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Alternating Current

155067 An alternating potential of frequency $f$ is applied on a circuit containing a resistance $R$ an a choke $L$ in series. The impedance of this current is

1

R + 2 π fL

2

\sqrt{R^{2} + 4 π^{2} f^{2} L^{2}}

3

\sqrt{R^{2} + L^{2}}

4

\sqrt{R^{2} + 2 π fL}

Explanation:

B Given, LR circuit,

We know that,
Impedance of LR circuit, $Z = \sqrt{R^{2} + {(X_{L})}^{2}}$
$Z = \sqrt{R^{2} + (ω L)^{2}} (∵ X_{L} = ω L)$
$Z = \sqrt{R^{2} + (2 π fL)^{2}} (∵ ω_{L} = 2 π fL)$
$Z = \sqrt{R^{2} + 4 π^{2} f^{2} L^{2}}$

CG PET- 2006

Alternating Current

155068 Dimensions of $\frac{R}{L}$ are

1

[T^{2}]

2

[T]

3

[T^{- 1}]

4

[T^{- 2}]

Explanation:

C We know that,
Time constant $(τ) = \frac{L}{R}$
So, $\frac{1}{τ} = \frac{R}{L}$
As we know, dimension of time constant is $(τ)$ is [T]
$∴ \frac{1}{[T]} = \frac{R}{L}$
Then, dimension of $\frac{R}{L} = [T^{- 1}]$

CG PET- 2006

Alternating Current

155069 The time constant of a $L - R$ circuit represents the time in which the current in the circuit

1 reaches a value equal to about

37 %

of its maximum value

2 reaches a value equal to about

63 %

of its maximum value

3 attains a constant value

4 attains

50 %

of the constant value

Explanation:

B We know that, In $L - R$ circuit\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)\tau= $t i m e c o n s t a n t W h e n,$ t=\tau\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\tau / \tau}\right)\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-1}\right)\mathrm{I}=\mathrm{I}_{0}\left(1-\frac{1}{2.7}\right)=\mathrm{I}_{0} \times 0.63=63 \%$

CG PET- 2005

Alternating Current

155070 In an $AC$ circuit, the current lags behind the voltage by $π / 3$ . The components in the circuit are

1

R

and

L

2

R

and

C

3

L

and

C

4 Only R

Explanation:

A
In a series $L - R$ circuit e.m.f
$E = E_{0} \sin ω t$ be applied to it
Let $i$ be the current in the circuit at any instant and $V_{L}$ and $V_{R}$ be the voltage across $L$ and $R$ respectively at that instant.
Then $V_{L} = i X_{L}$ and $V_{R}$ iR where $X_{L}$ is the inductive reactance. Now $V_{R}$ is in phase with the current $i$ while $V_{L}$ leads i by $90^{\circ}$

Above phase diagram shows that in $L - R$ circuit the voltage leads the currents by a phase angle by $ϕ$
$\tan ϕ = \frac{V_{L}}{V_{R}} = \frac{i X_{L}}{i R} = \frac{ω L}{R}$
$ϕ = \tan^{- 1} (\frac{ω L}{R})$
So if in an AC circuit the current lags behind the voltage by $π / 3$ . The components of the circuit are $L$ and $R$

CG PET- 2004

Alternating Current

155067 An alternating potential of frequency $f$ is applied on a circuit containing a resistance $R$ an a choke $L$ in series. The impedance of this current is

1

R + 2 π fL

2

\sqrt{R^{2} + 4 π^{2} f^{2} L^{2}}

3

\sqrt{R^{2} + L^{2}}

4

\sqrt{R^{2} + 2 π fL}

Explanation:

B Given, LR circuit,

We know that,
Impedance of LR circuit, $Z = \sqrt{R^{2} + {(X_{L})}^{2}}$
$Z = \sqrt{R^{2} + (ω L)^{2}} (∵ X_{L} = ω L)$
$Z = \sqrt{R^{2} + (2 π fL)^{2}} (∵ ω_{L} = 2 π fL)$
$Z = \sqrt{R^{2} + 4 π^{2} f^{2} L^{2}}$

CG PET- 2006

Alternating Current

155068 Dimensions of $\frac{R}{L}$ are

1

[T^{2}]

2

[T]

3

[T^{- 1}]

4

[T^{- 2}]

Explanation:

