274633
The equation of current in a purely inductive circuit is $5\text{sin}\left( 49\pi t-{{30}^{\circ }} \right)$. If the inductance is $30\text{mH}$ then the equation for the voltage across the inductor, will be :
Let $\left. \pi =\frac{22}{7} \right)$
(d) Voltage amplitude across the inductor
${{\text{v}}_{0}}={{\text{i}}_{0}}{{\text{x}}_{\text{L}}}={{\text{i}}_{0}}\left( \omega \text{L} \right)=\left( 5 \right)\left( 49\pi \right)\left( 30\times {{10}^{-3}} \right)=23.1\text{v}$
Voltage will lead current by ${{90}^{\circ }}$.
Therefore the equation for the voltage across the inductor $V=23.1\text{sin}\left( 49\pi t+{{60}^{\circ }} \right)$
NCERT Page-238 / N-182
AC (NCERT)
274628
In the case of an inductor
1 voltage lags the current by $\frac{\pi }{2}$
2 voltage leads the current by $\frac{\pi }{2}$
3 voltage leads the current by $\frac{\pi }{3}$
4 voltage leads the current by $\frac{\pi }{4}$
Explanation:
(b)
NCERT Page-239 / N-182
AC (NCERT)
274627
If the frequency of an A.C. is made 4 times of its initial value, the inductive reactance will
1 be 4 times
2 be half
3 be 2 times
4 romain the same
Explanation:
(a) In an inductor voltage leads the current by $\frac{\pi }{2}$ or current lags the voltage by $\frac{\pi }{2}$.
NCERT Page-239 / N-183
AC (NCERT)
274629
The instantaneous voltage through a device of impedance $20\text{ }\!\!\Omega\!\!\text{ }$ is $e=80\text{sin}100\pi \text{t}$. The effective value of the current is
1 $3\text{A}$
2 $2.828\text{A}$
3 $1.732\text{A}$
4 $4\text{A}$
Explanation:
(b) In a pure inductive circuit current always lags behind the emf by $\frac{\pi }{2}$.
If $v\left( t \right)={{v}_{0}}\text{sin}\omega t$ then $\text{I}={{\text{I}}_{0}}\text{sin}\left( \omega t-\frac{\pi }{2} \right)$
Now, given $v\left( t \right)=100\text{sin}\left( 500t \right)$
and ${{I}_{0}}=\frac{{{E}_{0}}}{\omega L}=\frac{100}{500\times 0.02}\left[ \because \text{L}=0.02\text{H} \right]$
${{I}_{0}}=10\text{sin}\left( 500t-\frac{\pi }{2} \right)=-10\text{cos}\left( 500t \right)$
NCERT Page-238 / N-182
AC (NCERT)
274630
A sinusoidal voltage $V\left( t \right)=100\text{sin}\left( 500t \right)$ is applied across a pure inductance of $\text{L}=0.02\text{H}$. The current through the coil is:
274633
The equation of current in a purely inductive circuit is $5\text{sin}\left( 49\pi t-{{30}^{\circ }} \right)$. If the inductance is $30\text{mH}$ then the equation for the voltage across the inductor, will be :
Let $\left. \pi =\frac{22}{7} \right)$
(d) Voltage amplitude across the inductor
${{\text{v}}_{0}}={{\text{i}}_{0}}{{\text{x}}_{\text{L}}}={{\text{i}}_{0}}\left( \omega \text{L} \right)=\left( 5 \right)\left( 49\pi \right)\left( 30\times {{10}^{-3}} \right)=23.1\text{v}$
Voltage will lead current by ${{90}^{\circ }}$.
Therefore the equation for the voltage across the inductor $V=23.1\text{sin}\left( 49\pi t+{{60}^{\circ }} \right)$
NCERT Page-238 / N-182
AC (NCERT)
274628
In the case of an inductor
1 voltage lags the current by $\frac{\pi }{2}$
2 voltage leads the current by $\frac{\pi }{2}$
3 voltage leads the current by $\frac{\pi }{3}$
4 voltage leads the current by $\frac{\pi }{4}$
Explanation:
(b)
NCERT Page-239 / N-182
AC (NCERT)
274627
If the frequency of an A.C. is made 4 times of its initial value, the inductive reactance will
1 be 4 times
2 be half
3 be 2 times
4 romain the same
Explanation:
(a) In an inductor voltage leads the current by $\frac{\pi }{2}$ or current lags the voltage by $\frac{\pi }{2}$.
NCERT Page-239 / N-183
AC (NCERT)
274629
The instantaneous voltage through a device of impedance $20\text{ }\!\!\Omega\!\!\text{ }$ is $e=80\text{sin}100\pi \text{t}$. The effective value of the current is
1 $3\text{A}$
2 $2.828\text{A}$
3 $1.732\text{A}$
4 $4\text{A}$
Explanation:
(b) In a pure inductive circuit current always lags behind the emf by $\frac{\pi }{2}$.
