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Alternating Current

155321 A current $I = I_{0} \sin (ω t - \frac{π}{2})$ flow in an $A C$ circuit, if potential $E = E_{0} \sin ω t$ has been applied. The power consumption in the circuit will be

1

\frac{E_{0} I_{0}}{\sqrt{2}}

2

\frac{EI}{\sqrt{2}}

3

\frac{E_{0} I_{0}}{2}

4 Zero

Explanation:

D Given,
$I = I_{0} \sin (ω t - \frac{π}{2})$
$E = E_{0} \sin ω t$
On comparing equation, $E = E_{0} \sin (ω t)$ and $I = I_{0} \sin (ω t$ $+ ϕ)$
Then, $ϕ = \frac{π}{2} = 90^{\circ}$
WE know that, power consumption is -
$P = \frac{I_{0} V_{0}}{2} \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos 90^{\circ}$
$P = 0$

TS EAMCET 08.05.2019

Alternating Current

155322 A resistor and an inductor are connected in series to an $A C$ source of voltage $150 \sin (100 π t + π)$ volt. If the current in the circuit is $5 \sin (100 π t + \frac{2 π}{3})$ ampere, then the average power dissipated and the resistance of the resistor are respectively

1

187.5 W, 30 Ω

2

187.5 W, 15 Ω

3

375 W, 30 Ω

4

375 W, 15 Ω

Explanation:

B Given,
$V = 150 \sin (100 π t + π) V$
Comparing this equation, $V = V_{o} \sin (ω t + ϕ_{1})$
Then, $V_{o} = 150, ϕ_{1} = π$
$I = 5 \sin (100 π t + \frac{2 π}{3}) A$
Comparing this equation, $I = I_{o} \sin (ω t + ϕ_{2})$
Then, $I_{o} = 5 A, ϕ_{2} = \frac{2 π}{3}$
$∴ Δ ϕ = ϕ = ϕ_{1} - ϕ_{2}$
$ϕ = \frac{π}{3}$
Now, average power in R-L circuit -
$P_{av} = V_{rms} \times I_{rms} \times \cos ϕ$
$P_{av} = \frac{V_{o}}{\sqrt{2}} \times \frac{I_{o}}{\sqrt{2}} \cos 60^{\circ}$
$P_{av} = \frac{V_{o}}{\sqrt{2}} \times \frac{I_{o}}{\sqrt{2}} \cos 60^{\circ} = \frac{150 \times 5}{2} \times \frac{1}{2}$
$P_{av} = 187.5 W$
Impendence
$Z = \frac{V_{0}}{I_{0}} = \frac{150}{5}$
$Z = 30 Ω$
$\frac{R}{Z} = \cos ϕ$
$R = 30 \cos 60^{\circ} = 30 \times \frac{1}{2}$
$R = 15 Ω$

AP EAMCET (20.04.2019) Shift-1

Alternating Current

155323 A small town is located $10 km$ away from a power plant. An average of $120 kW$ of electric power is sent to this town. The transmission lines have a total resistance of $0.40 Ω$ . Calculate the power loss, if the power is transmitted at $240 V$ .

1

100 W

2

10 W

3

100 kW

4

10 kW

Explanation:

C Given,
$P_{avg} = 120 kW$
$R = 0.40 Ω$
$V = 240 V$
RMS value of current
$I = \frac{P}{V} = \frac{120}{240} \times 10^{3}$
$I = 500 A$
Power loss if the power is transmitted at $240 V -$
$P_{loss} = I^{2} R$
$P_{loss} = (500)^{2} \times 0.40$
$P_{loss} = 100 \times 10^{3} W$
$P_{loss} = 100 kW$

J and K CET-2018

Alternating Current

155325 In a LCR series resonating circuit, the value of average power loss is :

1

\frac{V_{rms} I_{rms} R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}

2

\frac{V_{rms} I_{rms}}{R \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}

3

\frac{V_{rms} I_{rms}}{\sqrt{R^{2} + X_{L}^{2}}}

4

\frac{V_{rms} I_{rms}}{\sqrt{R^{2} + {(X_{L} + X_{C})}^{2}}}

Explanation:

A
We know that, impedance -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
Average power $(P_{avg}) = V_{rms} \times I_{rms} \cos ϕ$ $∵ \cos ϕ = \frac{R}{Z}$
$P_{avg} = \frac{V_{rms} I_{rms} R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}$

