QBManager

Alternating Current

155326 A resistor and an inductor are connected to an ac supply of $120 V$ and $50 Hz$ . The current in the circuit is $3 A$ . If the power consumed in the circuit is $108 W$ , then the resistance in the circuit is

1

12 Ω

2

40 Ω

3

\sqrt{(52 \times 25)} Ω

4

360 Ω

Explanation:

A Given that, $P = 108 W$

We know that,
$Power = I_{rms}^{2} R$
$R = \frac{108}{3^{2}}$
$R = 12 Ω$

BITSAT-2018

Alternating Current

155327 An alternating voltage $V = V_{0} \sin ω t$ is applied across a circuit. As a result, a current $I = I_{0} \sin$ $(ω t - π / 2)$ flows in it. The power consumed per cycle is

1 Zero

2

0.5 V_{0} I_{0}

3

0.707 V_{0} I_{0}

4

1.414 V_{0} I_{0}

Explanation:

A Given that, $V = V_{0} \sin ω t$

In the circuit,
$V = V_{o} \sin ω t$
$I = I_{0} \sin (ω t - \frac{π}{2})$
$ϕ = \frac{π}{2}$
So, power consumed,
$P = V_{rms} \times I_{rms} \times \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos \frac{π}{2}$
$P = 0$

Kerala CEE-2018

Alternating Current

155328 A 100 W electric bulb produces electromagnetic radiation with electric field amplitude of $\frac{2 V}{m}$ at a distance of $10 m$ . Assuming it as a point source, estimate the efficiency of the bulb.

1

4.9 %

2

2.5 %

3

6.6 %

4

19.7 %

Explanation:

C Given, $P = 100 W, E_{max} = 2 {Vm}^{- 1}, R = 10 m$ We know that, Intensity of radiation -
$I = \frac{Power}{Area} \times Efficiency$
$I = \frac{100}{4 \times π \times (10)^{2}} \times η$
Half by electric field,
$\frac{1}{2} I = \frac{1}{2} ε_{0} E_{ms}^{2} C$
We know that,
$E_{rms} = \frac{E_{max}}{\sqrt{2}}$
$I = ε_{0} E_{max}^{2} \times C$
$\frac{η}{4 π} = ε_{0} \times \frac{E_{max}^{2}}{2} \times C$
$η = \frac{4 π \times 8.85 \times 10^{- 12} \times 4 \times 3 \times 10^{8}}{2}$
$η = 0.0667 \times 100$
$η = 6.6 %$

TS- EAMCET-04.05.2018

Alternating Current

155329 An oscillating circuit consisting of a capacitor with capacitance $C = 10 μ F$ , a coil with inductance $L = 6.0 μ H$ and active resistance $R =$ $10 Ω$ . The mean power that should be fed to the circuit to maintain undamped harmonic oscillations with an external driving power with $50 Hz$ and $a V_{m}$ of $280 V$ is

1

3.8 W

2

48 W

3

3 mW

4

48 mW

Explanation:

A Given,
$R = 10 Ω$
$L = 6 H = 6 \times 10^{- 7} H$
$C = 10 μ F = 10 \times 10^{- 6} F$

We know that, impedance of the circuit -
$Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}$
Capacity reactance -
$X_{C} = \frac{1}{ω C} = \frac{1}{2 π fC}$
$X_{C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{- 6}}$
$= 318.4 Ω$
Inductive reactance $X_{L} = ω L$
$= 2 π f L$
$= 2 \times 3.14 \times 50 \times 6 \times 10^{- 6}$
$= 18.84 \times 10^{- 4} Ω$
$∴ Z = \sqrt{10^{2} + {(318.4 - 18.84 \times 10^{- 4})}^{2}}$
$Z = \sqrt{(100 + 318.39)^{2}}$
$Z = \sqrt{101477.36}$
$Z = 318.62 Ω$
$Therefore, I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{0} / \sqrt{2}}{Z}$
$I_{rms} = \frac{280}{\sqrt{2} \times 318.55}$
$I_{rms} = 0.621 A$
$So, P = I_{rms}^{2}, R$
$P = (0.621)^{2} \times 10$
$P = 3.8 W$

TS- EAMCET-04.05.2018

Alternating Current

155330 A coil has inductance of $0.4 H$ and resistance of $8 Ω$ . It is connected to an $AC$ source with peak emf $4 V$ and frequency $\frac{30}{π} Hz$ . The average power dissipated in the circuit is

