W$

UPSEE 2019

Alternating Current

155317 A resistance of $100 Ω$ , inductor of selfinductance $(\frac{4}{π^{2}}) H$ and a capacitor of unknown capacitance are connected in series to an a.c. source of $200 V$ and $50 Hz$ . When the current and voltage are in phase, the capacitance and power dissipated is respectively

1

2.5 \times 10^{- 5} F, 400 W

2

3.0 \times 10^{- 5} F, 50 W

3

2.0 \times 10^{- 5} F, 100 W

4

1.5 \times 10^{- 5} F, 200 W

Explanation:

A Given,
$R = 100 Ω$
$L = \frac{4}{π^{2}} H$
$V = 200 V$
$f = 50 Hz$

When voltage and current are in phase then
$X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
Capacitance $(C) = \frac{1}{ω^{2} L}$
$C = \frac{1}{L (2 π f)^{2}} = \frac{1 \times π^{2}}{4 (2 π \times 50)^{2}}$
$C = \frac{1}{40000}$
$C = 2.5 \times 10^{- 5} F$
We know that,
Power dissipated $(P) = \frac{V^{2}}{R}$
$P = \frac{40000}{100} = 400 W$

MHT-CET 2020

Alternating Current

155318 In an AC circuit, the current is given by $I = 5 \sin (100 t - π / 2)$ and the alternating potential is $V = 200 \sin (100) t$ volt. The power consumed in the circuit is

1

20 W

2

40 W

3

1000 W

4 zero

Explanation:

D Given, $I = 5 \sin (100 t - π / 2), V = 200$ $\sin (100) t$
We know that,
$I_{rms} = \frac{I_{p}}{\sqrt{2}} = \frac{5}{\sqrt{2}} A$
$V_{rms} = \frac{V_{p}}{\sqrt{2}} = \frac{200}{\sqrt{2}} V$
Phase difference $(ϕ) = \frac{π}{2}$
$∵ P = V_{rms} \times I_{rms} \times \cos ϕ$
$P = \frac{200}{\sqrt{2}} \times \frac{5}{\sqrt{2}} \times \cos \frac{π}{2}$
$P = 0$

AP EAMCET-23.09.2020

Alternating Current

155319 A lamp delivers a luminous flux of $100 W$ to an absorber of area $1 {cm}^{2}$ . The force due to radiation pressure is

1

3.3 \times 10^{- 4} N

2

16.5 \times 10^{- 7} N

3

3.3 \times 10^{- 6} N

4

3.3 \times 10^{- 7} N

Explanation:

D Given,
$A = 1 {cm}^{2} = 10^{- 4} m^{2}$
$P = 100 W$
$c = 3 \times 10^{8} m / \sec$
We know that,
Force $=$ Rate of change of linear momentum
$F = \frac{dp}{dt} = \frac{d}{dt} (\frac{v}{c})$
$F = \frac{1}{c} \frac{dv}{dt}$
Where, $v =$ Electromagnetic wave energy
$F = \frac{P}{c}$
$F = \frac{100}{3 \times 10^{8}}$
$F = 3.3 \times 10^{- 7} N$

AP EAMCET (20.04.2019) Shift-II

Alternating Current

155320 An electric bulb is rated as $200 V - 100 W$ . The power consumed by the bulb when operated at $100 V$ is

1

25 W

2

50 W

3

75 W

4

100 W

Explanation:

A Given that,
$P = 100 W$
$V = 200 V$
We know that, $P = \frac{V^{2}}{R}$
$R = \frac{(200)^{2}}{100} = \frac{200 \times 200}{100} = 400 Ω$
When bulb is operated at $100 V$
Then, power consumed
$P = \frac{V^{2}}{R} = \frac{(100)^{2}}{400} = \frac{100 \times 100}{400} = 25 W$
$P = 25 W$

UPSEE 2019