356054
The average current of a sinusoidally varying alternating current of peak value \(5\;A\) with initial phase zero, between the instants \(t=T / 8\) to \(t=T / 4\) is (where ' \(T\) ' is time period)
1 \(\frac{{10}}{\pi }\sqrt 2 \;A\)
2 \(\frac{5}{\pi }\sqrt 2 \;A\)
3 \(\frac{{20\sqrt 2 }}{\pi }A\)
4 \(\frac{{10}}{\pi }A\)
Explanation:
\( < i>=\dfrac{\int_{T / 8}^{T / 4} i d t}{\int_{T / 8}^{T / 4} d t}\) Then, \(i=5 \sin \omega t\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356055
Alternating current in circuit is given by \(I = {I_0}\sin 2\pi \,nt\). Then the time taken by the current to rise from zero to r.m.s. value is equal to
356056
An alternating voltage is given by \(E = 100\sin \left( {{\rm{\omega t}} + \frac{\pi }{6}} \right)V\).The voltage will be maximum for the first time when \(t\) is [\(T = \) periodic time]
1 \(\frac{T}{{12}}\)
2 \(\frac{T}{2}\)
3 \(\frac{T}{6}\)
4 \(\frac{T}{3}\)
Explanation:
Given, alternating voltage is \(E = 100\sin \left( {\omega t + \frac{\pi }{6}} \right)V\) The voltage will be maximum at time \(t\), when \(\sin \left( {\omega t + \frac{\pi }{6}} \right) = 1\) \( \Rightarrow \omega t + \frac{\pi }{6} = {\sin ^{ - 1}}\left( 1 \right) = \frac{\pi }{2}\) \( \Rightarrow \omega t = \frac{\pi }{3}\) \( \Rightarrow t = \frac{\pi }{{3\omega }} = \frac{{\pi \times T}}{{3 \times 2\pi }}\,\,\left[ {\because \omega = \frac{{2\pi }}{T}} \right]\) \( \Rightarrow t = \frac{T}{6}\)
MHTCET - 2019
PHXII07:ALTERNATING CURRENT
356057
An \(AC\) source is \(240 - V - 60\,Hz\). The value of voltage after \(\frac{1}{{720}}s\) starting from zero will be:
356058
In an \(a.c\) circuit, peak value of voltage is \(423\,volts,\) its effective voltage is:
1 \(400\;V\)
2 \(323\;V\)
3 \(300\;V\)
4 \(340\;V\)
Explanation:
We know, The \(r.m.s.\) or effective voltage is \(V_{m s}=\dfrac{V_{0}}{\sqrt{2}}\) \( \Rightarrow {V_{rms}} = \frac{{423}}{{\sqrt 2 }} = 300\;V\) So, correct option is (3).
356054
The average current of a sinusoidally varying alternating current of peak value \(5\;A\) with initial phase zero, between the instants \(t=T / 8\) to \(t=T / 4\) is (where ' \(T\) ' is time period)
1 \(\frac{{10}}{\pi }\sqrt 2 \;A\)
2 \(\frac{5}{\pi }\sqrt 2 \;A\)
3 \(\frac{{20\sqrt 2 }}{\pi }A\)
4 \(\frac{{10}}{\pi }A\)
Explanation:
\( < i>=\dfrac{\int_{T / 8}^{T / 4} i d t}{\int_{T / 8}^{T / 4} d t}\) Then, \(i=5 \sin \omega t\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356055
Alternating current in circuit is given by \(I = {I_0}\sin 2\pi \,nt\). Then the time taken by the current to rise from zero to r.m.s. value is equal to
356056
An alternating voltage is given by \(E = 100\sin \left( {{\rm{\omega t}} + \frac{\pi }{6}} \right)V\).The voltage will be maximum for the first time when \(t\) is [\(T = \) periodic time]
1 \(\frac{T}{{12}}\)
2 \(\frac{T}{2}\)
3 \(\frac{T}{6}\)
4 \(\frac{T}{3}\)
Explanation:
Given, alternating voltage is \(E = 100\sin \left( {\omega t + \frac{\pi }{6}} \right)V\) The voltage will be maximum at time \(t\), when \(\sin \left( {\omega t + \frac{\pi }{6}} \right) = 1\) \( \Rightarrow \omega t + \frac{\pi }{6} = {\sin ^{ - 1}}\left( 1 \right) = \frac{\pi }{2}\) \( \Rightarrow \omega t = \frac{\pi }{3}\) \( \Rightarrow t = \frac{\pi }{{3\omega }} = \frac{{\pi \times T}}{{3 \times 2\pi }}\,\,\left[ {\because \omega = \frac{{2\pi }}{T}} \right]\) \( \Rightarrow t = \frac{T}{6}\)
MHTCET - 2019
PHXII07:ALTERNATING CURRENT
356057
An \(AC\) source is \(240 - V - 60\,Hz\). The value of voltage after \(\frac{1}{{720}}s\) starting from zero will be:
356058
In an \(a.c\) circuit, peak value of voltage is \(423\,volts,\) its effective voltage is:
1 \(400\;V\)
2 \(323\;V\)
3 \(300\;V\)
4 \(340\;V\)
Explanation:
We know, The \(r.m.s.\) or effective voltage is \(V_{m s}=\dfrac{V_{0}}{\sqrt{2}}\) \( \Rightarrow {V_{rms}} = \frac{{423}}{{\sqrt 2 }} = 300\;V\) So, correct option is (3).
