QBManager

Alternating Current

155138 Which of the following components of an $L - C$ - $R$ circuit, with AC supply, dissipates energy?

1 Only L

2 Only R

3 Only C

4

L

and

C

Explanation:

B In an A.C. circuit, the power dissipated is -
$P = VI \cos ϕ$
Where, $ϕ =$ phase difference between voltage and current.
For inductor (L) and capacitor (C), the phase difference between $V$ and $I$ is $90^{\circ}$ .
Then, $P_{L}$ or $P_{C} = VI \cos 90^{\circ} = 0$
For resistor (R), the phase difference between $V$ and $I$ zero.
Then, $P_{R} = VI \cos 0^{\circ} = VI$
Hence, only due to resistor R energy is dissipated in an L-C-R circuit.

TS- EAMCET-09.09.2020

Alternating Current

155139 Which one of the following curves represents the variation of impedance $(Z)$ with frequency $f$ in a series L-C-R circuit, when connected to an AC source?

1

2

3

4

Explanation:

C We know,
Impedance for a series R-L-C circuit -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$
$Z = \sqrt{R^{2} + {(2 π f L - \frac{1}{2 π f C})}^{2}}$
At resonance, $X_{L} = X_{C}$
i.e. $Z_{min} = R$ and $f = f_{o} (f_{o} =$ resonance frequency $)$ Hence, the variation of $Z$ with frequency ( $f$ )in a series -

TS- EAMCET.14.09.2020

Alternating Current

155141 An L-C-R series circuit is connected to a source of alternating current. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of

1

π

2

\frac{π}{2}

3

\frac{π}{4}

4

0^{\circ}

Explanation:

D In LCR circuit
At resonance condition,
Inductive Reactance $=$ Capacitive reactance
$X_{L} = X_{C}$
Now, impedance in LCR circuit
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$Z = \sqrt{R^{2} + {(X_{L} - X_{L})}^{2}} [∵ X_{L} = X_{C}]$
$Z = \sqrt{R^{2}} = R$
So, At resonance, LCR circuit behave as pure resistive circuit. In purely resistive circuit, current and voltage are in same phase.
Then, phase difference, $ϕ = 0^{\circ}$ .

TS- EAMCET-14.09.2020

Alternating Current

155142 An alternating voltage $ε = 30 \sin 200 t$ (in volts) is applied to the circuit below. The amplitude of the current through the circuit is

1

3 A

2

2 A

3

1 A

4

0.5 A

Explanation:

A Given,
Alternating voltage, $ε = 30 \sin 200 t V$
.....(i)
Resistance, $(R) = 10 Ω$ ,
Inductance, $(L) = 0.05 H$ ,
Capacitance $(C) = 500 μ F = 500 \times 10^{- 6} F$

Standard equation of AC voltage, $ε = ε_{m} \sin ω t$
From equation (i) and (ii), we get-
$ε_{m} = 30 V, ω = 200 rad / \sec$
We know that,
Inductive reactance,
$X_{L} = ω L = 200 \times 0.05 = 10 Ω$
And, Capacitive reactance,
$X_{C} = \frac{I}{ω C} = \frac{1}{200 \times 500 \times 10^{- 6}} = 10 Ω$
Now, Impedance of the LCR circuit,
$Z = \sqrt{(R)^{2} + {(X_{L} - X_{C})}^{2}}$
$= \sqrt{(10)^{2} + (10 - 10)^{2}} = \sqrt{100 + 0}$
$Z = 10 Ω$
So, Amplitude of the current is,
$I_{max} = \frac{V_{m}}{Z} = \frac{ε_{m}}{Z} = \frac{30}{10} = 3 A$
$I_{max} = 3 A$

TS- EAMCET-10.09.2020

Alternating Current

155144 The $L - C - R$ circuit shown below is driven by an ideal AC voltage source.

At frequency $f = \frac{1}{2 π \sqrt{LC}}$ , choose the correct statement.

1 The current through

R

is zero.

2 The current through

R

is infinite.

3 The current through

R

depends on the value of

L

and

C

.

4 The current through

R

depends only on the value of

R

and not

L

and

C

.