If $v\left( t \right)={{v}_{0}}\text{sin}\omega t$ then $\text{I}={{\text{I}}_{0}}\text{sin}\left( \omega t-\frac{\pi }{2} \right)$
Now, given $v\left( t \right)=100\text{sin}\left( 500t \right)$
and ${{I}_{0}}=\frac{{{E}_{0}}}{\omega L}=\frac{100}{500\times 0.02}\left[ \because \text{L}=0.02\text{H} \right]$
${{I}_{0}}=10\text{sin}\left( 500t-\frac{\pi }{2} \right)=-10\text{cos}\left( 500t \right)$
NCERT Page-238 / N-182
AC (NCERT)
274630
A sinusoidal voltage $V\left( t \right)=100\text{sin}\left( 500t \right)$ is applied across a pure inductance of $\text{L}=0.02\text{H}$. The current through the coil is:
274633
The equation of current in a purely inductive circuit is $5\text{sin}\left( 49\pi t-{{30}^{\circ }} \right)$. If the inductance is $30\text{mH}$ then the equation for the voltage across the inductor, will be :
Let $\left. \pi =\frac{22}{7} \right)$
(d) Voltage amplitude across the inductor
${{\text{v}}_{0}}={{\text{i}}_{0}}{{\text{x}}_{\text{L}}}={{\text{i}}_{0}}\left( \omega \text{L} \right)=\left( 5 \right)\left( 49\pi \right)\left( 30\times {{10}^{-3}} \right)=23.1\text{v}$
Voltage will lead current by ${{90}^{\circ }}$.
Therefore the equation for the voltage across the inductor $V=23.1\text{sin}\left( 49\pi t+{{60}^{\circ }} \right)$
NCERT Page-238 / N-182
AC (NCERT)
274628
In the case of an inductor
1 voltage lags the current by $\frac{\pi }{2}$
2 voltage leads the current by $\frac{\pi }{2}$
3 voltage leads the current by $\frac{\pi }{3}$
4 voltage leads the current by $\frac{\pi }{4}$
Explanation:
(b)
NCERT Page-239 / N-182
AC (NCERT)
274627
If the frequency of an A.C. is made 4 times of its initial value, the inductive reactance will
1 be 4 times
2 be half
3 be 2 times
4 romain the same
Explanation:
(a) In an inductor voltage leads the current by $\frac{\pi }{2}$ or current lags the voltage by $\frac{\pi }{2}$.
NCERT Page-239 / N-183
AC (NCERT)
274629
The instantaneous voltage through a device of impedance $20\text{ }\!\!\Omega\!\!\text{ }$ is $e=80\text{sin}100\pi \text{t}$. The effective value of the current is
1 $3\text{A}$
2 $2.828\text{A}$
3 $1.732\text{A}$
4 $4\text{A}$
Explanation:
(b) In a pure inductive circuit current always lags behind the emf by $\frac{\pi }{2}$.
If $v\left( t \right)={{v}_{0}}\text{sin}\omega t$ then $\text{I}={{\text{I}}_{0}}\text{sin}\left( \omega t-\frac{\pi }{2} \right)$
Now, given $v\left( t \right)=100\text{sin}\left( 500t \right)$
and ${{I}_{0}}=\frac{{{E}_{0}}}{\omega L}=\frac{100}{500\times 0.02}\left[ \because \text{L}=0.02\text{H} \right]$
${{I}_{0}}=10\text{sin}\left( 500t-\frac{\pi }{2} \right)=-10\text{cos}\left( 500t \right)$
NCERT Page-238 / N-182
AC (NCERT)
274630
A sinusoidal voltage $V\left( t \right)=100\text{sin}\left( 500t \right)$ is applied across a pure inductance of $\text{L}=0.02\text{H}$. The current through the coil is:
NEET Test Series from KOTA - 10 Papers In MS WORD
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AC (NCERT)
274633
The equation of current in a purely inductive circuit is $5\text{sin}\left( 49\pi t-{{30}^{\circ }} \right)$. If the inductance is $30\text{mH}$ then the equation for the voltage across the inductor, will be :
Let $\left. \pi =\frac{22}{7} \right)$
(d) Voltage amplitude across the inductor
${{\text{v}}_{0}}={{\text{i}}_{0}}{{\text{x}}_{\text{L}}}={{\text{i}}_{0}}\left( \omega \text{L} \right)=\left( 5 \right)\left( 49\pi \right)\left( 30\times {{10}^{-3}} \right)=23.1\text{v}$
Voltage will lead current by ${{90}^{\circ }}$.