C We know that,
Time constant $(τ) = \frac{L}{R}$
So, $\frac{1}{τ} = \frac{R}{L}$
As we know, dimension of time constant is $(τ)$ is [T]
$∴ \frac{1}{[T]} = \frac{R}{L}$
Then, dimension of $\frac{R}{L} = [T^{- 1}]$

CG PET- 2006

Alternating Current

155069 The time constant of a $L - R$ circuit represents the time in which the current in the circuit

1 reaches a value equal to about

37 %

of its maximum value

2 reaches a value equal to about

63 %

of its maximum value

3 attains a constant value

4 attains

50 %

of the constant value

Explanation:

B We know that, In $L - R$ circuit\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)\tau= $t i m e c o n s t a n t W h e n,$ t=\tau\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\tau / \tau}\right)\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-1}\right)\mathrm{I}=\mathrm{I}_{0}\left(1-\frac{1}{2.7}\right)=\mathrm{I}_{0} \times 0.63=63 \%$

CG PET- 2005

Alternating Current

155070 In an $AC$ circuit, the current lags behind the voltage by $π / 3$ . The components in the circuit are

1

R

and

L

2

R

and

C

3

L

and

C

4 Only R

Explanation:

A
In a series $L - R$ circuit e.m.f
$E = E_{0} \sin ω t$ be applied to it
Let $i$ be the current in the circuit at any instant and $V_{L}$ and $V_{R}$ be the voltage across $L$ and $R$ respectively at that instant.
Then $V_{L} = i X_{L}$ and $V_{R}$ iR where $X_{L}$ is the inductive reactance. Now $V_{R}$ is in phase with the current $i$ while $V_{L}$ leads i by $90^{\circ}$

Above phase diagram shows that in $L - R$ circuit the voltage leads the currents by a phase angle by $ϕ$
$\tan ϕ = \frac{V_{L}}{V_{R}} = \frac{i X_{L}}{i R} = \frac{ω L}{R}$
$ϕ = \tan^{- 1} (\frac{ω L}{R})$
So if in an AC circuit the current lags behind the voltage by $π / 3$ . The components of the circuit are $L$ and $R$

CG PET- 2004

Alternating Current

155067 An alternating potential of frequency $f$ is applied on a circuit containing a resistance $R$ an a choke $L$ in series. The impedance of this current is

1

R + 2 π fL

2

\sqrt{R^{2} + 4 π^{2} f^{2} L^{2}}

3

\sqrt{R^{2} + L^{2}}

4

\sqrt{R^{2} + 2 π fL}

Explanation:

B Given, LR circuit,

We know that,
Impedance of LR circuit, $Z = \sqrt{R^{2} + {(X_{L})}^{2}}$
$Z = \sqrt{R^{2} + (ω L)^{2}} (∵ X_{L} = ω L)$
$Z = \sqrt{R^{2} + (2 π fL)^{2}} (∵ ω_{L} = 2 π fL)$
$Z = \sqrt{R^{2} + 4 π^{2} f^{2} L^{2}}$

CG PET- 2006

Alternating Current

155068 Dimensions of $\frac{R}{L}$ are

1

[T^{2}]

2

[T]

3

[T^{- 1}]

4

[T^{- 2}]

Explanation:

C We know that,
Time constant $(τ) = \frac{L}{R}$
So, $\frac{1}{τ} = \frac{R}{L}$
As we know, dimension of time constant is $(τ)$ is [T]
$∴ \frac{1}{[T]} = \frac{R}{L}$
Then, dimension of $\frac{R}{L} = [T^{- 1}]$

CG PET- 2006

Alternating Current

155069 The time constant of a $L - R$ circuit represents the time in which the current in the circuit

1 reaches a value equal to about

37 %

of its maximum value

2 reaches a value equal to about

63 %

of its maximum value

3 attains a constant value

4 attains

50 %

of the constant value

Explanation:

B We know that, In $L - R$ circuit\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)\tau= $t i m e c o n s t a n t W h e n,$ t=\tau\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\tau / \tau}\right)\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-1}\right)\mathrm{I}=\mathrm{I}_{0}\left(1-\frac{1}{2.7}\right)=\mathrm{I}_{0} \times 0.63=63 \%$

CG PET- 2005

Alternating Current

155070 In an $AC$ circuit, the current lags behind the voltage by $π / 3$ . The components in the circuit are

1

R

and

L

2

R

and

C

3

L

and

C

4 Only R

Explanation:

A
In a series $L - R$ circuit e.m.f
$E = E_{0} \sin ω t$ be applied to it
Let $i$ be the current in the circuit at any instant and $V_{L}$ and $V_{R}$ be the voltage across $L$ and $R$ respectively at that instant.
Then $V_{L} = i X_{L}$ and $V_{R}$ iR where $X_{L}$ is the inductive reactance. Now $V_{R}$ is in phase with the current $i$ while $V_{L}$ leads i by $90^{\circ}$

Above phase diagram shows that in $L - R$ circuit the voltage leads the currents by a phase angle by $ϕ$
$\tan ϕ = \frac{V_{L}}{V_{R}} = \frac{i X_{L}}{i R} = \frac{ω L}{R}$
$ϕ = \tan^{- 1} (\frac{ω L}{R})$
So if in an AC circuit the current lags behind the voltage by $π / 3$ . The components of the circuit are $L$ and $R$

CG PET- 2004

Alternating Current

155067 An alternating potential of frequency $f$ is applied on a circuit containing a resistance $R$ an a choke $L$ in series. The impedance of this current is

1

R + 2 π fL

2

\sqrt{R^{2} + 4 π^{2} f^{2} L^{2}}

3

\sqrt{R^{2} + L^{2}}

4

\sqrt{R^{2} + 2 π fL}

Explanation:

B Given, LR circuit,

We know that,
Impedance of LR circuit, $Z = \sqrt{R^{2} + {(X_{L})}^{2}}$
$Z = \sqrt{R^{2} + (ω L)^{2}} (∵ X_{L} = ω L)$
$Z = \sqrt{R^{2} + (2 π fL)^{2}} (∵ ω_{L} = 2 π fL)$
$Z = \sqrt{R^{2} + 4 π^{2} f^{2} L^{2}}$

CG PET- 2006

Alternating Current

155068 Dimensions of $\frac{R}{L}$ are

1

[T^{2}]

2

[T]

3

[T^{- 1}]

4

[T^{- 2}]

Explanation:

C We know that,
Time constant $(τ) = \frac{L}{R}$
So, $\frac{1}{τ} = \frac{R}{L}$
As we know, dimension of time constant is $(τ)$ is [T]
$∴ \frac{1}{[T]} = \frac{R}{L}$
Then, dimension of $\frac{R}{L} = [T^{- 1}]$

CG PET- 2006

Alternating Current

155069 The time constant of a $L - R$ circuit represents the time in which the current in the circuit

1 reaches a value equal to about

37 %

of its maximum value

2 reaches a value equal to about

63 %

of its maximum value

3 attains a constant value

4 attains

50 %

of the constant value

Explanation:

B We know that, In $L - R$ circuit\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)\tau= $t i m e c o n s t a n t W h e n,$ t=\tau\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\tau / \tau}\right)\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-1}\right)\mathrm{I}=\mathrm{I}_{0}\left(1-\frac{1}{2.7}\right)=\mathrm{I}_{0} \times 0.63=63 \%$

CG PET- 2005

Alternating Current

155070 In an $AC$ circuit, the current lags behind the voltage by $π / 3$ . The components in the circuit are

1

R

and

L

2

R

and

C

3

L

and

C

4 Only R

Explanation:

A
In a series $L - R$ circuit e.m.f
$E = E_{0} \sin ω t$ be applied to it
Let $i$ be the current in the circuit at any instant and $V_{L}$ and $V_{R}$ be the voltage across $L$ and $R$ respectively at that instant.
Then $V_{L} = i X_{L}$ and $V_{R}$ iR where $X_{L}$ is the inductive reactance. Now $V_{R}$ is in phase with the current $i$ while $V_{L}$ leads i by $90^{\circ}$

Above phase diagram shows that in $L - R$ circuit the voltage leads the currents by a phase angle by $ϕ$
$\tan ϕ = \frac{V_{L}}{V_{R}} = \frac{i X_{L}}{i R} = \frac{ω L}{R}$
$ϕ = \tan^{- 1} (\frac{ω L}{R})$
So if in an AC circuit the current lags behind the voltage by $π / 3$ . The components of the circuit are $L$ and $R$

CG PET- 2004

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