Explanation:

A Given circuit diagram -

And, $f = \frac{1}{2 π \sqrt{LC}}$
We know, $X_{L} = ω L$
$X_{L} = 2 π f L = 2 π \times \frac{1}{2 π \sqrt{L C}} \times L$
$X_{L} = \sqrt{\frac{L}{C}} Ω$
And, $X_{C} = \frac{1}{2 π f C} = \frac{1}{2 π \times \frac{1}{2 π \sqrt{L C}} \times C} = \sqrt{\frac{L}{C}} Ω$
Voltage across capacitor $V_{C} (+) = V_{L} (+)$
Current through resistor $(I_{R}) = I_{L} - I_{C}$
$= \frac{V_{L}}{X_{L}} - \frac{V_{C}}{X_{C}}$
$I_{R} = \frac{V_{L}}{\sqrt{\frac{L}{C}}} - \frac{V_{L}}{\sqrt{\frac{L}{C}}}$
${∵ V_{L} = V_{C}}$
So,
$I_{R} = 0 A$
Hence the current through $R$ is zero.

TS- EAMCET-11.09.2020

Alternating Current

155138 Which of the following components of an $L - C$ - $R$ circuit, with AC supply, dissipates energy?

1 Only L

2 Only R

3 Only C

4

L

and

C

Explanation:

B In an A.C. circuit, the power dissipated is -
$P = VI \cos ϕ$
Where, $ϕ =$ phase difference between voltage and current.
For inductor (L) and capacitor (C), the phase difference between $V$ and $I$ is $90^{\circ}$ .
Then, $P_{L}$ or $P_{C} = VI \cos 90^{\circ} = 0$
For resistor (R), the phase difference between $V$ and $I$ zero.
Then, $P_{R} = VI \cos 0^{\circ} = VI$
Hence, only due to resistor R energy is dissipated in an L-C-R circuit.

TS- EAMCET-09.09.2020

Alternating Current

155139 Which one of the following curves represents the variation of impedance $(Z)$ with frequency $f$ in a series L-C-R circuit, when connected to an AC source?

1

2

3

4

Explanation:

C We know,
Impedance for a series R-L-C circuit -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$
$Z = \sqrt{R^{2} + {(2 π f L - \frac{1}{2 π f C})}^{2}}$
At resonance, $X_{L} = X_{C}$
i.e. $Z_{min} = R$ and $f = f_{o} (f_{o} =$ resonance frequency $)$ Hence, the variation of $Z$ with frequency ( $f$ )in a series -

TS- EAMCET.14.09.2020

Alternating Current

155141 An L-C-R series circuit is connected to a source of alternating current. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of

1

π

2

\frac{π}{2}

3

\frac{π}{4}

4

0^{\circ}

Explanation:

D In LCR circuit
At resonance condition,
Inductive Reactance $=$ Capacitive reactance
$X_{L} = X_{C}$
Now, impedance in LCR circuit
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$Z = \sqrt{R^{2} + {(X_{L} - X_{L})}^{2}} [∵ X_{L} = X_{C}]$
$Z = \sqrt{R^{2}} = R$
So, At resonance, LCR circuit behave as pure resistive circuit. In purely resistive circuit, current and voltage are in same phase.
Then, phase difference, $ϕ = 0^{\circ}$ .

TS- EAMCET-14.09.2020

Alternating Current

155142 An alternating voltage $ε = 30 \sin 200 t$ (in volts) is applied to the circuit below. The amplitude of the current through the circuit is

1

3 A

2

2 A

3

1 A

4

0.5 A

Explanation:

A Given,
Alternating voltage, $ε = 30 \sin 200 t V$
.....(i)
Resistance, $(R) = 10 Ω$ ,
Inductance, $(L) = 0.05 H$ ,
Capacitance $(C) = 500 μ F = 500 \times 10^{- 6} F$

Standard equation of AC voltage, $ε = ε_{m} \sin ω t$
From equation (i) and (ii), we get-
$ε_{m} = 30 V, ω = 200 rad / \sec$
We know that,
Inductive reactance,
$X_{L} = ω L = 200 \times 0.05 = 10 Ω$
And, Capacitive reactance,
$X_{C} = \frac{I}{ω C} = \frac{1}{200 \times 500 \times 10^{- 6}} = 10 Ω$
Now, Impedance of the LCR circuit,
$Z = \sqrt{(R)^{2} + {(X_{L} - X_{C})}^{2}}$
$= \sqrt{(10)^{2} + (10 - 10)^{2}} = \sqrt{100 + 0}$
$Z = 10 Ω$
So, Amplitude of the current is,
$I_{max} = \frac{V_{m}}{Z} = \frac{ε_{m}}{Z} = \frac{30}{10} = 3 A$
$I_{max} = 3 A$

TS- EAMCET-10.09.2020

Alternating Current

155144 The $L - C - R$ circuit shown below is driven by an ideal AC voltage source.

At frequency $f = \frac{1}{2 π \sqrt{LC}}$ , choose the correct statement.