Therefore the equation for the voltage across the inductor $V=23.1\text{sin}\left( 49\pi t+{{60}^{\circ }} \right)$
NCERT Page-238 / N-182
AC (NCERT)
274628
In the case of an inductor
1 voltage lags the current by $\frac{\pi }{2}$
2 voltage leads the current by $\frac{\pi }{2}$
3 voltage leads the current by $\frac{\pi }{3}$
4 voltage leads the current by $\frac{\pi }{4}$
Explanation:
(b)
NCERT Page-239 / N-182
AC (NCERT)
274627
If the frequency of an A.C. is made 4 times of its initial value, the inductive reactance will
1 be 4 times
2 be half
3 be 2 times
4 romain the same
Explanation:
(a) In an inductor voltage leads the current by $\frac{\pi }{2}$ or current lags the voltage by $\frac{\pi }{2}$.
NCERT Page-239 / N-183
AC (NCERT)
274629
The instantaneous voltage through a device of impedance $20\text{ }\!\!\Omega\!\!\text{ }$ is $e=80\text{sin}100\pi \text{t}$. The effective value of the current is
1 $3\text{A}$
2 $2.828\text{A}$
3 $1.732\text{A}$
4 $4\text{A}$
Explanation:
(b) In a pure inductive circuit current always lags behind the emf by $\frac{\pi }{2}$.
If $v\left( t \right)={{v}_{0}}\text{sin}\omega t$ then $\text{I}={{\text{I}}_{0}}\text{sin}\left( \omega t-\frac{\pi }{2} \right)$
Now, given $v\left( t \right)=100\text{sin}\left( 500t \right)$
and ${{I}_{0}}=\frac{{{E}_{0}}}{\omega L}=\frac{100}{500\times 0.02}\left[ \because \text{L}=0.02\text{H} \right]$
${{I}_{0}}=10\text{sin}\left( 500t-\frac{\pi }{2} \right)=-10\text{cos}\left( 500t \right)$
NCERT Page-238 / N-182
AC (NCERT)
274630
A sinusoidal voltage $V\left( t \right)=100\text{sin}\left( 500t \right)$ is applied across a pure inductance of $\text{L}=0.02\text{H}$. The current through the coil is:
274633
The equation of current in a purely inductive circuit is $5\text{sin}\left( 49\pi t-{{30}^{\circ }} \right)$. If the inductance is $30\text{mH}$ then the equation for the voltage across the inductor, will be :
Let $\left. \pi =\frac{22}{7} \right)$
(d) Voltage amplitude across the inductor
${{\text{v}}_{0}}={{\text{i}}_{0}}{{\text{x}}_{\text{L}}}={{\text{i}}_{0}}\left( \omega \text{L} \right)=\left( 5 \right)\left( 49\pi \right)\left( 30\times {{10}^{-3}} \right)=23.1\text{v}$
Voltage will lead current by ${{90}^{\circ }}$.
Therefore the equation for the voltage across the inductor $V=23.1\text{sin}\left( 49\pi t+{{60}^{\circ }} \right)$
NCERT Page-238 / N-182
AC (NCERT)
274628
In the case of an inductor
1 voltage lags the current by $\frac{\pi }{2}$
2 voltage leads the current by $\frac{\pi }{2}$
3 voltage leads the current by $\frac{\pi }{3}$
4 voltage leads the current by $\frac{\pi }{4}$
Explanation:
(b)
NCERT Page-239 / N-182
AC (NCERT)
274627
If the frequency of an A.C. is made 4 times of its initial value, the inductive reactance will
1 be 4 times
2 be half
3 be 2 times
4 romain the same
Explanation:
(a) In an inductor voltage leads the current by $\frac{\pi }{2}$ or current lags the voltage by $\frac{\pi }{2}$.
NCERT Page-239 / N-183
AC (NCERT)
274629
The instantaneous voltage through a device of impedance $20\text{ }\!\!\Omega\!\!\text{ }$ is $e=80\text{sin}100\pi \text{t}$. The effective value of the current is
1 $3\text{A}$
2 $2.828\text{A}$
3 $1.732\text{A}$
4 $4\text{A}$
Explanation:
(b) In a pure inductive circuit current always lags behind the emf by $\frac{\pi }{2}$.
If $v\left( t \right)={{v}_{0}}\text{sin}\omega t$ then $\text{I}={{\text{I}}_{0}}\text{sin}\left( \omega t-\frac{\pi }{2} \right)$
Now, given $v\left( t \right)=100\text{sin}\left( 500t \right)$
and ${{I}_{0}}=\frac{{{E}_{0}}}{\omega L}=\frac{100}{500\times 0.02}\left[ \because \text{L}=0.02\text{H} \right]$
${{I}_{0}}=10\text{sin}\left( 500t-\frac{\pi }{2} \right)=-10\text{cos}\left( 500t \right)$
NCERT Page-238 / N-182
AC (NCERT)
274630
A sinusoidal voltage $V\left( t \right)=100\text{sin}\left( 500t \right)$ is applied across a pure inductance of $\text{L}=0.02\text{H}$. The current through the coil is: