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
B Given, LR circuit, We know that, Impedance of LR circuit, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}\right)^{2}}$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+(\omega \mathrm{L})^{2}} \left(\because \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}\right)$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+(2 \pi \mathrm{fL})^{2}} \left(\because \omega_{\mathrm{L}}=2 \pi \mathrm{fL}\right)$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+4 \pi^{2} \mathrm{f}^{2} \mathrm{~L}^{2}}$
CG PET- 2006
Alternating Current
155068
Dimensions of $\frac{R}{L}$ are
1 $\left[\mathrm{T}^{2}\right]$
2 $[\mathrm{T}]$
3 $\left[\mathrm{T}^{-1}\right]$
4 $\left[\mathrm{T}^{-2}\right]$
Explanation:
C We know that, Time constant $(\tau)=\frac{\mathrm{L}}{\mathrm{R}}$ So, $\quad \frac{1}{\tau}=\frac{\mathrm{R}}{\mathrm{L}}$ As we know, dimension of time constant is $(\tau)$ is [T] $\therefore \quad \frac{1}{[\mathrm{~T}]}=\frac{\mathrm{R}}{\mathrm{L}}$ Then, dimension of $\frac{\mathrm{R}}{\mathrm{L}}=\left[\mathrm{T}^{-1}\right]$
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 $\mathrm{L}-\mathrm{R}$ circuit$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)$ $\tau=$ time constant When, $t=\tau$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\tau / \tau}\right)$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-1}\right)$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\frac{1}{2.7}\right)=\mathrm{I}_{0} \times 0.63=63 \%$
CG PET- 2005
Alternating Current
155070
In an $\mathrm{AC}$ circuit, the current lags behind the voltage by $\pi / 3$. The components in the circuit are
1 $\mathrm{R}$ and $\mathrm{L}$
2 $\mathrm{R}$ and $\mathrm{C}$
3 $\mathrm{L}$ and $\mathrm{C}$
4 Only R
Explanation:
A In a series $L-R$ circuit e.m.f $\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}$ 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 $\mathrm{V}_{\mathrm{L}}$ leads i by $90^{\circ}$ Above phase diagram shows that in $\mathrm{L}-\mathrm{R}$ circuit the voltage leads the currents by a phase angle by $\phi$ $\tan \phi=\frac{V_{L}}{V_{R}}=\frac{i X_{L}}{i R}=\frac{\omega L}{R}$ $\phi=\tan ^{-1}\left(\frac{\omega L}{R}\right)$ So if in an AC circuit the current lags behind the voltage by $\pi / 3$. The components of the circuit are $\mathrm{L}$ and $\mathrm{R}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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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
B Given, LR circuit, We know that, Impedance of LR circuit, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}\right)^{2}}$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+(\omega \mathrm{L})^{2}} \left(\because \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}\right)$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+(2 \pi \mathrm{fL})^{2}} \left(\because \omega_{\mathrm{L}}=2 \pi \mathrm{fL}\right)$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+4 \pi^{2} \mathrm{f}^{2} \mathrm{~L}^{2}}$
CG PET- 2006
Alternating Current
155068
Dimensions of $\frac{R}{L}$ are
1 $\left[\mathrm{T}^{2}\right]$
2 $[\mathrm{T}]$
3 $\left[\mathrm{T}^{-1}\right]$
4 $\left[\mathrm{T}^{-2}\right]$
Explanation:
C We know that, Time constant $(\tau)=\frac{\mathrm{L}}{\mathrm{R}}$ So, $\quad \frac{1}{\tau}=\frac{\mathrm{R}}{\mathrm{L}}$ As we know, dimension of time constant is $(\tau)$ is [T] $\therefore \quad \frac{1}{[\mathrm{~T}]}=\frac{\mathrm{R}}{\mathrm{L}}$ Then, dimension of $\frac{\mathrm{R}}{\mathrm{L}}=\left[\mathrm{T}^{-1}\right]$
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 $\mathrm{L}-\mathrm{R}$ circuit$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)$ $\tau=$ time constant When, $t=\tau$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\tau / \tau}\right)$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-1}\right)$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\frac{1}{2.7}\right)=\mathrm{I}_{0} \times 0.63=63 \%$
CG PET- 2005
Alternating Current
155070
In an $\mathrm{AC}$ circuit, the current lags behind the voltage by $\pi / 3$. The components in the circuit are
1 $\mathrm{R}$ and $\mathrm{L}$
2 $\mathrm{R}$ and $\mathrm{C}$
3 $\mathrm{L}$ and $\mathrm{C}$
4 Only R
Explanation:
A In a series $L-R$ circuit e.m.f $\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}$ 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 $\mathrm{V}_{\mathrm{L}}$ leads i by $90^{\circ}$ Above phase diagram shows that in $\mathrm{L}-\mathrm{R}$ circuit the voltage leads the currents by a phase angle by $\phi$ $\tan \phi=\frac{V_{L}}{V_{R}}=\frac{i X_{L}}{i R}=\frac{\omega L}{R}$ $\phi=\tan ^{-1}\left(\frac{\omega L}{R}\right)$ So if in an AC circuit the current lags behind the voltage by $\pi / 3$. The components of the circuit are $\mathrm{L}$ and $\mathrm{R}$
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
B Given, LR circuit, We know that, Impedance of LR circuit, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}\right)^{2}}$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+(\omega \mathrm{L})^{2}} \left(\because \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}\right)$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+(2 \pi \mathrm{fL})^{2}} \left(\because \omega_{\mathrm{L}}=2 \pi \mathrm{fL}\right)$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+4 \pi^{2} \mathrm{f}^{2} \mathrm{~L}^{2}}$
CG PET- 2006
Alternating Current
155068
Dimensions of $\frac{R}{L}$ are
1 $\left[\mathrm{T}^{2}\right]$
2 $[\mathrm{T}]$
3 $\left[\mathrm{T}^{-1}\right]$
4 $\left[\mathrm{T}^{-2}\right]$
Explanation:
C We know that, Time constant $(\tau)=\frac{\mathrm{L}}{\mathrm{R}}$ So, $\quad \frac{1}{\tau}=\frac{\mathrm{R}}{\mathrm{L}}$ As we know, dimension of time constant is $(\tau)$ is [T] $\therefore \quad \frac{1}{[\mathrm{~T}]}=\frac{\mathrm{R}}{\mathrm{L}}$ Then, dimension of $\frac{\mathrm{R}}{\mathrm{L}}=\left[\mathrm{T}^{-1}\right]$
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 $\mathrm{L}-\mathrm{R}$ circuit$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)$ $\tau=$ time constant When, $t=\tau$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\tau / \tau}\right)$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-1}\right)$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\frac{1}{2.7}\right)=\mathrm{I}_{0} \times 0.63=63 \%$
CG PET- 2005
Alternating Current
155070
In an $\mathrm{AC}$ circuit, the current lags behind the voltage by $\pi / 3$. The components in the circuit are
1 $\mathrm{R}$ and $\mathrm{L}$
2 $\mathrm{R}$ and $\mathrm{C}$
3 $\mathrm{L}$ and $\mathrm{C}$
4 Only R
Explanation:
A In a series $L-R$ circuit e.m.f $\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}$ 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 $\mathrm{V}_{\mathrm{L}}$ leads i by $90^{\circ}$ Above phase diagram shows that in $\mathrm{L}-\mathrm{R}$ circuit the voltage leads the currents by a phase angle by $\phi$ $\tan \phi=\frac{V_{L}}{V_{R}}=\frac{i X_{L}}{i R}=\frac{\omega L}{R}$ $\phi=\tan ^{-1}\left(\frac{\omega L}{R}\right)$ So if in an AC circuit the current lags behind the voltage by $\pi / 3$. The components of the circuit are $\mathrm{L}$ and $\mathrm{R}$
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
B Given, LR circuit, We know that, Impedance of LR circuit, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}\right)^{2}}$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+(\omega \mathrm{L})^{2}} \left(\because \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}\right)$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+(2 \pi \mathrm{fL})^{2}} \left(\because \omega_{\mathrm{L}}=2 \pi \mathrm{fL}\right)$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+4 \pi^{2} \mathrm{f}^{2} \mathrm{~L}^{2}}$
CG PET- 2006
Alternating Current
155068
Dimensions of $\frac{R}{L}$ are
1 $\left[\mathrm{T}^{2}\right]$
2 $[\mathrm{T}]$
3 $\left[\mathrm{T}^{-1}\right]$
4 $\left[\mathrm{T}^{-2}\right]$
Explanation:
C We know that, Time constant $(\tau)=\frac{\mathrm{L}}{\mathrm{R}}$ So, $\quad \frac{1}{\tau}=\frac{\mathrm{R}}{\mathrm{L}}$ As we know, dimension of time constant is $(\tau)$ is [T] $\therefore \quad \frac{1}{[\mathrm{~T}]}=\frac{\mathrm{R}}{\mathrm{L}}$ Then, dimension of $\frac{\mathrm{R}}{\mathrm{L}}=\left[\mathrm{T}^{-1}\right]$
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 $\mathrm{L}-\mathrm{R}$ circuit$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\mathrm{t} / \tau}\right)$ $\tau=$ time constant When, $t=\tau$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-\tau / \tau}\right)$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\mathrm{e}^{-1}\right)$ $\mathrm{I}=\mathrm{I}_{0}\left(1-\frac{1}{2.7}\right)=\mathrm{I}_{0} \times 0.63=63 \%$
CG PET- 2005
Alternating Current
155070
In an $\mathrm{AC}$ circuit, the current lags behind the voltage by $\pi / 3$. The components in the circuit are
1 $\mathrm{R}$ and $\mathrm{L}$
2 $\mathrm{R}$ and $\mathrm{C}$
3 $\mathrm{L}$ and $\mathrm{C}$
4 Only R
Explanation:
A In a series $L-R$ circuit e.m.f $\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}$ 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 $\mathrm{V}_{\mathrm{L}}$ leads i by $90^{\circ}$ Above phase diagram shows that in $\mathrm{L}-\mathrm{R}$ circuit the voltage leads the currents by a phase angle by $\phi$ $\tan \phi=\frac{V_{L}}{V_{R}}=\frac{i X_{L}}{i R}=\frac{\omega L}{R}$ $\phi=\tan ^{-1}\left(\frac{\omega L}{R}\right)$ So if in an AC circuit the current lags behind the voltage by $\pi / 3$. The components of the circuit are $\mathrm{L}$ and $\mathrm{R}$