1 The current through

R

is zero.

2 The current through

R

is infinite.

3 The current through

R

depends on the value of

L

and

C

.

4 The current through

R

depends only on the value of

R

and not

L

and

C

.

Explanation:

A Given circuit diagram -

And, $f = \frac{1}{2 π \sqrt{LC}}$
We know, $X_{L} = ω L$
$X_{L} = 2 π f L = 2 π \times \frac{1}{2 π \sqrt{L C}} \times L$
$X_{L} = \sqrt{\frac{L}{C}} Ω$
And, $X_{C} = \frac{1}{2 π f C} = \frac{1}{2 π \times \frac{1}{2 π \sqrt{L C}} \times C} = \sqrt{\frac{L}{C}} Ω$
Voltage across capacitor $V_{C} (+) = V_{L} (+)$
Current through resistor $(I_{R}) = I_{L} - I_{C}$
$= \frac{V_{L}}{X_{L}} - \frac{V_{C}}{X_{C}}$
$I_{R} = \frac{V_{L}}{\sqrt{\frac{L}{C}}} - \frac{V_{L}}{\sqrt{\frac{L}{C}}}$
${∵ V_{L} = V_{C}}$
So,
$I_{R} = 0 A$
Hence the current through $R$ is zero.

TS- EAMCET-11.09.2020

Alternating Current

155138 Which of the following components of an $L - C$ - $R$ circuit, with AC supply, dissipates energy?

1 Only L

2 Only R

3 Only C

4

L

and

C

Explanation:

B In an A.C. circuit, the power dissipated is -
$P = VI \cos ϕ$
Where, $ϕ =$ phase difference between voltage and current.
For inductor (L) and capacitor (C), the phase difference between $V$ and $I$ is $90^{\circ}$ .
Then, $P_{L}$ or $P_{C} = VI \cos 90^{\circ} = 0$
For resistor (R), the phase difference between $V$ and $I$ zero.
Then, $P_{R} = VI \cos 0^{\circ} = VI$
Hence, only due to resistor R energy is dissipated in an L-C-R circuit.

TS- EAMCET-09.09.2020

Alternating Current

155139 Which one of the following curves represents the variation of impedance $(Z)$ with frequency $f$ in a series L-C-R circuit, when connected to an AC source?

1

2

3

4

Explanation:

C We know,
Impedance for a series R-L-C circuit -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$
$Z = \sqrt{R^{2} + {(2 π f L - \frac{1}{2 π f C})}^{2}}$
At resonance, $X_{L} = X_{C}$
i.e. $Z_{min} = R$ and $f = f_{o} (f_{o} =$ resonance frequency $)$ Hence, the variation of $Z$ with frequency ( $f$ )in a series -

TS- EAMCET.14.09.2020

Alternating Current

155141 An L-C-R series circuit is connected to a source of alternating current. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of

1

π

2

\frac{π}{2}

3

\frac{π}{4}

4

0^{\circ}

Explanation:

D In LCR circuit
At resonance condition,
Inductive Reactance $=$ Capacitive reactance
$X_{L} = X_{C}$
Now, impedance in LCR circuit
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$Z = \sqrt{R^{2} + {(X_{L} - X_{L})}^{2}} [∵ X_{L} = X_{C}]$
$Z = \sqrt{R^{2}} = R$
So, At resonance, LCR circuit behave as pure resistive circuit. In purely resistive circuit, current and voltage are in same phase.
Then, phase difference, $ϕ = 0^{\circ}$ .