356054
The average current of a sinusoidally varying alternating current of peak value \(5\;A\) with initial phase zero, between the instants \(t=T / 8\) to \(t=T / 4\) is (where ' \(T\) ' is time period)
1 \(\frac{{10}}{\pi }\sqrt 2 \;A\)
2 \(\frac{5}{\pi }\sqrt 2 \;A\)
3 \(\frac{{20\sqrt 2 }}{\pi }A\)
4 \(\frac{{10}}{\pi }A\)
Explanation:
\( < i>=\dfrac{\int_{T / 8}^{T / 4} i d t}{\int_{T / 8}^{T / 4} d t}\) Then, \(i=5 \sin \omega t\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356055
Alternating current in circuit is given by \(I = {I_0}\sin 2\pi \,nt\). Then the time taken by the current to rise from zero to r.m.s. value is equal to
356056
An alternating voltage is given by \(E = 100\sin \left( {{\rm{\omega t}} + \frac{\pi }{6}} \right)V\).The voltage will be maximum for the first time when \(t\) is [\(T = \) periodic time]
1 \(\frac{T}{{12}}\)
2 \(\frac{T}{2}\)
3 \(\frac{T}{6}\)
4 \(\frac{T}{3}\)
Explanation:
Given, alternating voltage is \(E = 100\sin \left( {\omega t + \frac{\pi }{6}} \right)V\) The voltage will be maximum at time \(t\), when \(\sin \left( {\omega t + \frac{\pi }{6}} \right) = 1\) \( \Rightarrow \omega t + \frac{\pi }{6} = {\sin ^{ - 1}}\left( 1 \right) = \frac{\pi }{2}\) \( \Rightarrow \omega t = \frac{\pi }{3}\) \( \Rightarrow t = \frac{\pi }{{3\omega }} = \frac{{\pi \times T}}{{3 \times 2\pi }}\,\,\left[ {\because \omega = \frac{{2\pi }}{T}} \right]\) \( \Rightarrow t = \frac{T}{6}\)
MHTCET - 2019
PHXII07:ALTERNATING CURRENT
356057
An \(AC\) source is \(240 - V - 60\,Hz\). The value of voltage after \(\frac{1}{{720}}s\) starting from zero will be:
356058
In an \(a.c\) circuit, peak value of voltage is \(423\,volts,\) its effective voltage is:
1 \(400\;V\)
2 \(323\;V\)
3 \(300\;V\)
4 \(340\;V\)
Explanation:
We know, The \(r.m.s.\) or effective voltage is \(V_{m s}=\dfrac{V_{0}}{\sqrt{2}}\) \( \Rightarrow {V_{rms}} = \frac{{423}}{{\sqrt 2 }} = 300\;V\) So, correct option is (3).