1

1 W

2

0.5 W

3

0.3 W

4

0.1 W

Explanation:

D Given that,
$X_{L} = 0.4 H$
$R = 8 Ω$
$f = \frac{30}{π} Hz$
$V_{Peak} = 4 V$
Average power dissipated is
$P_{avg} = V_{rms} I_{rms} \cos ϕ$
$P_{avg} = \frac{V_{max}^{2}}{Z} \times \frac{R}{Z}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{R^{2} + X_{L}^{2}})}^{2}}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{R^{2} + (2 π {fL}^{2})})}^{2}}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{8^{2} + {(2 π \frac{30}{π} \times 0.4)}^{2}})}^{2}}$
$P_{avg} = \frac{8 \times 8}{64 + 576}$
$P_{avg} = \frac{64}{640} = \frac{1}{10} = 0.1 W$

TS- EAMCET-05.05.2018

Alternating Current

155326 A resistor and an inductor are connected to an ac supply of $120 V$ and $50 Hz$ . The current in the circuit is $3 A$ . If the power consumed in the circuit is $108 W$ , then the resistance in the circuit is

1

12 Ω

2

40 Ω

3

\sqrt{(52 \times 25)} Ω

4

360 Ω

Explanation:

A Given that, $P = 108 W$

We know that,
$Power = I_{rms}^{2} R$
$R = \frac{108}{3^{2}}$
$R = 12 Ω$

BITSAT-2018

Alternating Current

155327 An alternating voltage $V = V_{0} \sin ω t$ is applied across a circuit. As a result, a current $I = I_{0} \sin$ $(ω t - π / 2)$ flows in it. The power consumed per cycle is

1 Zero

2

0.5 V_{0} I_{0}

3

0.707 V_{0} I_{0}

4

1.414 V_{0} I_{0}

Explanation:

A Given that, $V = V_{0} \sin ω t$

In the circuit,
$V = V_{o} \sin ω t$
$I = I_{0} \sin (ω t - \frac{π}{2})$
$ϕ = \frac{π}{2}$
So, power consumed,
$P = V_{rms} \times I_{rms} \times \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos \frac{π}{2}$
$P = 0$

Kerala CEE-2018

Alternating Current

155328 A 100 W electric bulb produces electromagnetic radiation with electric field amplitude of $\frac{2 V}{m}$ at a distance of $10 m$ . Assuming it as a point source, estimate the efficiency of the bulb.

1

4.9 %

2

2.5 %

3

6.6 %

4

19.7 %

Explanation:

C Given, $P = 100 W, E_{max} = 2 {Vm}^{- 1}, R = 10 m$ We know that, Intensity of radiation -
$I = \frac{Power}{Area} \times Efficiency$
$I = \frac{100}{4 \times π \times (10)^{2}} \times η$
Half by electric field,
$\frac{1}{2} I = \frac{1}{2} ε_{0} E_{ms}^{2} C$
We know that,
$E_{rms} = \frac{E_{max}}{\sqrt{2}}$
$I = ε_{0} E_{max}^{2} \times C$
$\frac{η}{4 π} = ε_{0} \times \frac{E_{max}^{2}}{2} \times C$
$η = \frac{4 π \times 8.85 \times 10^{- 12} \times 4 \times 3 \times 10^{8}}{2}$
$η = 0.0667 \times 100$
$η = 6.6 %$

TS- EAMCET-04.05.2018

Alternating Current

155329 An oscillating circuit consisting of a capacitor with capacitance $C = 10 μ F$ , a coil with inductance $L = 6.0 μ H$ and active resistance $R =$ $10 Ω$ . The mean power that should be fed to the circuit to maintain undamped harmonic oscillations with an external driving power with $50 Hz$ and $a V_{m}$ of $280 V$ is

1

3.8 W

2

48 W

3

3 mW

4

48 mW

Explanation:

A Given,
$R = 10 Ω$
$L = 6 H = 6 \times 10^{- 7} H$
$C = 10 μ F = 10 \times 10^{- 6} F$