TS- EAMCET-14.09.2020

Alternating Current

155142 An alternating voltage $ε = 30 \sin 200 t$ (in volts) is applied to the circuit below. The amplitude of the current through the circuit is

1

3 A

2

2 A

3

1 A

4

0.5 A

Explanation:

A Given,
Alternating voltage, $ε = 30 \sin 200 t V$
.....(i)
Resistance, $(R) = 10 Ω$ ,
Inductance, $(L) = 0.05 H$ ,
Capacitance $(C) = 500 μ F = 500 \times 10^{- 6} F$

Standard equation of AC voltage, $ε = ε_{m} \sin ω t$
From equation (i) and (ii), we get-
$ε_{m} = 30 V, ω = 200 rad / \sec$
We know that,
Inductive reactance,
$X_{L} = ω L = 200 \times 0.05 = 10 Ω$
And, Capacitive reactance,
$X_{C} = \frac{I}{ω C} = \frac{1}{200 \times 500 \times 10^{- 6}} = 10 Ω$
Now, Impedance of the LCR circuit,
$Z = \sqrt{(R)^{2} + {(X_{L} - X_{C})}^{2}}$
$= \sqrt{(10)^{2} + (10 - 10)^{2}} = \sqrt{100 + 0}$
$Z = 10 Ω$
So, Amplitude of the current is,
$I_{max} = \frac{V_{m}}{Z} = \frac{ε_{m}}{Z} = \frac{30}{10} = 3 A$
$I_{max} = 3 A$

TS- EAMCET-10.09.2020

Alternating Current

155144 The $L - C - R$ circuit shown below is driven by an ideal AC voltage source.

At frequency $f = \frac{1}{2 π \sqrt{LC}}$ , choose the correct statement.

1 The current through

R

is zero.

2 The current through

R

is infinite.

3 The current through

R

depends on the value of

L

and

C

.

4 The current through

R

depends only on the value of

R

and not

L

and

C

.

Explanation:

A Given circuit diagram -

And, $f = \frac{1}{2 π \sqrt{LC}}$
We know, $X_{L} = ω L$
$X_{L} = 2 π f L = 2 π \times \frac{1}{2 π \sqrt{L C}} \times L$
$X_{L} = \sqrt{\frac{L}{C}} Ω$
And, $X_{C} = \frac{1}{2 π f C} = \frac{1}{2 π \times \frac{1}{2 π \sqrt{L C}} \times C} = \sqrt{\frac{L}{C}} Ω$
Voltage across capacitor $V_{C} (+) = V_{L} (+)$
Current through resistor $(I_{R}) = I_{L} - I_{C}$
$= \frac{V_{L}}{X_{L}} - \frac{V_{C}}{X_{C}}$
$I_{R} = \frac{V_{L}}{\sqrt{\frac{L}{C}}} - \frac{V_{L}}{\sqrt{\frac{L}{C}}}$
${∵ V_{L} = V_{C}}$
So,
$I_{R} = 0 A$
Hence the current through $R$ is zero.

TS- EAMCET-11.09.2020

Alternating Current

155138 Which of the following components of an $L - C$ - $R$ circuit, with AC supply, dissipates energy?

1 Only L

2 Only R

3 Only C

4

L

and

C

Explanation:

B In an A.C. circuit, the power dissipated is -
$P = VI \cos ϕ$
Where, $ϕ =$ phase difference between voltage and current.
For inductor (L) and capacitor (C), the phase difference between $V$ and $I$ is $90^{\circ}$ .
Then, $P_{L}$ or $P_{C} = VI \cos 90^{\circ} = 0$
For resistor (R), the phase difference between $V$ and $I$ zero.
Then, $P_{R} = VI \cos 0^{\circ} = VI$
Hence, only due to resistor R energy is dissipated in an L-C-R circuit.

TS- EAMCET-09.09.2020

Alternating Current

155139 Which one of the following curves represents the variation of impedance $(Z)$ with frequency $f$ in a series L-C-R circuit, when connected to an AC source?

1

2

3

4

Explanation:

C We know,
Impedance for a series R-L-C circuit -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$
$Z = \sqrt{R^{2} + {(2 π f L - \frac{1}{2 π f C})}^{2}}$
At resonance, $X_{L} = X_{C}$
i.e. $Z_{min} = R$ and $f = f_{o} (f_{o} =$ resonance frequency $)$ Hence, the variation of $Z$ with frequency ( $f$ )in a series -

TS- EAMCET.14.09.2020

Alternating Current

155141 An L-C-R series circuit is connected to a source of alternating current. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of

1

π

2

\frac{π}{2}

3

\frac{π}{4}

4

0^{\circ}

Explanation:

D In LCR circuit
At resonance condition,
Inductive Reactance $=$ Capacitive reactance
$X_{L} = X_{C}$
Now, impedance in LCR circuit
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$Z = \sqrt{R^{2} + {(X_{L} - X_{L})}^{2}} [∵ X_{L} = X_{C}]$
$Z = \sqrt{R^{2}} = R$
So, At resonance, LCR circuit behave as pure resistive circuit. In purely resistive circuit, current and voltage are in same phase.
Then, phase difference, $ϕ = 0^{\circ}$ .

