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Alternating Current

155331 An inductor and a resistor are connected in series to an ac source of variable frequency. the power factor of the circuit is $\frac{\sqrt{3}}{2}$ . If the frequency of the ac is increased by $200 %$ , the power factor of the circuit is

1 0.8

2 0.9

3 0.7

4 0.5

Explanation:

D According to question,

Given, power factor $(\cos ϕ) = \frac{\sqrt{3}}{2}$
$f = 50 Hz$
$f^{'} = f + 200 % of f$
$f^{'} = f + \frac{200}{100} \times f$
$f^{'} = 3 f$
For R, L circuit
$\cos ϕ = \frac{R}{Z} = \frac{R}{\sqrt{R^{2} + X_{L}^{2}}}$
$\cos^{2} ϕ = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
${(\frac{\sqrt{3}}{2})}^{2} = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
$\frac{3}{4} = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
$3 R^{2} + 3 X_{L}^{2} = 4 R^{2}$
$R^{2} = 3 \times X_{L}^{2}$
$X_{L} = \frac{R}{\sqrt{3}}$
$∵ X_{L} \propto f$
$∴$ After change in frequency, $X_{L}^{'} = 3 L$
$X_{L}^{'} = 3 \times \frac{R}{\sqrt{3}}$
$∴ \cos ϕ^{'} = \frac{R}{\sqrt{R^{2} + X_{L}^{' 2}}}$
$\cos ϕ^{'} = \frac{R}{\sqrt{R^{2} + (\sqrt{3} R)^{2}}}$
When the frequency of the applied ac is $50 Hz$ ,
$\cos ϕ^{'} = \frac{R}{\sqrt{4} R^{2}}$
$\cos ϕ^{'} = \frac{R}{2 R}$
$\cos ϕ^{'} = 0.5$

AP EAMCET-24.04.2017

Alternating Current

155332 What is power dissipation in an A.C. circuit in which voltage and current are given by $V = 300 \sin (ω t + π / 2), I = 5 \sin ω t$ ?

1 zero

2 300 units

3 150 units

4 75 units

Explanation:

A Given,
$V = 300 \sin (ω t + π / 2)$
$I = 5 \sin ω t$
On comparing equation, $V = V_{0} \sin (ω t + ϕ)$ and $I = I_{0} \sin$ $(ω t)$
Then, $I_{0} = 5 A, V_{0} = 300 V$
We know that, $(P) = \frac{1}{2} V_{0} I_{0} \cos ϕ$
$P = \frac{1}{2} \times 300 \times 5 \cos (π / 2)$
$P = 0$

AMU-2017

Alternating Current

155333 In an AC circuit, current is $3 A$ and voltage 210 $V$ and power is $63 W$ . The power factor is

1 0.11

2 0.09

3 0.08

4 0.10

Explanation:

D Given,
$I = 3 A$
$V = 210 V$
$P = 63 W$
$∵$ Power drawn by the circuit is -
$P = IV \cos ϕ$
Power factor, $(\cos ϕ) = \frac{P}{IV}$
$\cos ϕ = \frac{63}{3 \times 210}$
$\cos ϕ = 0.10$

GUJCET 2017

Alternating Current

155334 In an $AC$ circuit the potential difference and current $i$ are represented respectively by $V =$ $100 \sin (100 t)$ volt, and
$I = 100 \sin (100 t + \frac{π}{3})$ milli ampere. The power in the circuit is

1

2.5 W

2

5 W

3

10 W

4

10^{4} W

Explanation:

A Given,
$V = 100 \sin (100 t) V$
$I = 100 \sin (100 t + \frac{π}{3}) mA$
$I = 100 \times 10^{- 3} \sin (100 t + \frac{π}{3}) A$
Comparing equation (i) with, $V = V_{0} \sin (ω t)$
Then, $V_{0} = 100, ω = 100$
$ϕ = \frac{π}{3}$
Now, power $(P) = V_{rms} I_{rms} \cos θ$
$P = \frac{V_{0}}{\sqrt{2}} \times \frac{I_{0}}{\sqrt{2}} \cos \frac{π}{3}$
$P = \frac{100 \times 100 \times 10^{- 3}}{2} \times \frac{1}{2} = 2.5 W$