We know that, impedance of the circuit -
$Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}$
Capacity reactance -
$X_{C} = \frac{1}{ω C} = \frac{1}{2 π fC}$
$X_{C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{- 6}}$
$= 318.4 Ω$
Inductive reactance $X_{L} = ω L$
$= 2 π f L$
$= 2 \times 3.14 \times 50 \times 6 \times 10^{- 6}$
$= 18.84 \times 10^{- 4} Ω$
$∴ Z = \sqrt{10^{2} + {(318.4 - 18.84 \times 10^{- 4})}^{2}}$
$Z = \sqrt{(100 + 318.39)^{2}}$
$Z = \sqrt{101477.36}$
$Z = 318.62 Ω$
$Therefore, I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{0} / \sqrt{2}}{Z}$
$I_{rms} = \frac{280}{\sqrt{2} \times 318.55}$
$I_{rms} = 0.621 A$
$So, P = I_{rms}^{2}, R$
$P = (0.621)^{2} \times 10$
$P = 3.8 W$

TS- EAMCET-04.05.2018

Alternating Current

155330 A coil has inductance of $0.4 H$ and resistance of $8 Ω$ . It is connected to an $AC$ source with peak emf $4 V$ and frequency $\frac{30}{π} Hz$ . The average power dissipated in the circuit is

1

1 W

2

0.5 W

3

0.3 W

4

0.1 W

Explanation:

D Given that,
$X_{L} = 0.4 H$
$R = 8 Ω$
$f = \frac{30}{π} Hz$
$V_{Peak} = 4 V$
Average power dissipated is
$P_{avg} = V_{rms} I_{rms} \cos ϕ$
$P_{avg} = \frac{V_{max}^{2}}{Z} \times \frac{R}{Z}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{R^{2} + X_{L}^{2}})}^{2}}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{R^{2} + (2 π {fL}^{2})})}^{2}}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{8^{2} + {(2 π \frac{30}{π} \times 0.4)}^{2}})}^{2}}$
$P_{avg} = \frac{8 \times 8}{64 + 576}$
$P_{avg} = \frac{64}{640} = \frac{1}{10} = 0.1 W$

TS- EAMCET-05.05.2018

Alternating Current

155326 A resistor and an inductor are connected to an ac supply of $120 V$ and $50 Hz$ . The current in the circuit is $3 A$ . If the power consumed in the circuit is $108 W$ , then the resistance in the circuit is

1

12 Ω

2

40 Ω

3

\sqrt{(52 \times 25)} Ω

4

360 Ω

Explanation:

A Given that, $P = 108 W$

We know that,
$Power = I_{rms}^{2} R$
$R = \frac{108}{3^{2}}$
$R = 12 Ω$

BITSAT-2018

Alternating Current

155327 An alternating voltage $V = V_{0} \sin ω t$ is applied across a circuit. As a result, a current $I = I_{0} \sin$ $(ω t - π / 2)$ flows in it. The power consumed per cycle is

1 Zero

2

0.5 V_{0} I_{0}

3

0.707 V_{0} I_{0}

4

1.414 V_{0} I_{0}

Explanation:

A Given that, $V = V_{0} \sin ω t$

In the circuit,
$V = V_{o} \sin ω t$
$I = I_{0} \sin (ω t - \frac{π}{2})$
$ϕ = \frac{π}{2}$
So, power consumed,
$P = V_{rms} \times I_{rms} \times \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos \frac{π}{2}$
$P = 0$

Kerala CEE-2018

Alternating Current

155328 A 100 W electric bulb produces electromagnetic radiation with electric field amplitude of $\frac{2 V}{m}$ at a distance of $10 m$ . Assuming it as a point source, estimate the efficiency of the bulb.

1

4.9 %

2

2.5 %

3

6.6 %

4

19.7 %

Explanation:

C Given, $P = 100 W, E_{max} = 2 {Vm}^{- 1}, R = 10 m$ We know that, Intensity of radiation -
$I = \frac{Power}{Area} \times Efficiency$
$I = \frac{100}{4 \times π \times (10)^{2}} \times η$
Half by electric field,
$\frac{1}{2} I = \frac{1}{2} ε_{0} E_{ms}^{2} C$
We know that,
$E_{rms} = \frac{E_{max}}{\sqrt{2}}$
$I = ε_{0} E_{max}^{2} \times C$
$\frac{η}{4 π} = ε_{0} \times \frac{E_{max}^{2}}{2} \times C$
$η = \frac{4 π \times 8.85 \times 10^{- 12} \times 4 \times 3 \times 10^{8}}{2}$
$η = 0.0667 \times 100$
$η = 6.6 %$

TS- EAMCET-04.05.2018

Alternating Current

155329 An oscillating circuit consisting of a capacitor with capacitance $C = 10 μ F$ , a coil with inductance $L = 6.0 μ H$ and active resistance $R =$ $10 Ω$ . The mean power that should be fed to the circuit to maintain undamped harmonic oscillations with an external driving power with $50 Hz$ and $a V_{m}$ of $280 V$ is

