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Alternating Current

155047 An inductor coil is connected to a capacitor and an $AC$ source of rms voltage $8 V$ in series. The rms current in the circuit is $16 A$ and is in phase with emf. If this inductor coil is connected to $6 V$ DC battery, the magnitude of steady current is

1

8 A

2

10 A

3

12 A

4

16 A

Explanation:

C Given,
rms voltage of AC source, $V_{rms} = 8 V$ ,
rms current in Circuit, $I_{rms} = 16 A$

Then
Impedance $Z = \frac{V_{rms}}{I_{rms}} = \frac{8}{16} = 0.5 Ω$
When Inductor coil connected with $6 V$ DC battery
$V_{DC} = 6$ volt, $Z = 0.5 Ω$
Then,
$Current, I = \frac{V_{DC}}{Z} = \frac{6}{0.5}$
$I = 12 A$

TS- EAMCET-03.05.2019

Alternating Current

155048 An inductor of inductance $0.2 H$ is connected in series with a resistance, a capacitance and an AC source of frequency $\frac{1}{π} \times 10^{4} Hz$ . For what value of capacitance, the current will be maximum in the circuit

1

1.25 \times 10^{- 8} F

2

6.5 \times 10^{- 9} F

3

5.0 \times 10^{- 8} F

4

3.0 \times 10^{- 9} F

Explanation:

A Given,
$L = 0.2 H, f = \frac{1}{π} \times 10^{4} Hz$
$ω = 2 π f = 2 π \times \frac{1}{π} \times 10^{4}$
$= 2 \times 10^{4} rad / s$
For maximum current in circuit,
$X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
$C = \frac{1}{ω^{2} L} = \frac{1}{{(2 \times 10^{4})}^{2} \times 0.2}$
$= \frac{1}{4 \times 10^{8} \times 0.2} = 1.25 \times 10^{- 8} F$

TS EAMCET 08.05.2019

Alternating Current

155049 An inductor of inductance $L$ and resistor $R$ are joined together in series and connected by a source of frequency $ω$ . The power dissipated in the circuit is :

1

\frac{R^{2} + ω^{2} L^{2}}{V}

2

\frac{V^{2} R}{R^{2} + ω^{2} L^{2}}

3

\frac{V}{R^{2} + ω^{2} L^{2}}

4

\frac{V^{2} R}{\sqrt{R^{2} + ω^{2} L^{2}}}

Explanation:

B The power dissipated in the circuits, $P = VI \cos ϕ$
Power factor, $\cos ϕ = \frac{R}{Z}$
$I = I_{rms} = \frac{V}{Z}$
So the power, $P = VI \cos ϕ$
$= V \frac{V}{Z} \cdot \frac{R}{Z} = \frac{V^{2} R}{Z^{2}}$
As we know that, $Z^{2} = R^{2} + X_{L}^{2} = R^{2} + (ω L)^{2}$
$∴$ Power, $P = \frac{V^{2} R}{R^{2} + ω^{2} L^{2}}$

Karnataka CET-2019

Alternating Current

155050 A coil has inductance $2 H$ . The ratio of its reactance, when it is connected first to an a.c. source and then to d.c. source, is

1 1

2 less than 1

3 infinity

4 zero

Explanation:

C Given,
Inductance of coil $= 2 H$
When Reactance of coil is connected to AC source
${(X_{L})}_{AC} = ω L = 2 π fL$
Case-II: When reactance connected across DC,
Then, $X_{L} = ω L = 2 π fL$
$∵ f = 0$ for $DC$
$∴ X_{L} = 0$
The ratio of reactance of $AC$ source to $DC$ source $[\frac{ω L}{0}] = \infty$ .

MHT-CET 2019

Alternating Current

155047 An inductor coil is connected to a capacitor and an $AC$ source of rms voltage $8 V$ in series. The rms current in the circuit is $16 A$ and is in phase with emf. If this inductor coil is connected to $6 V$ DC battery, the magnitude of steady current is

1

8 A

2

10 A

3

12 A

4

16 A

Explanation:

C Given,
rms voltage of AC source, $V_{rms} = 8 V$ ,
rms current in Circuit, $I_{rms} = 16 A$

Then
Impedance $Z = \frac{V_{rms}}{I_{rms}} = \frac{8}{16} = 0.5 Ω$
When Inductor coil connected with $6 V$ DC battery
$V_{DC} = 6$ volt, $Z = 0.5 Ω$
Then,
$Current, I = \frac{V_{DC}}{Z} = \frac{6}{0.5}$
$I = 12 A$

TS- EAMCET-03.05.2019

Alternating Current

155048 An inductor of inductance $0.2 H$ is connected in series with a resistance, a capacitance and an AC source of frequency $\frac{1}{π} \times 10^{4} Hz$ . For what value of capacitance, the current will be maximum in the circuit