Manipal UGET-2011

Alternating Current

155331 An inductor and a resistor are connected in series to an ac source of variable frequency. the power factor of the circuit is $\frac{\sqrt{3}}{2}$ . If the frequency of the ac is increased by $200 %$ , the power factor of the circuit is

1 0.8

2 0.9

3 0.7

4 0.5

Explanation:

D According to question,

Given, power factor $(\cos ϕ) = \frac{\sqrt{3}}{2}$
$f = 50 Hz$
$f^{'} = f + 200 % of f$
$f^{'} = f + \frac{200}{100} \times f$
$f^{'} = 3 f$
For R, L circuit
$\cos ϕ = \frac{R}{Z} = \frac{R}{\sqrt{R^{2} + X_{L}^{2}}}$
$\cos^{2} ϕ = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
${(\frac{\sqrt{3}}{2})}^{2} = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
$\frac{3}{4} = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
$3 R^{2} + 3 X_{L}^{2} = 4 R^{2}$
$R^{2} = 3 \times X_{L}^{2}$
$X_{L} = \frac{R}{\sqrt{3}}$
$∵ X_{L} \propto f$
$∴$ After change in frequency, $X_{L}^{'} = 3 L$
$X_{L}^{'} = 3 \times \frac{R}{\sqrt{3}}$
$∴ \cos ϕ^{'} = \frac{R}{\sqrt{R^{2} + X_{L}^{' 2}}}$
$\cos ϕ^{'} = \frac{R}{\sqrt{R^{2} + (\sqrt{3} R)^{2}}}$
When the frequency of the applied ac is $50 Hz$ ,
$\cos ϕ^{'} = \frac{R}{\sqrt{4} R^{2}}$
$\cos ϕ^{'} = \frac{R}{2 R}$
$\cos ϕ^{'} = 0.5$

AP EAMCET-24.04.2017

Alternating Current

155332 What is power dissipation in an A.C. circuit in which voltage and current are given by $V = 300 \sin (ω t + π / 2), I = 5 \sin ω t$ ?

1 zero

2 300 units

3 150 units

4 75 units

Explanation:

A Given,
$V = 300 \sin (ω t + π / 2)$
$I = 5 \sin ω t$
On comparing equation, $V = V_{0} \sin (ω t + ϕ)$ and $I = I_{0} \sin$ $(ω t)$
Then, $I_{0} = 5 A, V_{0} = 300 V$
We know that, $(P) = \frac{1}{2} V_{0} I_{0} \cos ϕ$
$P = \frac{1}{2} \times 300 \times 5 \cos (π / 2)$
$P = 0$

AMU-2017

Alternating Current

155333 In an AC circuit, current is $3 A$ and voltage 210 $V$ and power is $63 W$ . The power factor is

1 0.11

2 0.09

3 0.08

4 0.10

Explanation:

D Given,
$I = 3 A$
$V = 210 V$
$P = 63 W$
$∵$ Power drawn by the circuit is -
$P = IV \cos ϕ$
Power factor, $(\cos ϕ) = \frac{P}{IV}$
$\cos ϕ = \frac{63}{3 \times 210}$
$\cos ϕ = 0.10$

GUJCET 2017

Alternating Current

155334 In an $AC$ circuit the potential difference and current $i$ are represented respectively by $V =$ $100 \sin (100 t)$ volt, and
$I = 100 \sin (100 t + \frac{π}{3})$ milli ampere. The power in the circuit is

1

2.5 W

2

5 W

3

10 W

4

10^{4} W

Explanation:

A Given,
$V = 100 \sin (100 t) V$
$I = 100 \sin (100 t + \frac{π}{3}) mA$
$I = 100 \times 10^{- 3} \sin (100 t + \frac{π}{3}) A$
Comparing equation (i) with, $V = V_{0} \sin (ω t)$
Then, $V_{0} = 100, ω = 100$
$ϕ = \frac{π}{3}$
Now, power $(P) = V_{rms} I_{rms} \cos θ$
$P = \frac{V_{0}}{\sqrt{2}} \times \frac{I_{0}}{\sqrt{2}} \cos \frac{π}{3}$
$P = \frac{100 \times 100 \times 10^{- 3}}{2} \times \frac{1}{2} = 2.5 W$

