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PHXII07:ALTERNATING CURRENT

356162 A sinusoidal $A C$ current flows through a resistor of resistance $R$ . If the peak current is $I_{p,}$ then the power dissipated is

1

\frac{4}{π} I_{p}^{2} R

2

\frac{1}{π} I_{p}^{2} R

3

\frac{1}{2} I_{p}^{2} R

4

I_{p}^{2} R \cos θ

Explanation:

$P o w e r = I_{r m s}^{2} R = {(\frac{I_{p}}{\sqrt{2}})}^{2} R = \frac{I_{p}^{2} R}{2}$

PHXII07:ALTERNATING CURRENT

356163 A choke coil has negligible resistance. The alternating potential drop across it is $220 v o l t s$ and the current is $5 m A .$ The power consumed is

1

220 \times \frac{5}{1000} W

2

\frac{220}{5} W

3 Zero

4

2.20 \times 5 W

Explanation:

Phase angle $= \frac{π}{2}$ (between $V$ and $I$ )
$\Rightarrow \cos ϕ = 0$
$\Rightarrow \cos ϕ = 0$
$\Rightarrow ⟨ P ⟩ = 0$
Since $⟨ P ⟩ = V I \cos ϕ$
Average power is zero.
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356164 An electric heater rated $225 V$ and $550 W$ is connected to $A C$ mains. The current drawn by it is:

1

0.8 A

2

2.0 A

3

0.4 A

4

1.25 A

Explanation:

$∵ P = V I \Rightarrow I = \frac{P}{V} = \frac{550}{225} = 2.0 A$

PHXII07:ALTERNATING CURRENT

356165 A 100 $W$ /200 $V$ bulb and an inductor are connected in series to a $220 V / 50 H z$ supply. Find the power consumed by the bulb. $(L = 1 H)$

1

74 W

2

84 W

3

100 W

4

92 W

Explanation:

Given $R = \frac{{(200)}^{2}}{100} = 400 Ω$
$P = I^{2} R = {(\frac{V}{| Z |})}^{2} R = {(\frac{220}{\sqrt{{(400)}^{2} + {(100 π)}^{2}}})}^{2} \times 400$
$= 74.46 W$

PHXII07:ALTERNATING CURRENT

356162 A sinusoidal $A C$ current flows through a resistor of resistance $R$ . If the peak current is $I_{p,}$ then the power dissipated is

1

\frac{4}{π} I_{p}^{2} R

2

\frac{1}{π} I_{p}^{2} R

3

\frac{1}{2} I_{p}^{2} R

4

I_{p}^{2} R \cos θ

Explanation:

$P o w e r = I_{r m s}^{2} R = {(\frac{I_{p}}{\sqrt{2}})}^{2} R = \frac{I_{p}^{2} R}{2}$

PHXII07:ALTERNATING CURRENT

356163 A choke coil has negligible resistance. The alternating potential drop across it is $220 v o l t s$ and the current is $5 m A .$ The power consumed is

1

220 \times \frac{5}{1000} W

2

\frac{220}{5} W

3 Zero

4

2.20 \times 5 W

Explanation:

Phase angle $= \frac{π}{2}$ (between $V$ and $I$ )
$\Rightarrow \cos ϕ = 0$
$\Rightarrow \cos ϕ = 0$
$\Rightarrow ⟨ P ⟩ = 0$
Since $⟨ P ⟩ = V I \cos ϕ$
Average power is zero.
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356164 An electric heater rated $225 V$ and $550 W$ is connected to $A C$ mains. The current drawn by it is:

1

0.8 A

2

2.0 A

3

0.4 A

4

1.25 A

Explanation:

$∵ P = V I \Rightarrow I = \frac{P}{V} = \frac{550}{225} = 2.0 A$

PHXII07:ALTERNATING CURRENT

356165 A 100 $W$ /200 $V$ bulb and an inductor are connected in series to a $220 V / 50 H z$ supply. Find the power consumed by the bulb. $(L = 1 H)$

1

74 W

2

84 W

3

100 W

4

92 W

Explanation:

Given $R = \frac{{(200)}^{2}}{100} = 400 Ω$
$P = I^{2} R = {(\frac{V}{| Z |})}^{2} R = {(\frac{220}{\sqrt{{(400)}^{2} + {(100 π)}^{2}}})}^{2} \times 400$
$= 74.46 W$