TS- EAMCET-14.09.2020

Alternating Current

155142 An alternating voltage $ε = 30 \sin 200 t$ (in volts) is applied to the circuit below. The amplitude of the current through the circuit is

1

3 A

2

2 A

3

1 A

4

0.5 A

Explanation:

A Given,
Alternating voltage, $ε = 30 \sin 200 t V$
.....(i)
Resistance, $(R) = 10 Ω$ ,
Inductance, $(L) = 0.05 H$ ,
Capacitance $(C) = 500 μ F = 500 \times 10^{- 6} F$

Standard equation of AC voltage, $ε = ε_{m} \sin ω t$
From equation (i) and (ii), we get-
$ε_{m} = 30 V, ω = 200 rad / \sec$
We know that,
Inductive reactance,
$X_{L} = ω L = 200 \times 0.05 = 10 Ω$
And, Capacitive reactance,
$X_{C} = \frac{I}{ω C} = \frac{1}{200 \times 500 \times 10^{- 6}} = 10 Ω$
Now, Impedance of the LCR circuit,
$Z = \sqrt{(R)^{2} + {(X_{L} - X_{C})}^{2}}$
$= \sqrt{(10)^{2} + (10 - 10)^{2}} = \sqrt{100 + 0}$
$Z = 10 Ω$
So, Amplitude of the current is,
$I_{max} = \frac{V_{m}}{Z} = \frac{ε_{m}}{Z} = \frac{30}{10} = 3 A$
$I_{max} = 3 A$

TS- EAMCET-10.09.2020

Alternating Current

155144 The $L - C - R$ circuit shown below is driven by an ideal AC voltage source.

At frequency $f = \frac{1}{2 π \sqrt{LC}}$ , choose the correct statement.

1 The current through

R

is zero.

2 The current through

R

is infinite.

3 The current through

R

depends on the value of

L

and

C

.

4 The current through

R

depends only on the value of

R

and not

L

and

C

.

Explanation:

A Given circuit diagram -

And, $f = \frac{1}{2 π \sqrt{LC}}$
We know, $X_{L} = ω L$
$X_{L} = 2 π f L = 2 π \times \frac{1}{2 π \sqrt{L C}} \times L$
$X_{L} = \sqrt{\frac{L}{C}} Ω$
And, $X_{C} = \frac{1}{2 π f C} = \frac{1}{2 π \times \frac{1}{2 π \sqrt{L C}} \times C} = \sqrt{\frac{L}{C}} Ω$
Voltage across capacitor $V_{C} (+) = V_{L} (+)$
Current through resistor $(I_{R}) = I_{L} - I_{C}$
$= \frac{V_{L}}{X_{L}} - \frac{V_{C}}{X_{C}}$
$I_{R} = \frac{V_{L}}{\sqrt{\frac{L}{C}}} - \frac{V_{L}}{\sqrt{\frac{L}{C}}}$
${∵ V_{L} = V_{C}}$
So,
$I_{R} = 0 A$
Hence the current through $R$ is zero.

TS- EAMCET-11.09.2020

Alternating Current

155138 Which of the following components of an $L - C$ - $R$ circuit, with AC supply, dissipates energy?

1 Only L

2 Only R

3 Only C

4

L

and

C

Explanation:

B In an A.C. circuit, the power dissipated is -
$P = VI \cos ϕ$
Where, $ϕ =$ phase difference between voltage and current.
For inductor (L) and capacitor (C), the phase difference between $V$ and $I$ is $90^{\circ}$ .
Then, $P_{L}$ or $P_{C} = VI \cos 90^{\circ} = 0$
For resistor (R), the phase difference between $V$ and $I$ zero.
Then, $P_{R} = VI \cos 0^{\circ} = VI$
Hence, only due to resistor R energy is dissipated in an L-C-R circuit.

TS- EAMCET-09.09.2020

Alternating Current

155139 Which one of the following curves represents the variation of impedance $(Z)$ with frequency $f$ in a series L-C-R circuit, when connected to an AC source?