1

3.8 W

2

48 W

3

3 mW

4

48 mW

Explanation:

A Given,
$R = 10 Ω$
$L = 6 H = 6 \times 10^{- 7} H$
$C = 10 μ F = 10 \times 10^{- 6} F$

We know that, impedance of the circuit -
$Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}$
Capacity reactance -
$X_{C} = \frac{1}{ω C} = \frac{1}{2 π fC}$
$X_{C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{- 6}}$
$= 318.4 Ω$
Inductive reactance $X_{L} = ω L$
$= 2 π f L$
$= 2 \times 3.14 \times 50 \times 6 \times 10^{- 6}$
$= 18.84 \times 10^{- 4} Ω$
$∴ Z = \sqrt{10^{2} + {(318.4 - 18.84 \times 10^{- 4})}^{2}}$
$Z = \sqrt{(100 + 318.39)^{2}}$
$Z = \sqrt{101477.36}$
$Z = 318.62 Ω$
$Therefore, I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{0} / \sqrt{2}}{Z}$
$I_{rms} = \frac{280}{\sqrt{2} \times 318.55}$
$I_{rms} = 0.621 A$
$So, P = I_{rms}^{2}, R$
$P = (0.621)^{2} \times 10$
$P = 3.8 W$

TS- EAMCET-04.05.2018

Alternating Current

155330 A coil has inductance of $0.4 H$ and resistance of $8 Ω$ . It is connected to an $AC$ source with peak emf $4 V$ and frequency $\frac{30}{π} Hz$ . The average power dissipated in the circuit is

1

1 W

2

0.5 W

3

0.3 W

4

0.1 W

Explanation:

D Given that,
$X_{L} = 0.4 H$
$R = 8 Ω$
$f = \frac{30}{π} Hz$
$V_{Peak} = 4 V$
Average power dissipated is
$P_{avg} = V_{rms} I_{rms} \cos ϕ$
$P_{avg} = \frac{V_{max}^{2}}{Z} \times \frac{R}{Z}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{R^{2} + X_{L}^{2}})}^{2}}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{R^{2} + (2 π {fL}^{2})})}^{2}}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{8^{2} + {(2 π \frac{30}{π} \times 0.4)}^{2}})}^{2}}$
$P_{avg} = \frac{8 \times 8}{64 + 576}$
$P_{avg} = \frac{64}{640} = \frac{1}{10} = 0.1 W$

TS- EAMCET-05.05.2018

Alternating Current

155326 A resistor and an inductor are connected to an ac supply of $120 V$ and $50 Hz$ . The current in the circuit is $3 A$ . If the power consumed in the circuit is $108 W$ , then the resistance in the circuit is

1

12 Ω

2

40 Ω

3

\sqrt{(52 \times 25)} Ω

4

360 Ω

Explanation:

A Given that, $P = 108 W$

We know that,
$Power = I_{rms}^{2} R$
$R = \frac{108}{3^{2}}$
$R = 12 Ω$

BITSAT-2018

Alternating Current

155327 An alternating voltage $V = V_{0} \sin ω t$ is applied across a circuit. As a result, a current $I = I_{0} \sin$ $(ω t - π / 2)$ flows in it. The power consumed per cycle is

1 Zero

2

0.5 V_{0} I_{0}

3

0.707 V_{0} I_{0}

4

1.414 V_{0} I_{0}

Explanation:

A Given that, $V = V_{0} \sin ω t$

In the circuit,
$V = V_{o} \sin ω t$
$I = I_{0} \sin (ω t - \frac{π}{2})$
$ϕ = \frac{π}{2}$
So, power consumed,
$P = V_{rms} \times I_{rms} \times \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos \frac{π}{2}$
$P = 0$

Kerala CEE-2018

Alternating Current

155328 A 100 W electric bulb produces electromagnetic radiation with electric field amplitude of $\frac{2 V}{m}$ at a distance of $10 m$ . Assuming it as a point source, estimate the efficiency of the bulb.