Manipal UGET-2011

Alternating Current

155331 An inductor and a resistor are connected in series to an ac source of variable frequency. the power factor of the circuit is $\frac{\sqrt{3}}{2}$ . If the frequency of the ac is increased by $200 %$ , the power factor of the circuit is

1 0.8

2 0.9

3 0.7

4 0.5

Explanation:

D According to question,

Given, power factor $(\cos ϕ) = \frac{\sqrt{3}}{2}$
$f = 50 Hz$
$f^{'} = f + 200 % of f$
$f^{'} = f + \frac{200}{100} \times f$
$f^{'} = 3 f$
For R, L circuit
$\cos ϕ = \frac{R}{Z} = \frac{R}{\sqrt{R^{2} + X_{L}^{2}}}$
$\cos^{2} ϕ = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
${(\frac{\sqrt{3}}{2})}^{2} = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
$\frac{3}{4} = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
$3 R^{2} + 3 X_{L}^{2} = 4 R^{2}$
$R^{2} = 3 \times X_{L}^{2}$
$X_{L} = \frac{R}{\sqrt{3}}$
$∵ X_{L} \propto f$
$∴$ After change in frequency, $X_{L}^{'} = 3 L$
$X_{L}^{'} = 3 \times \frac{R}{\sqrt{3}}$
$∴ \cos ϕ^{'} = \frac{R}{\sqrt{R^{2} + X_{L}^{' 2}}}$
$\cos ϕ^{'} = \frac{R}{\sqrt{R^{2} + (\sqrt{3} R)^{2}}}$
When the frequency of the applied ac is $50 Hz$ ,
$\cos ϕ^{'} = \frac{R}{\sqrt{4} R^{2}}$
$\cos ϕ^{'} = \frac{R}{2 R}$
$\cos ϕ^{'} = 0.5$

AP EAMCET-24.04.2017

Alternating Current

155332 What is power dissipation in an A.C. circuit in which voltage and current are given by $V = 300 \sin (ω t + π / 2), I = 5 \sin ω t$ ?

1 zero

2 300 units

3 150 units

4 75 units

Explanation:

A Given,
$V = 300 \sin (ω t + π / 2)$
$I = 5 \sin ω t$
On comparing equation, $V = V_{0} \sin (ω t + ϕ)$ and $I = I_{0} \sin$ $(ω t)$
Then, $I_{0} = 5 A, V_{0} = 300 V$
We know that, $(P) = \frac{1}{2} V_{0} I_{0} \cos ϕ$
$P = \frac{1}{2} \times 300 \times 5 \cos (π / 2)$
$P = 0$

AMU-2017

Alternating Current

155333 In an AC circuit, current is $3 A$ and voltage 210 $V$ and power is $63 W$ . The power factor is

1 0.11

2 0.09

3 0.08

4 0.10

Explanation:

D Given,
$I = 3 A$
$V = 210 V$
$P = 63 W$
$∵$ Power drawn by the circuit is -
$P = IV \cos ϕ$
Power factor, $(\cos ϕ) = \frac{P}{IV}$
$\cos ϕ = \frac{63}{3 \times 210}$
$\cos ϕ = 0.10$

GUJCET 2017

Alternating Current

155334 In an $AC$ circuit the potential difference and current $i$ are represented respectively by $V =$ $100 \sin (100 t)$ volt, and
$I = 100 \sin (100 t + \frac{π}{3})$ milli ampere. The power in the circuit is

1

2.5 W

2

5 W

3

10 W

4

10^{4} W

Explanation:

A Given,
$V = 100 \sin (100 t) V$
$I = 100 \sin (100 t + \frac{π}{3}) mA$
$I = 100 \times 10^{- 3} \sin (100 t + \frac{π}{3}) A$
Comparing equation (i) with, $V = V_{0} \sin (ω t)$
Then, $V_{0} = 100, ω = 100$
$ϕ = \frac{π}{3}$
Now, power $(P) = V_{rms} I_{rms} \cos θ$
$P = \frac{V_{0}}{\sqrt{2}} \times \frac{I_{0}}{\sqrt{2}} \cos \frac{π}{3}$
$P = \frac{100 \times 100 \times 10^{- 3}}{2} \times \frac{1}{2} = 2.5 W$