1

2

3

4

Explanation:

C We know,
Impedance for a series R-L-C circuit -
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$
$Z = \sqrt{R^{2} + {(2 π f L - \frac{1}{2 π f C})}^{2}}$
At resonance, $X_{L} = X_{C}$
i.e. $Z_{min} = R$ and $f = f_{o} (f_{o} =$ resonance frequency $)$ Hence, the variation of $Z$ with frequency ( $f$ )in a series -

TS- EAMCET.14.09.2020

Alternating Current

155141 An L-C-R series circuit is connected to a source of alternating current. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of

1

π

2

\frac{π}{2}

3

\frac{π}{4}

4

0^{\circ}

Explanation:

D In LCR circuit
At resonance condition,
Inductive Reactance $=$ Capacitive reactance
$X_{L} = X_{C}$
Now, impedance in LCR circuit
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$
$Z = \sqrt{R^{2} + {(X_{L} - X_{L})}^{2}} [∵ X_{L} = X_{C}]$
$Z = \sqrt{R^{2}} = R$
So, At resonance, LCR circuit behave as pure resistive circuit. In purely resistive circuit, current and voltage are in same phase.
Then, phase difference, $ϕ = 0^{\circ}$ .

TS- EAMCET-14.09.2020

Alternating Current

155142 An alternating voltage $ε = 30 \sin 200 t$ (in volts) is applied to the circuit below. The amplitude of the current through the circuit is

1

3 A

2

2 A

3

1 A

4

0.5 A

Explanation:

A Given,
Alternating voltage, $ε = 30 \sin 200 t V$
.....(i)
Resistance, $(R) = 10 Ω$ ,
Inductance, $(L) = 0.05 H$ ,
Capacitance $(C) = 500 μ F = 500 \times 10^{- 6} F$

Standard equation of AC voltage, $ε = ε_{m} \sin ω t$
From equation (i) and (ii), we get-
$ε_{m} = 30 V, ω = 200 rad / \sec$
We know that,
Inductive reactance,
$X_{L} = ω L = 200 \times 0.05 = 10 Ω$
And, Capacitive reactance,
$X_{C} = \frac{I}{ω C} = \frac{1}{200 \times 500 \times 10^{- 6}} = 10 Ω$
Now, Impedance of the LCR circuit,
$Z = \sqrt{(R)^{2} + {(X_{L} - X_{C})}^{2}}$
$= \sqrt{(10)^{2} + (10 - 10)^{2}} = \sqrt{100 + 0}$
$Z = 10 Ω$
So, Amplitude of the current is,
$I_{max} = \frac{V_{m}}{Z} = \frac{ε_{m}}{Z} = \frac{30}{10} = 3 A$
$I_{max} = 3 A$

TS- EAMCET-10.09.2020

Alternating Current

155144 The $L - C - R$ circuit shown below is driven by an ideal AC voltage source.

At frequency $f = \frac{1}{2 π \sqrt{LC}}$ , choose the correct statement.

1 The current through

R

is zero.

2 The current through

R

is infinite.

3 The current through

R

depends on the value of

L

and

C

.

4 The current through

R

depends only on the value of

R

and not

L

and

C

.

Explanation:

A Given circuit diagram -

And, $f = \frac{1}{2 π \sqrt{LC}}$
We know, $X_{L} = ω L$
$X_{L} = 2 π f L = 2 π \times \frac{1}{2 π \sqrt{L C}} \times L$
$X_{L} = \sqrt{\frac{L}{C}} Ω$
And, $X_{C} = \frac{1}{2 π f C} = \frac{1}{2 π \times \frac{1}{2 π \sqrt{L C}} \times C} = \sqrt{\frac{L}{C}} Ω$
Voltage across capacitor $V_{C} (+) = V_{L} (+)$
Current through resistor $(I_{R}) = I_{L} - I_{C}$
$= \frac{V_{L}}{X_{L}} - \frac{V_{C}}{X_{C}}$
$I_{R} = \frac{V_{L}}{\sqrt{\frac{L}{C}}} - \frac{V_{L}}{\sqrt{\frac{L}{C}}}$
${∵ V_{L} = V_{C}}$
So,
$I_{R} = 0 A$
Hence the current through $R$ is zero.

TS- EAMCET-11.09.2020

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