356060
If reading of an ammeter is 20\(A\), then the peak value of current is:
1 \(\frac{{20}}{{\sqrt 2 }}A\)
2 \(\frac{5}{{\sqrt 2 }}A\)
3 \(20\sqrt 2 A\)
4 \(20\sqrt 2 \)
Explanation:
Ammeter reads the root mean square value of current \(({I_{rms}})\) is related to the peak value of current \(({I_0})\) by the relation \({I_{rms}} = \frac{{{I_0}}}{{\sqrt 2 }} \Rightarrow {I_0} = \sqrt 2 \times {I_{rms}} = 20\sqrt 2 A\)
PHXII07:ALTERNATING CURRENT
356061
The peak value of the alternating current given by \({I=4 \sin \omega t+4 \sin (\omega t+2 \pi / 3)}\) is
356060
If reading of an ammeter is 20\(A\), then the peak value of current is:
1 \(\frac{{20}}{{\sqrt 2 }}A\)
2 \(\frac{5}{{\sqrt 2 }}A\)
3 \(20\sqrt 2 A\)
4 \(20\sqrt 2 \)
Explanation:
Ammeter reads the root mean square value of current \(({I_{rms}})\) is related to the peak value of current \(({I_0})\) by the relation \({I_{rms}} = \frac{{{I_0}}}{{\sqrt 2 }} \Rightarrow {I_0} = \sqrt 2 \times {I_{rms}} = 20\sqrt 2 A\)
PHXII07:ALTERNATING CURRENT
356061
The peak value of the alternating current given by \({I=4 \sin \omega t+4 \sin (\omega t+2 \pi / 3)}\) is
356060
If reading of an ammeter is 20\(A\), then the peak value of current is:
1 \(\frac{{20}}{{\sqrt 2 }}A\)
2 \(\frac{5}{{\sqrt 2 }}A\)
3 \(20\sqrt 2 A\)
4 \(20\sqrt 2 \)
Explanation:
Ammeter reads the root mean square value of current \(({I_{rms}})\) is related to the peak value of current \(({I_0})\) by the relation \({I_{rms}} = \frac{{{I_0}}}{{\sqrt 2 }} \Rightarrow {I_0} = \sqrt 2 \times {I_{rms}} = 20\sqrt 2 A\)
PHXII07:ALTERNATING CURRENT
356061
The peak value of the alternating current given by \({I=4 \sin \omega t+4 \sin (\omega t+2 \pi / 3)}\) is
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PHXII07:ALTERNATING CURRENT
356059
The frequency of \(ac\) mains in India is
1 \(30\,c/s\,{\mathop{\rm or}\nolimits} \,Hz\)
2 \(50\,c/s\,{\mathop{\rm or}\nolimits} \,Hz\)
3 \(60\,c/s\,{\mathop{\rm or}\nolimits} \,Hz\)
4 \(120\,c/s\,{\mathop{\rm or}\nolimits} \,Hz\)
Explanation:
Conceptual Question
PHXII07:ALTERNATING CURRENT
356060
If reading of an ammeter is 20\(A\), then the peak value of current is:
1 \(\frac{{20}}{{\sqrt 2 }}A\)
2 \(\frac{5}{{\sqrt 2 }}A\)
3 \(20\sqrt 2 A\)
4 \(20\sqrt 2 \)
Explanation:
Ammeter reads the root mean square value of current \(({I_{rms}})\) is related to the peak value of current \(({I_0})\) by the relation \({I_{rms}} = \frac{{{I_0}}}{{\sqrt 2 }} \Rightarrow {I_0} = \sqrt 2 \times {I_{rms}} = 20\sqrt 2 A\)
PHXII07:ALTERNATING CURRENT
356061
The peak value of the alternating current given by \({I=4 \sin \omega t+4 \sin (\omega t+2 \pi / 3)}\) is