1

4.9 %

2

2.5 %

3

6.6 %

4

19.7 %

Explanation:

C Given, $P = 100 W, E_{max} = 2 {Vm}^{- 1}, R = 10 m$ We know that, Intensity of radiation -
$I = \frac{Power}{Area} \times Efficiency$
$I = \frac{100}{4 \times π \times (10)^{2}} \times η$
Half by electric field,
$\frac{1}{2} I = \frac{1}{2} ε_{0} E_{ms}^{2} C$
We know that,
$E_{rms} = \frac{E_{max}}{\sqrt{2}}$
$I = ε_{0} E_{max}^{2} \times C$
$\frac{η}{4 π} = ε_{0} \times \frac{E_{max}^{2}}{2} \times C$
$η = \frac{4 π \times 8.85 \times 10^{- 12} \times 4 \times 3 \times 10^{8}}{2}$
$η = 0.0667 \times 100$
$η = 6.6 %$

TS- EAMCET-04.05.2018

Alternating Current

155329 An oscillating circuit consisting of a capacitor with capacitance $C = 10 μ F$ , a coil with inductance $L = 6.0 μ H$ and active resistance $R =$ $10 Ω$ . The mean power that should be fed to the circuit to maintain undamped harmonic oscillations with an external driving power with $50 Hz$ and $a V_{m}$ of $280 V$ is

1

3.8 W

2

48 W

3

3 mW

4

48 mW

Explanation:

A Given,
$R = 10 Ω$
$L = 6 H = 6 \times 10^{- 7} H$
$C = 10 μ F = 10 \times 10^{- 6} F$

We know that, impedance of the circuit -
$Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}$
Capacity reactance -
$X_{C} = \frac{1}{ω C} = \frac{1}{2 π fC}$
$X_{C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{- 6}}$
$= 318.4 Ω$
Inductive reactance $X_{L} = ω L$
$= 2 π f L$
$= 2 \times 3.14 \times 50 \times 6 \times 10^{- 6}$
$= 18.84 \times 10^{- 4} Ω$
$∴ Z = \sqrt{10^{2} + {(318.4 - 18.84 \times 10^{- 4})}^{2}}$
$Z = \sqrt{(100 + 318.39)^{2}}$
$Z = \sqrt{101477.36}$
$Z = 318.62 Ω$
$Therefore, I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{0} / \sqrt{2}}{Z}$
$I_{rms} = \frac{280}{\sqrt{2} \times 318.55}$
$I_{rms} = 0.621 A$
$So, P = I_{rms}^{2}, R$
$P = (0.621)^{2} \times 10$
$P = 3.8 W$

TS- EAMCET-04.05.2018

Alternating Current

155330 A coil has inductance of $0.4 H$ and resistance of $8 Ω$ . It is connected to an $AC$ source with peak emf $4 V$ and frequency $\frac{30}{π} Hz$ . The average power dissipated in the circuit is

1

1 W

2

0.5 W

3

0.3 W

4

0.1 W

Explanation:

D Given that,
$X_{L} = 0.4 H$
$R = 8 Ω$
$f = \frac{30}{π} Hz$
$V_{Peak} = 4 V$
Average power dissipated is
$P_{avg} = V_{rms} I_{rms} \cos ϕ$
$P_{avg} = \frac{V_{max}^{2}}{Z} \times \frac{R}{Z}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{R^{2} + X_{L}^{2}})}^{2}}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{R^{2} + (2 π {fL}^{2})})}^{2}}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{8^{2} + {(2 π \frac{30}{π} \times 0.4)}^{2}})}^{2}}$
$P_{avg} = \frac{8 \times 8}{64 + 576}$
$P_{avg} = \frac{64}{640} = \frac{1}{10} = 0.1 W$

TS- EAMCET-05.05.2018

Alternating Current

155326 A resistor and an inductor are connected to an ac supply of $120 V$ and $50 Hz$ . The current in the circuit is $3 A$ . If the power consumed in the circuit is $108 W$ , then the resistance in the circuit is

1

12 Ω

2

40 Ω

3

\sqrt{(52 \times 25)} Ω

4

360 Ω

Explanation:

A Given that, $P = 108 W$

We know that,
$Power = I_{rms}^{2} R$
$R = \frac{108}{3^{2}}$
$R = 12 Ω$

BITSAT-2018

Alternating Current

155327 An alternating voltage $V = V_{0} \sin ω t$ is applied across a circuit. As a result, a current $I = I_{0} \sin$ $(ω t - π / 2)$ flows in it. The power consumed per cycle is

1 Zero

2

0.5 V_{0} I_{0}

3

0.707 V_{0} I_{0}

4

1.414 V_{0} I_{0}

Explanation:

A Given that, $V = V_{0} \sin ω t$

In the circuit,
$V = V_{o} \sin ω t$
$I = I_{0} \sin (ω t - \frac{π}{2})$
$ϕ = \frac{π}{2}$
So, power consumed,
$P = V_{rms} \times I_{rms} \times \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos ϕ$
$P = \frac{I_{0} V_{0}}{2} \cos \frac{π}{2}$
$P = 0$

Kerala CEE-2018

Alternating Current

155328 A 100 W electric bulb produces electromagnetic radiation with electric field amplitude of $\frac{2 V}{m}$ at a distance of $10 m$ . Assuming it as a point source, estimate the efficiency of the bulb.