PHXII07:ALTERNATING CURRENT

356162 A sinusoidal $A C$ current flows through a resistor of resistance $R$ . If the peak current is $I_{p,}$ then the power dissipated is

1

\frac{4}{π} I_{p}^{2} R

2

\frac{1}{π} I_{p}^{2} R

3

\frac{1}{2} I_{p}^{2} R

4

I_{p}^{2} R \cos θ

Explanation:

$P o w e r = I_{r m s}^{2} R = {(\frac{I_{p}}{\sqrt{2}})}^{2} R = \frac{I_{p}^{2} R}{2}$

PHXII07:ALTERNATING CURRENT

356163 A choke coil has negligible resistance. The alternating potential drop across it is $220 v o l t s$ and the current is $5 m A .$ The power consumed is

1

220 \times \frac{5}{1000} W

2

\frac{220}{5} W

3 Zero

4

2.20 \times 5 W

Explanation:

Phase angle $= \frac{π}{2}$ (between $V$ and $I$ )
$\Rightarrow \cos ϕ = 0$
$\Rightarrow \cos ϕ = 0$
$\Rightarrow ⟨ P ⟩ = 0$
Since $⟨ P ⟩ = V I \cos ϕ$
Average power is zero.
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356164 An electric heater rated $225 V$ and $550 W$ is connected to $A C$ mains. The current drawn by it is:

1

0.8 A

2

2.0 A

3

0.4 A

4

1.25 A

Explanation:

$∵ P = V I \Rightarrow I = \frac{P}{V} = \frac{550}{225} = 2.0 A$

PHXII07:ALTERNATING CURRENT

356165 A 100 $W$ /200 $V$ bulb and an inductor are connected in series to a $220 V / 50 H z$ supply. Find the power consumed by the bulb. $(L = 1 H)$

1

74 W

2

84 W

3

100 W

4

92 W

Explanation:

Given $R = \frac{{(200)}^{2}}{100} = 400 Ω$
$P = I^{2} R = {(\frac{V}{| Z |})}^{2} R = {(\frac{220}{\sqrt{{(400)}^{2} + {(100 π)}^{2}}})}^{2} \times 400$
$= 74.46 W$

PHXII07:ALTERNATING CURRENT

356162 A sinusoidal $A C$ current flows through a resistor of resistance $R$ . If the peak current is $I_{p,}$ then the power dissipated is

1

\frac{4}{π} I_{p}^{2} R

2

\frac{1}{π} I_{p}^{2} R

3

\frac{1}{2} I_{p}^{2} R

4

I_{p}^{2} R \cos θ

Explanation:

$P o w e r = I_{r m s}^{2} R = {(\frac{I_{p}}{\sqrt{2}})}^{2} R = \frac{I_{p}^{2} R}{2}$

PHXII07:ALTERNATING CURRENT

356163 A choke coil has negligible resistance. The alternating potential drop across it is $220 v o l t s$ and the current is $5 m A .$ The power consumed is

1

220 \times \frac{5}{1000} W

2

\frac{220}{5} W

3 Zero

4

2.20 \times 5 W

Explanation:

Phase angle $= \frac{π}{2}$ (between $V$ and $I$ )
$\Rightarrow \cos ϕ = 0$
$\Rightarrow \cos ϕ = 0$
$\Rightarrow ⟨ P ⟩ = 0$
Since $⟨ P ⟩ = V I \cos ϕ$
Average power is zero.
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356164 An electric heater rated $225 V$ and $550 W$ is connected to $A C$ mains. The current drawn by it is:

1

0.8 A

2

2.0 A

3

0.4 A

4

1.25 A

Explanation:

$∵ P = V I \Rightarrow I = \frac{P}{V} = \frac{550}{225} = 2.0 A$

PHXII07:ALTERNATING CURRENT

356165 A 100 $W$ /200 $V$ bulb and an inductor are connected in series to a $220 V / 50 H z$ supply. Find the power consumed by the bulb. $(L = 1 H)$

1

74 W

2

84 W

3

100 W

4

92 W

Explanation:

Given $R = \frac{{(200)}^{2}}{100} = 400 Ω$
$P = I^{2} R = {(\frac{V}{| Z |})}^{2} R = {(\frac{220}{\sqrt{{(400)}^{2} + {(100 π)}^{2}}})}^{2} \times 400$
$= 74.46 W$