AIIMS-26.05.2018(M)]**#

Alternating Current

155321 A current $I = I_{0} \sin (ω t - \frac{π}{2})$ flow in an $A C$ circuit, if potential $E = E_{0} \sin ω t$ has been applied. The power consumption in the circuit will be

1

\frac{E_{0} I_{0}}{\sqrt{2}}

2

\frac{EI}{\sqrt{2}}

3

\frac{E_{0} I_{0}}{2}

4 Zero

Explanation:

D Given,
$I = I_{0} \sin (ω t - \frac{π}{2})$
$E = E_{0} \sin ω t$
On comparing equation, $E = E_{0} \sin (ω t)$ and $I = I_{0} \sin (ω t$ $+ ϕ)$
Then, $ϕ = \frac{π}{2} = 90^{\circ}$
WE know that, power consumption is -
$P = \frac{I_{0} V_{0}}{2} \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos 90^{\circ}$
$P = 0$

TS EAMCET 08.05.2019

Alternating Current

155322 A resistor and an inductor are connected in series to an $A C$ source of voltage $150 \sin (100 π t + π)$ volt. If the current in the circuit is $5 \sin (100 π t + \frac{2 π}{3})$ ampere, then the average power dissipated and the resistance of the resistor are respectively

1

187.5 W, 30 Ω

2

187.5 W, 15 Ω

3

375 W, 30 Ω

4

375 W, 15 Ω

Explanation:

B Given,
$V = 150 \sin (100 π t + π) V$
Comparing this equation, $V = V_{o} \sin (ω t + ϕ_{1})$
Then, $V_{o} = 150, ϕ_{1} = π$
$I = 5 \sin (100 π t + \frac{2 π}{3}) A$
Comparing this equation, $I = I_{o} \sin (ω t + ϕ_{2})$
Then, $I_{o} = 5 A, ϕ_{2} = \frac{2 π}{3}$
$∴ Δ ϕ = ϕ = ϕ_{1} - ϕ_{2}$
$ϕ = \frac{π}{3}$
Now, average power in R-L circuit -
$P_{av} = V_{rms} \times I_{rms} \times \cos ϕ$
$P_{av} = \frac{V_{o}}{\sqrt{2}} \times \frac{I_{o}}{\sqrt{2}} \cos 60^{\circ}$
$P_{av} = \frac{V_{o}}{\sqrt{2}} \times \frac{I_{o}}{\sqrt{2}} \cos 60^{\circ} = \frac{150 \times 5}{2} \times \frac{1}{2}$
$P_{av} = 187.5 W$
Impendence
$Z = \frac{V_{0}}{I_{0}} = \frac{150}{5}$
$Z = 30 Ω$
$\frac{R}{Z} = \cos ϕ$
$R = 30 \cos 60^{\circ} = 30 \times \frac{1}{2}$
$R = 15 Ω$

AP EAMCET (20.04.2019) Shift-1

Alternating Current

155323 A small town is located $10 km$ away from a power plant. An average of $120 kW$ of electric power is sent to this town. The transmission lines have a total resistance of $0.40 Ω$ . Calculate the power loss, if the power is transmitted at $240 V$ .

1

100 W

2

10 W

3

100 kW

4

10 kW

Explanation:

C Given,
$P_{avg} = 120 kW$
$R = 0.40 Ω$
$V = 240 V$
RMS value of current
$I = \frac{P}{V} = \frac{120}{240} \times 10^{3}$
$I = 500 A$
Power loss if the power is transmitted at $240 V -$
$P_{loss} = I^{2} R$
$P_{loss} = (500)^{2} \times 0.40$
$P_{loss} = 100 \times 10^{3} W$
$P_{loss} = 100 kW$

J and K CET-2018

Alternating Current

155325 In a LCR series resonating circuit, the value of average power loss is :

1

\frac{V_{rms} I_{rms} R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}

2

\frac{V_{rms} I_{rms}}{R \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}

3

\frac{V_{rms} I_{rms}}{\sqrt{R^{2} + X_{L}^{2}}}

4

\frac{V_{rms} I_{rms}}{\sqrt{R^{2} + {(X_{L} + X_{C})}^{2}}}

Explanation:

A
We know that, impedance -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
Average power $(P_{avg}) = V_{rms} \times I_{rms} \cos ϕ$ $∵ \cos ϕ = \frac{R}{Z}$
$P_{avg} = \frac{V_{rms} I_{rms} R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}$