1

4.9 %

2

2.5 %

3

6.6 %

4

19.7 %

Explanation:

C Given, $P = 100 W, E_{max} = 2 {Vm}^{- 1}, R = 10 m$ We know that, Intensity of radiation -
$I = \frac{Power}{Area} \times Efficiency$
$I = \frac{100}{4 \times π \times (10)^{2}} \times η$
Half by electric field,
$\frac{1}{2} I = \frac{1}{2} ε_{0} E_{ms}^{2} C$
We know that,
$E_{rms} = \frac{E_{max}}{\sqrt{2}}$
$I = ε_{0} E_{max}^{2} \times C$
$\frac{η}{4 π} = ε_{0} \times \frac{E_{max}^{2}}{2} \times C$
$η = \frac{4 π \times 8.85 \times 10^{- 12} \times 4 \times 3 \times 10^{8}}{2}$
$η = 0.0667 \times 100$
$η = 6.6 %$

TS- EAMCET-04.05.2018

Alternating Current

155329 An oscillating circuit consisting of a capacitor with capacitance $C = 10 μ F$ , a coil with inductance $L = 6.0 μ H$ and active resistance $R =$ $10 Ω$ . The mean power that should be fed to the circuit to maintain undamped harmonic oscillations with an external driving power with $50 Hz$ and $a V_{m}$ of $280 V$ is

1

3.8 W

2

48 W

3

3 mW

4

48 mW

Explanation:

A Given,
$R = 10 Ω$
$L = 6 H = 6 \times 10^{- 7} H$
$C = 10 μ F = 10 \times 10^{- 6} F$

We know that, impedance of the circuit -
$Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}$
Capacity reactance -
$X_{C} = \frac{1}{ω C} = \frac{1}{2 π fC}$
$X_{C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{- 6}}$
$= 318.4 Ω$
Inductive reactance $X_{L} = ω L$
$= 2 π f L$
$= 2 \times 3.14 \times 50 \times 6 \times 10^{- 6}$
$= 18.84 \times 10^{- 4} Ω$
$∴ Z = \sqrt{10^{2} + {(318.4 - 18.84 \times 10^{- 4})}^{2}}$
$Z = \sqrt{(100 + 318.39)^{2}}$
$Z = \sqrt{101477.36}$
$Z = 318.62 Ω$
$Therefore, I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{0} / \sqrt{2}}{Z}$
$I_{rms} = \frac{280}{\sqrt{2} \times 318.55}$
$I_{rms} = 0.621 A$
$So, P = I_{rms}^{2}, R$
$P = (0.621)^{2} \times 10$
$P = 3.8 W$

TS- EAMCET-04.05.2018

Alternating Current

155330 A coil has inductance of $0.4 H$ and resistance of $8 Ω$ . It is connected to an $AC$ source with peak emf $4 V$ and frequency $\frac{30}{π} Hz$ . The average power dissipated in the circuit is

1

1 W

2

0.5 W

3

0.3 W

4

0.1 W

Explanation:

D Given that,
$X_{L} = 0.4 H$
$R = 8 Ω$
$f = \frac{30}{π} Hz$
$V_{Peak} = 4 V$
Average power dissipated is
$P_{avg} = V_{rms} I_{rms} \cos ϕ$
$P_{avg} = \frac{V_{max}^{2}}{Z} \times \frac{R}{Z}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{R^{2} + X_{L}^{2}})}^{2}}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{R^{2} + (2 π {fL}^{2})})}^{2}}$
$P_{avg} = \frac{(4 / \sqrt{2})^{2} \times 8}{{(\sqrt{8^{2} + {(2 π \frac{30}{π} \times 0.4)}^{2}})}^{2}}$
$P_{avg} = \frac{8 \times 8}{64 + 576}$
$P_{avg} = \frac{64}{640} = \frac{1}{10} = 0.1 W$

TS- EAMCET-05.05.2018