356054
The average current of a sinusoidally varying alternating current of peak value \(5\;A\) with initial phase zero, between the instants \(t=T / 8\) to \(t=T / 4\) is (where ' \(T\) ' is time period)
1 \(\frac{{10}}{\pi }\sqrt 2 \;A\)
2 \(\frac{5}{\pi }\sqrt 2 \;A\)
3 \(\frac{{20\sqrt 2 }}{\pi }A\)
4 \(\frac{{10}}{\pi }A\)
Explanation:
\( < i>=\dfrac{\int_{T / 8}^{T / 4} i d t}{\int_{T / 8}^{T / 4} d t}\) Then, \(i=5 \sin \omega t\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356055
Alternating current in circuit is given by \(I = {I_0}\sin 2\pi \,nt\). Then the time taken by the current to rise from zero to r.m.s. value is equal to
356056
An alternating voltage is given by \(E = 100\sin \left( {{\rm{\omega t}} + \frac{\pi }{6}} \right)V\).The voltage will be maximum for the first time when \(t\) is [\(T = \) periodic time]
1 \(\frac{T}{{12}}\)
2 \(\frac{T}{2}\)
3 \(\frac{T}{6}\)
4 \(\frac{T}{3}\)
Explanation:
Given, alternating voltage is \(E = 100\sin \left( {\omega t + \frac{\pi }{6}} \right)V\) The voltage will be maximum at time \(t\), when \(\sin \left( {\omega t + \frac{\pi }{6}} \right) = 1\) \( \Rightarrow \omega t + \frac{\pi }{6} = {\sin ^{ - 1}}\left( 1 \right) = \frac{\pi }{2}\) \( \Rightarrow \omega t = \frac{\pi }{3}\) \( \Rightarrow t = \frac{\pi }{{3\omega }} = \frac{{\pi \times T}}{{3 \times 2\pi }}\,\,\left[ {\because \omega = \frac{{2\pi }}{T}} \right]\) \( \Rightarrow t = \frac{T}{6}\)
MHTCET - 2019
PHXII07:ALTERNATING CURRENT
356057
An \(AC\) source is \(240 - V - 60\,Hz\). The value of voltage after \(\frac{1}{{720}}s\) starting from zero will be:
356058
In an \(a.c\) circuit, peak value of voltage is \(423\,volts,\) its effective voltage is:
1 \(400\;V\)
2 \(323\;V\)
3 \(300\;V\)
4 \(340\;V\)
Explanation:
We know, The \(r.m.s.\) or effective voltage is \(V_{m s}=\dfrac{V_{0}}{\sqrt{2}}\) \( \Rightarrow {V_{rms}} = \frac{{423}}{{\sqrt 2 }} = 300\;V\) So, correct option is (3).
356054
The average current of a sinusoidally varying alternating current of peak value \(5\;A\) with initial phase zero, between the instants \(t=T / 8\) to \(t=T / 4\) is (where ' \(T\) ' is time period)
1 \(\frac{{10}}{\pi }\sqrt 2 \;A\)
2 \(\frac{5}{\pi }\sqrt 2 \;A\)
3 \(\frac{{20\sqrt 2 }}{\pi }A\)
4 \(\frac{{10}}{\pi }A\)
Explanation:
\( < i>=\dfrac{\int_{T / 8}^{T / 4} i d t}{\int_{T / 8}^{T / 4} d t}\) Then, \(i=5 \sin \omega t\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356055
Alternating current in circuit is given by \(I = {I_0}\sin 2\pi \,nt\). Then the time taken by the current to rise from zero to r.m.s. value is equal to
356056
An alternating voltage is given by \(E = 100\sin \left( {{\rm{\omega t}} + \frac{\pi }{6}} \right)V\).The voltage will be maximum for the first time when \(t\) is [\(T = \) periodic time]
1 \(\frac{T}{{12}}\)
2 \(\frac{T}{2}\)
3 \(\frac{T}{6}\)
4 \(\frac{T}{3}\)
Explanation:
Given, alternating voltage is \(E = 100\sin \left( {\omega t + \frac{\pi }{6}} \right)V\) The voltage will be maximum at time \(t\), when \(\sin \left( {\omega t + \frac{\pi }{6}} \right) = 1\) \( \Rightarrow \omega t + \frac{\pi }{6} = {\sin ^{ - 1}}\left( 1 \right) = \frac{\pi }{2}\) \( \Rightarrow \omega t = \frac{\pi }{3}\) \( \Rightarrow t = \frac{\pi }{{3\omega }} = \frac{{\pi \times T}}{{3 \times 2\pi }}\,\,\left[ {\because \omega = \frac{{2\pi }}{T}} \right]\) \( \Rightarrow t = \frac{T}{6}\)
MHTCET - 2019
PHXII07:ALTERNATING CURRENT
356057
An \(AC\) source is \(240 - V - 60\,Hz\). The value of voltage after \(\frac{1}{{720}}s\) starting from zero will be:
356058
In an \(a.c\) circuit, peak value of voltage is \(423\,volts,\) its effective voltage is:
1 \(400\;V\)
2 \(323\;V\)
3 \(300\;V\)
4 \(340\;V\)
Explanation:
We know, The \(r.m.s.\) or effective voltage is \(V_{m s}=\dfrac{V_{0}}{\sqrt{2}}\) \( \Rightarrow {V_{rms}} = \frac{{423}}{{\sqrt 2 }} = 300\;V\) So, correct option is (3).