AIIMS-26.05.2018(M)]**#

Alternating Current

155321 A current $I = I_{0} \sin (ω t - \frac{π}{2})$ flow in an $A C$ circuit, if potential $E = E_{0} \sin ω t$ has been applied. The power consumption in the circuit will be

1

\frac{E_{0} I_{0}}{\sqrt{2}}

2

\frac{EI}{\sqrt{2}}

3

\frac{E_{0} I_{0}}{2}

4 Zero

Explanation:

D Given,
$I = I_{0} \sin (ω t - \frac{π}{2})$
$E = E_{0} \sin ω t$
On comparing equation, $E = E_{0} \sin (ω t)$ and $I = I_{0} \sin (ω t$ $+ ϕ)$
Then, $ϕ = \frac{π}{2} = 90^{\circ}$
WE know that, power consumption is -
$P = \frac{I_{0} V_{0}}{2} \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos 90^{\circ}$
$P = 0$

TS EAMCET 08.05.2019

Alternating Current

155322 A resistor and an inductor are connected in series to an $A C$ source of voltage $150 \sin (100 π t + π)$ volt. If the current in the circuit is $5 \sin (100 π t + \frac{2 π}{3})$ ampere, then the average power dissipated and the resistance of the resistor are respectively

1

187.5 W, 30 Ω

2

187.5 W, 15 Ω

3

375 W, 30 Ω

4

375 W, 15 Ω

Explanation:

B Given,
$V = 150 \sin (100 π t + π) V$
Comparing this equation, $V = V_{o} \sin (ω t + ϕ_{1})$
Then, $V_{o} = 150, ϕ_{1} = π$
$I = 5 \sin (100 π t + \frac{2 π}{3}) A$
Comparing this equation, $I = I_{o} \sin (ω t + ϕ_{2})$
Then, $I_{o} = 5 A, ϕ_{2} = \frac{2 π}{3}$
$∴ Δ ϕ = ϕ = ϕ_{1} - ϕ_{2}$
$ϕ = \frac{π}{3}$
Now, average power in R-L circuit -
$P_{av} = V_{rms} \times I_{rms} \times \cos ϕ$
$P_{av} = \frac{V_{o}}{\sqrt{2}} \times \frac{I_{o}}{\sqrt{2}} \cos 60^{\circ}$
$P_{av} = \frac{V_{o}}{\sqrt{2}} \times \frac{I_{o}}{\sqrt{2}} \cos 60^{\circ} = \frac{150 \times 5}{2} \times \frac{1}{2}$
$P_{av} = 187.5 W$
Impendence
$Z = \frac{V_{0}}{I_{0}} = \frac{150}{5}$
$Z = 30 Ω$
$\frac{R}{Z} = \cos ϕ$
$R = 30 \cos 60^{\circ} = 30 \times \frac{1}{2}$
$R = 15 Ω$

AP EAMCET (20.04.2019) Shift-1

Alternating Current

155323 A small town is located $10 km$ away from a power plant. An average of $120 kW$ of electric power is sent to this town. The transmission lines have a total resistance of $0.40 Ω$ . Calculate the power loss, if the power is transmitted at $240 V$ .

1

100 W

2

10 W

3

100 kW

4

10 kW

Explanation:

C Given,
$P_{avg} = 120 kW$
$R = 0.40 Ω$
$V = 240 V$
RMS value of current
$I = \frac{P}{V} = \frac{120}{240} \times 10^{3}$
$I = 500 A$
Power loss if the power is transmitted at $240 V -$
$P_{loss} = I^{2} R$
$P_{loss} = (500)^{2} \times 0.40$
$P_{loss} = 100 \times 10^{3} W$
$P_{loss} = 100 kW$

J and K CET-2018

Alternating Current

155325 In a LCR series resonating circuit, the value of average power loss is :

1

\frac{V_{rms} I_{rms} R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}

2

\frac{V_{rms} I_{rms}}{R \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}

3

\frac{V_{rms} I_{rms}}{\sqrt{R^{2} + X_{L}^{2}}}

4

\frac{V_{rms} I_{rms}}{\sqrt{R^{2} + {(X_{L} + X_{C})}^{2}}}

Explanation:

A
We know that, impedance -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
Average power $(P_{avg}) = V_{rms} \times I_{rms} \cos ϕ$ $∵ \cos ϕ = \frac{R}{Z}$
$P_{avg} = \frac{V_{rms} I_{rms} R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}$

