356076
Determine the virtual value of \(a c\) current shown in figure.
1 \(3\,A\)
2 \(4\,A\)
3 \(5\,A\)
4 \(2\,A\)
Explanation:
\(I_{r. m . s}=\sqrt{\dfrac{I_{1}{ }^{2}+I_{2}{ }^{2}+I_{3}{ }^{2}}{N}}=\sqrt{\dfrac{2^{2}+2^{2}+2^{2}}{3}}\) \({I_{r.m.s}} = 2\;A\) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356077
The current vs time graph is shown in the figure. The average and rms currents are. (for one cycle)
356078
The \(r.m.s\) voltage of the wave from shown is:
1 \(10\;V\)
2 \(7\;V\)
3 \(6.37\;V\)
4 \(11\;V\)
Explanation:
\(V_{r m s}=\sqrt{\dfrac{\left(V_{1}\right)^{2}+\left(V_{2}\right)^{3}+\left(V_{3}\right)^{2}}{\text { Number of loops }}}\) \(V_{r m s}=\sqrt{\dfrac{(10)^{2}+(10)^{3}+(10)^{2}}{3}}\) \({V_{rms}} = \sqrt {\frac{{300}}{3}} = \sqrt {100} = 10\;V\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356079
If the instantaneous current in a circuit is given by \(I=2 \cos (\omega t+\phi)\) ampere, the \(r m s\) value of the current is
1 \(2\;A\)
2 \(\sqrt 2 \,A\)
3 \(2\sqrt 2 \,A\)
4 Zero
Explanation:
\({I_{rms}} = \frac{{{I_0}}}{{\sqrt 2 }} = \frac{2}{{\sqrt 2 }} = \sqrt 2 \;A\) So, correct option is (2).
356076
Determine the virtual value of \(a c\) current shown in figure.
1 \(3\,A\)
2 \(4\,A\)
3 \(5\,A\)
4 \(2\,A\)
Explanation:
\(I_{r. m . s}=\sqrt{\dfrac{I_{1}{ }^{2}+I_{2}{ }^{2}+I_{3}{ }^{2}}{N}}=\sqrt{\dfrac{2^{2}+2^{2}+2^{2}}{3}}\) \({I_{r.m.s}} = 2\;A\) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356077
The current vs time graph is shown in the figure. The average and rms currents are. (for one cycle)
356078
The \(r.m.s\) voltage of the wave from shown is:
1 \(10\;V\)
2 \(7\;V\)
3 \(6.37\;V\)
4 \(11\;V\)
Explanation:
\(V_{r m s}=\sqrt{\dfrac{\left(V_{1}\right)^{2}+\left(V_{2}\right)^{3}+\left(V_{3}\right)^{2}}{\text { Number of loops }}}\) \(V_{r m s}=\sqrt{\dfrac{(10)^{2}+(10)^{3}+(10)^{2}}{3}}\) \({V_{rms}} = \sqrt {\frac{{300}}{3}} = \sqrt {100} = 10\;V\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356079
If the instantaneous current in a circuit is given by \(I=2 \cos (\omega t+\phi)\) ampere, the \(r m s\) value of the current is
1 \(2\;A\)
2 \(\sqrt 2 \,A\)
3 \(2\sqrt 2 \,A\)
4 Zero
Explanation:
\({I_{rms}} = \frac{{{I_0}}}{{\sqrt 2 }} = \frac{2}{{\sqrt 2 }} = \sqrt 2 \;A\) So, correct option is (2).
356076
Determine the virtual value of \(a c\) current shown in figure.
1 \(3\,A\)
2 \(4\,A\)
3 \(5\,A\)
4 \(2\,A\)
Explanation:
\(I_{r. m . s}=\sqrt{\dfrac{I_{1}{ }^{2}+I_{2}{ }^{2}+I_{3}{ }^{2}}{N}}=\sqrt{\dfrac{2^{2}+2^{2}+2^{2}}{3}}\) \({I_{r.m.s}} = 2\;A\) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356077
The current vs time graph is shown in the figure. The average and rms currents are. (for one cycle)
356078
The \(r.m.s\) voltage of the wave from shown is:
1 \(10\;V\)
2 \(7\;V\)
3 \(6.37\;V\)
4 \(11\;V\)
Explanation:
\(V_{r m s}=\sqrt{\dfrac{\left(V_{1}\right)^{2}+\left(V_{2}\right)^{3}+\left(V_{3}\right)^{2}}{\text { Number of loops }}}\) \(V_{r m s}=\sqrt{\dfrac{(10)^{2}+(10)^{3}+(10)^{2}}{3}}\) \({V_{rms}} = \sqrt {\frac{{300}}{3}} = \sqrt {100} = 10\;V\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356079
If the instantaneous current in a circuit is given by \(I=2 \cos (\omega t+\phi)\) ampere, the \(r m s\) value of the current is
1 \(2\;A\)
2 \(\sqrt 2 \,A\)
3 \(2\sqrt 2 \,A\)
4 Zero
Explanation:
\({I_{rms}} = \frac{{{I_0}}}{{\sqrt 2 }} = \frac{2}{{\sqrt 2 }} = \sqrt 2 \;A\) So, correct option is (2).
356076
Determine the virtual value of \(a c\) current shown in figure.
1 \(3\,A\)
2 \(4\,A\)
3 \(5\,A\)
4 \(2\,A\)
Explanation:
\(I_{r. m . s}=\sqrt{\dfrac{I_{1}{ }^{2}+I_{2}{ }^{2}+I_{3}{ }^{2}}{N}}=\sqrt{\dfrac{2^{2}+2^{2}+2^{2}}{3}}\) \({I_{r.m.s}} = 2\;A\) So, correct option is (4).
PHXII07:ALTERNATING CURRENT
356077
The current vs time graph is shown in the figure. The average and rms currents are. (for one cycle)
356078
The \(r.m.s\) voltage of the wave from shown is:
1 \(10\;V\)
2 \(7\;V\)
3 \(6.37\;V\)
4 \(11\;V\)
Explanation:
\(V_{r m s}=\sqrt{\dfrac{\left(V_{1}\right)^{2}+\left(V_{2}\right)^{3}+\left(V_{3}\right)^{2}}{\text { Number of loops }}}\) \(V_{r m s}=\sqrt{\dfrac{(10)^{2}+(10)^{3}+(10)^{2}}{3}}\) \({V_{rms}} = \sqrt {\frac{{300}}{3}} = \sqrt {100} = 10\;V\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356079
If the instantaneous current in a circuit is given by \(I=2 \cos (\omega t+\phi)\) ampere, the \(r m s\) value of the current is
1 \(2\;A\)
2 \(\sqrt 2 \,A\)
3 \(2\sqrt 2 \,A\)
4 Zero
Explanation:
\({I_{rms}} = \frac{{{I_0}}}{{\sqrt 2 }} = \frac{2}{{\sqrt 2 }} = \sqrt 2 \;A\) So, correct option is (2).