Manipal UGET-2011

Alternating Current

155331 An inductor and a resistor are connected in series to an ac source of variable frequency. the power factor of the circuit is $\frac{\sqrt{3}}{2}$ . If the frequency of the ac is increased by $200 %$ , the power factor of the circuit is

1 0.8

2 0.9

3 0.7

4 0.5

Explanation:

D According to question,

Given, power factor $(\cos ϕ) = \frac{\sqrt{3}}{2}$
$f = 50 Hz$
$f^{'} = f + 200 % of f$
$f^{'} = f + \frac{200}{100} \times f$
$f^{'} = 3 f$
For R, L circuit
$\cos ϕ = \frac{R}{Z} = \frac{R}{\sqrt{R^{2} + X_{L}^{2}}}$
$\cos^{2} ϕ = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
${(\frac{\sqrt{3}}{2})}^{2} = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
$\frac{3}{4} = \frac{R^{2}}{R^{2} + X_{L}^{2}}$
$3 R^{2} + 3 X_{L}^{2} = 4 R^{2}$
$R^{2} = 3 \times X_{L}^{2}$
$X_{L} = \frac{R}{\sqrt{3}}$
$∵ X_{L} \propto f$
$∴$ After change in frequency, $X_{L}^{'} = 3 L$
$X_{L}^{'} = 3 \times \frac{R}{\sqrt{3}}$
$∴ \cos ϕ^{'} = \frac{R}{\sqrt{R^{2} + X_{L}^{' 2}}}$
$\cos ϕ^{'} = \frac{R}{\sqrt{R^{2} + (\sqrt{3} R)^{2}}}$
When the frequency of the applied ac is $50 Hz$ ,
$\cos ϕ^{'} = \frac{R}{\sqrt{4} R^{2}}$
$\cos ϕ^{'} = \frac{R}{2 R}$
$\cos ϕ^{'} = 0.5$

AP EAMCET-24.04.2017

Alternating Current

155332 What is power dissipation in an A.C. circuit in which voltage and current are given by $V = 300 \sin (ω t + π / 2), I = 5 \sin ω t$ ?

1 zero

2 300 units

3 150 units

4 75 units

Explanation:

A Given,
$V = 300 \sin (ω t + π / 2)$
$I = 5 \sin ω t$
On comparing equation, $V = V_{0} \sin (ω t + ϕ)$ and $I = I_{0} \sin$ $(ω t)$
Then, $I_{0} = 5 A, V_{0} = 300 V$
We know that, $(P) = \frac{1}{2} V_{0} I_{0} \cos ϕ$
$P = \frac{1}{2} \times 300 \times 5 \cos (π / 2)$
$P = 0$

AMU-2017

Alternating Current

155333 In an AC circuit, current is $3 A$ and voltage 210 $V$ and power is $63 W$ . The power factor is

1 0.11

2 0.09

3 0.08

4 0.10

Explanation:

D Given,
$I = 3 A$
$V = 210 V$
$P = 63 W$
$∵$ Power drawn by the circuit is -
$P = IV \cos ϕ$
Power factor, $(\cos ϕ) = \frac{P}{IV}$
$\cos ϕ = \frac{63}{3 \times 210}$
$\cos ϕ = 0.10$

GUJCET 2017

Alternating Current

155334 In an $AC$ circuit the potential difference and current $i$ are represented respectively by $V =$ $100 \sin (100 t)$ volt, and
$I = 100 \sin (100 t + \frac{π}{3})$ milli ampere. The power in the circuit is

1

2.5 W

2

5 W

3

10 W

4

10^{4} W

Explanation:

A Given,
$V = 100 \sin (100 t) V$
$I = 100 \sin (100 t + \frac{π}{3}) mA$
$I = 100 \times 10^{- 3} \sin (100 t + \frac{π}{3}) A$
Comparing equation (i) with, $V = V_{0} \sin (ω t)$
Then, $V_{0} = 100, ω = 100$
$ϕ = \frac{π}{3}$
Now, power $(P) = V_{rms} I_{rms} \cos θ$
$P = \frac{V_{0}}{\sqrt{2}} \times \frac{I_{0}}{\sqrt{2}} \cos \frac{π}{3}$
$P = \frac{100 \times 100 \times 10^{- 3}}{2} \times \frac{1}{2} = 2.5 W$

Manipal UGET-2011

Question Bank Series

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