AIIMS-26.05.2018(M)]**#

Alternating Current

155321 A current $I = I_{0} \sin (ω t - \frac{π}{2})$ flow in an $A C$ circuit, if potential $E = E_{0} \sin ω t$ has been applied. The power consumption in the circuit will be

1

\frac{E_{0} I_{0}}{\sqrt{2}}

2

\frac{EI}{\sqrt{2}}

3

\frac{E_{0} I_{0}}{2}

4 Zero

Explanation:

D Given,
$I = I_{0} \sin (ω t - \frac{π}{2})$
$E = E_{0} \sin ω t$
On comparing equation, $E = E_{0} \sin (ω t)$ and $I = I_{0} \sin (ω t$ $+ ϕ)$
Then, $ϕ = \frac{π}{2} = 90^{\circ}$
WE know that, power consumption is -
$P = \frac{I_{0} V_{0}}{2} \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos 90^{\circ}$
$P = 0$

TS EAMCET 08.05.2019

Alternating Current

155322 A resistor and an inductor are connected in series to an $A C$ source of voltage $150 \sin (100 π t + π)$ volt. If the current in the circuit is $5 \sin (100 π t + \frac{2 π}{3})$ ampere, then the average power dissipated and the resistance of the resistor are respectively

1

187.5 W, 30 Ω

2

187.5 W, 15 Ω

3

375 W, 30 Ω

4

375 W, 15 Ω

Explanation:

B Given,
$V = 150 \sin (100 π t + π) V$
Comparing this equation, $V = V_{o} \sin (ω t + ϕ_{1})$
Then, $V_{o} = 150, ϕ_{1} = π$
$I = 5 \sin (100 π t + \frac{2 π}{3}) A$
Comparing this equation, $I = I_{o} \sin (ω t + ϕ_{2})$
Then, $I_{o} = 5 A, ϕ_{2} = \frac{2 π}{3}$
$∴ Δ ϕ = ϕ = ϕ_{1} - ϕ_{2}$
$ϕ = \frac{π}{3}$
Now, average power in R-L circuit -
$P_{av} = V_{rms} \times I_{rms} \times \cos ϕ$
$P_{av} = \frac{V_{o}}{\sqrt{2}} \times \frac{I_{o}}{\sqrt{2}} \cos 60^{\circ}$
$P_{av} = \frac{V_{o}}{\sqrt{2}} \times \frac{I_{o}}{\sqrt{2}} \cos 60^{\circ} = \frac{150 \times 5}{2} \times \frac{1}{2}$
$P_{av} = 187.5 W$
Impendence
$Z = \frac{V_{0}}{I_{0}} = \frac{150}{5}$
$Z = 30 Ω$
$\frac{R}{Z} = \cos ϕ$
$R = 30 \cos 60^{\circ} = 30 \times \frac{1}{2}$
$R = 15 Ω$

AP EAMCET (20.04.2019) Shift-1

Alternating Current

155323 A small town is located $10 km$ away from a power plant. An average of $120 kW$ of electric power is sent to this town. The transmission lines have a total resistance of $0.40 Ω$ . Calculate the power loss, if the power is transmitted at $240 V$ .

1

100 W

2

10 W

3

100 kW

4

10 kW

Explanation:

C Given,
$P_{avg} = 120 kW$
$R = 0.40 Ω$
$V = 240 V$
RMS value of current
$I = \frac{P}{V} = \frac{120}{240} \times 10^{3}$
$I = 500 A$
Power loss if the power is transmitted at $240 V -$
$P_{loss} = I^{2} R$
$P_{loss} = (500)^{2} \times 0.40$
$P_{loss} = 100 \times 10^{3} W$
$P_{loss} = 100 kW$

J and K CET-2018

Alternating Current

155325 In a LCR series resonating circuit, the value of average power loss is :

1

\frac{V_{rms} I_{rms} R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}

2

\frac{V_{rms} I_{rms}}{R \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}

3

\frac{V_{rms} I_{rms}}{\sqrt{R^{2} + X_{L}^{2}}}

4

\frac{V_{rms} I_{rms}}{\sqrt{R^{2} + {(X_{L} + X_{C})}^{2}}}

Explanation:

A
We know that, impedance -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
Average power $(P_{avg}) = V_{rms} \times I_{rms} \cos ϕ$ $∵ \cos ϕ = \frac{R}{Z}$
$P_{avg} = \frac{V_{rms} I_{rms} R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}}$

AIIMS-26.05.2018(M)]**#