1

1.25 \times 10^{- 8} F

2

6.5 \times 10^{- 9} F

3

5.0 \times 10^{- 8} F

4

3.0 \times 10^{- 9} F

Explanation:

A Given,
$L = 0.2 H, f = \frac{1}{π} \times 10^{4} Hz$
$ω = 2 π f = 2 π \times \frac{1}{π} \times 10^{4}$
$= 2 \times 10^{4} rad / s$
For maximum current in circuit,
$X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
$C = \frac{1}{ω^{2} L} = \frac{1}{{(2 \times 10^{4})}^{2} \times 0.2}$
$= \frac{1}{4 \times 10^{8} \times 0.2} = 1.25 \times 10^{- 8} F$

TS EAMCET 08.05.2019

Alternating Current

155049 An inductor of inductance $L$ and resistor $R$ are joined together in series and connected by a source of frequency $ω$ . The power dissipated in the circuit is :

1

\frac{R^{2} + ω^{2} L^{2}}{V}

2

\frac{V^{2} R}{R^{2} + ω^{2} L^{2}}

3

\frac{V}{R^{2} + ω^{2} L^{2}}

4

\frac{V^{2} R}{\sqrt{R^{2} + ω^{2} L^{2}}}

Explanation:

B The power dissipated in the circuits, $P = VI \cos ϕ$
Power factor, $\cos ϕ = \frac{R}{Z}$
$I = I_{rms} = \frac{V}{Z}$
So the power, $P = VI \cos ϕ$
$= V \frac{V}{Z} \cdot \frac{R}{Z} = \frac{V^{2} R}{Z^{2}}$
As we know that, $Z^{2} = R^{2} + X_{L}^{2} = R^{2} + (ω L)^{2}$
$∴$ Power, $P = \frac{V^{2} R}{R^{2} + ω^{2} L^{2}}$

Karnataka CET-2019

Alternating Current

155050 A coil has inductance $2 H$ . The ratio of its reactance, when it is connected first to an a.c. source and then to d.c. source, is

1 1

2 less than 1

3 infinity

4 zero

Explanation:

C Given,
Inductance of coil $= 2 H$
When Reactance of coil is connected to AC source
${(X_{L})}_{AC} = ω L = 2 π fL$
Case-II: When reactance connected across DC,
Then, $X_{L} = ω L = 2 π fL$
$∵ f = 0$ for $DC$
$∴ X_{L} = 0$
The ratio of reactance of $AC$ source to $DC$ source $[\frac{ω L}{0}] = \infty$ .

MHT-CET 2019

Alternating Current

155047 An inductor coil is connected to a capacitor and an $AC$ source of rms voltage $8 V$ in series. The rms current in the circuit is $16 A$ and is in phase with emf. If this inductor coil is connected to $6 V$ DC battery, the magnitude of steady current is

1

8 A

2

10 A

3

12 A

4

16 A

Explanation:

C Given,
rms voltage of AC source, $V_{rms} = 8 V$ ,
rms current in Circuit, $I_{rms} = 16 A$

Then
Impedance $Z = \frac{V_{rms}}{I_{rms}} = \frac{8}{16} = 0.5 Ω$
When Inductor coil connected with $6 V$ DC battery
$V_{DC} = 6$ volt, $Z = 0.5 Ω$
Then,
$Current, I = \frac{V_{DC}}{Z} = \frac{6}{0.5}$
$I = 12 A$

TS- EAMCET-03.05.2019

Alternating Current

155048 An inductor of inductance $0.2 H$ is connected in series with a resistance, a capacitance and an AC source of frequency $\frac{1}{π} \times 10^{4} Hz$ . For what value of capacitance, the current will be maximum in the circuit

1

1.25 \times 10^{- 8} F

2

6.5 \times 10^{- 9} F

3

5.0 \times 10^{- 8} F

4

3.0 \times 10^{- 9} F

Explanation:

A Given,
$L = 0.2 H, f = \frac{1}{π} \times 10^{4} Hz$
$ω = 2 π f = 2 π \times \frac{1}{π} \times 10^{4}$
$= 2 \times 10^{4} rad / s$
For maximum current in circuit,
$X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
$C = \frac{1}{ω^{2} L} = \frac{1}{{(2 \times 10^{4})}^{2} \times 0.2}$
$= \frac{1}{4 \times 10^{8} \times 0.2} = 1.25 \times 10^{- 8} F$

TS EAMCET 08.05.2019

Alternating Current

155049 An inductor of inductance $L$ and resistor $R$ are joined together in series and connected by a source of frequency $ω$ . The power dissipated in the circuit is :

1

\frac{R^{2} + ω^{2} L^{2}}{V}

2

\frac{V^{2} R}{R^{2} + ω^{2} L^{2}}

3

\frac{V}{R^{2} + ω^{2} L^{2}}

4

\frac{V^{2} R}{\sqrt{R^{2} + ω^{2} L^{2}}}

Explanation:

B The power dissipated in the circuits, $P = VI \cos ϕ$
Power factor, $\cos ϕ = \frac{R}{Z}$
$I = I_{rms} = \frac{V}{Z}$
So the power, $P = VI \cos ϕ$
$= V \frac{V}{Z} \cdot \frac{R}{Z} = \frac{V^{2} R}{Z^{2}}$
As we know that, $Z^{2} = R^{2} + X_{L}^{2} = R^{2} + (ω L)^{2}$
$∴$ Power, $P = \frac{V^{2} R}{R^{2} + ω^{2} L^{2}}$

Karnataka CET-2019

Alternating Current

155050 A coil has inductance $2 H$ . The ratio of its reactance, when it is connected first to an a.c. source and then to d.c. source, is

1 1

2 less than 1

3 infinity

4 zero

Explanation:

C Given,
Inductance of coil $= 2 H$
When Reactance of coil is connected to AC source
${(X_{L})}_{AC} = ω L = 2 π fL$
Case-II: When reactance connected across DC,
Then, $X_{L} = ω L = 2 π fL$
$∵ f = 0$ for $DC$
$∴ X_{L} = 0$
The ratio of reactance of $AC$ source to $DC$ source $[\frac{ω L}{0}] = \infty$ .

MHT-CET 2019

Alternating Current

155047 An inductor coil is connected to a capacitor and an $AC$ source of rms voltage $8 V$ in series. The rms current in the circuit is $16 A$ and is in phase with emf. If this inductor coil is connected to $6 V$ DC battery, the magnitude of steady current is

1

8 A

2

10 A

3

12 A

4

16 A

Explanation:

C Given,
rms voltage of AC source, $V_{rms} = 8 V$ ,
rms current in Circuit, $I_{rms} = 16 A$

Then
Impedance $Z = \frac{V_{rms}}{I_{rms}} = \frac{8}{16} = 0.5 Ω$
When Inductor coil connected with $6 V$ DC battery
$V_{DC} = 6$ volt, $Z = 0.5 Ω$
Then,
$Current, I = \frac{V_{DC}}{Z} = \frac{6}{0.5}$
$I = 12 A$

TS- EAMCET-03.05.2019

Alternating Current

155048 An inductor of inductance $0.2 H$ is connected in series with a resistance, a capacitance and an AC source of frequency $\frac{1}{π} \times 10^{4} Hz$ . For what value of capacitance, the current will be maximum in the circuit

1

1.25 \times 10^{- 8} F

2

6.5 \times 10^{- 9} F

3

5.0 \times 10^{- 8} F

4

3.0 \times 10^{- 9} F

Explanation:

A Given,
$L = 0.2 H, f = \frac{1}{π} \times 10^{4} Hz$
$ω = 2 π f = 2 π \times \frac{1}{π} \times 10^{4}$
$= 2 \times 10^{4} rad / s$
For maximum current in circuit,
$X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$
$C = \frac{1}{ω^{2} L} = \frac{1}{{(2 \times 10^{4})}^{2} \times 0.2}$
$= \frac{1}{4 \times 10^{8} \times 0.2} = 1.25 \times 10^{- 8} F$

TS EAMCET 08.05.2019

Alternating Current

155049 An inductor of inductance $L$ and resistor $R$ are joined together in series and connected by a source of frequency $ω$ . The power dissipated in the circuit is :

1

\frac{R^{2} + ω^{2} L^{2}}{V}

2

\frac{V^{2} R}{R^{2} + ω^{2} L^{2}}

3

\frac{V}{R^{2} + ω^{2} L^{2}}

4

\frac{V^{2} R}{\sqrt{R^{2} + ω^{2} L^{2}}}

Explanation:

B The power dissipated in the circuits, $P = VI \cos ϕ$
Power factor, $\cos ϕ = \frac{R}{Z}$
$I = I_{rms} = \frac{V}{Z}$
So the power, $P = VI \cos ϕ$
$= V \frac{V}{Z} \cdot \frac{R}{Z} = \frac{V^{2} R}{Z^{2}}$
As we know that, $Z^{2} = R^{2} + X_{L}^{2} = R^{2} + (ω L)^{2}$
$∴$ Power, $P = \frac{V^{2} R}{R^{2} + ω^{2} L^{2}}$

Karnataka CET-2019

Alternating Current

155050 A coil has inductance $2 H$ . The ratio of its reactance, when it is connected first to an a.c. source and then to d.c. source, is

1 1

2 less than 1

3 infinity

4 zero

Explanation:

C Given,
Inductance of coil $= 2 H$
When Reactance of coil is connected to AC source
${(X_{L})}_{AC} = ω L = 2 π fL$
Case-II: When reactance connected across DC,
Then, $X_{L} = ω L = 2 π fL$
$∵ f = 0$ for $DC$
$∴ X_{L} = 0$
The ratio of reactance of $AC$ source to $DC$ source $[\frac{ω L}{0}] = \infty$ .

MHT-CET 2019