356093
In an \(A C\) circuit, the \(r m s\) value of the current, \(I_{r m s}\), is related to the peak current \(I_{0}\) as
1 \(I_{r m s}=\dfrac{1}{\pi} I_{0}\)
2 \(I_{r m s}=\dfrac{1}{\sqrt{2}} I_{0}\)
3 \(I_{r m s}=\sqrt{2} I_{0}\)
4 \(I_{r m s}=\pi I_{0}\)
Explanation:
\(I_{r m s}=\dfrac{I_{0}}{\sqrt{2}}\) So, correct option is (2).
PHXII07:ALTERNATING CURRENT
356094
The voltage time \((V - t)\) graph for triangular wave having peak value \(V_{o}\) is as shown in figure. The rms value of \(V\) in time interval from \(t = 0\) to \(T/4\) is
1 \(\dfrac{V_{o}}{\sqrt{3}}\)
2 \(\dfrac{V_{o}}{2}\)
3 \(\dfrac{V_{o}}{\sqrt{2}}\)
4 None of these
Explanation:
The equation of the line from 0 to \(\dfrac{T}{4}\) is \(\begin{gathered}V=\dfrac{V_{o}}{T / 4} t \Rightarrow V=\dfrac{4 V_{o}}{T} t \\V_{r m s}=\sqrt{ < V^{2}>} \\ < V^{2}>=\left(\dfrac{4 V_{o}}{T}\right)^{2}\left\{\dfrac{\int_{0}^{T / 4} t^{2} d t}{\int_{0}^{T / 4} d t}\right\} \Rightarrow V_{r m s}=\dfrac{V_{o}}{\sqrt{3}}\end{gathered}\)
PHXII07:ALTERNATING CURRENT
356095
A periodic voltage \({V}\) varies with time \({t}\), as shown in the figure. \({T}\) is the time period. The rms value of the voltage is
1 \({\dfrac{V_{0}}{8}}\)
2 \({\dfrac{V_{0}}{2}}\)
3 \({V_{0}}\)
4 \({\dfrac{V_{0}}{4}}\)
Explanation:
Root mean square value \({\langle V\rangle=\sqrt{\dfrac{\int_{0}^{T / 4} V_{0}^{2} d t}{\int_{0}^{T} d t}}=\sqrt{\dfrac{V_{0}^{2}\left(\dfrac{T}{4}\right)}{T}}}\) \({=\sqrt{\dfrac{V_{0}^{2}}{4}}=\dfrac{V_{0}}{2}}\)
PHXII07:ALTERNATING CURRENT
356096
A direct current of 5\(A\) is superimposed on an alternating current \(I = 10\sin \omega t\) flowing through a wire. The effective value of the resulting current will be
356093
In an \(A C\) circuit, the \(r m s\) value of the current, \(I_{r m s}\), is related to the peak current \(I_{0}\) as
1 \(I_{r m s}=\dfrac{1}{\pi} I_{0}\)
2 \(I_{r m s}=\dfrac{1}{\sqrt{2}} I_{0}\)
3 \(I_{r m s}=\sqrt{2} I_{0}\)
4 \(I_{r m s}=\pi I_{0}\)
Explanation:
\(I_{r m s}=\dfrac{I_{0}}{\sqrt{2}}\) So, correct option is (2).
PHXII07:ALTERNATING CURRENT
356094
The voltage time \((V - t)\) graph for triangular wave having peak value \(V_{o}\) is as shown in figure. The rms value of \(V\) in time interval from \(t = 0\) to \(T/4\) is
1 \(\dfrac{V_{o}}{\sqrt{3}}\)
2 \(\dfrac{V_{o}}{2}\)
3 \(\dfrac{V_{o}}{\sqrt{2}}\)
4 None of these
Explanation:
The equation of the line from 0 to \(\dfrac{T}{4}\) is \(\begin{gathered}V=\dfrac{V_{o}}{T / 4} t \Rightarrow V=\dfrac{4 V_{o}}{T} t \\V_{r m s}=\sqrt{ < V^{2}>} \\ < V^{2}>=\left(\dfrac{4 V_{o}}{T}\right)^{2}\left\{\dfrac{\int_{0}^{T / 4} t^{2} d t}{\int_{0}^{T / 4} d t}\right\} \Rightarrow V_{r m s}=\dfrac{V_{o}}{\sqrt{3}}\end{gathered}\)
PHXII07:ALTERNATING CURRENT
356095
A periodic voltage \({V}\) varies with time \({t}\), as shown in the figure. \({T}\) is the time period. The rms value of the voltage is
1 \({\dfrac{V_{0}}{8}}\)
2 \({\dfrac{V_{0}}{2}}\)
3 \({V_{0}}\)
4 \({\dfrac{V_{0}}{4}}\)
Explanation:
Root mean square value \({\langle V\rangle=\sqrt{\dfrac{\int_{0}^{T / 4} V_{0}^{2} d t}{\int_{0}^{T} d t}}=\sqrt{\dfrac{V_{0}^{2}\left(\dfrac{T}{4}\right)}{T}}}\) \({=\sqrt{\dfrac{V_{0}^{2}}{4}}=\dfrac{V_{0}}{2}}\)
PHXII07:ALTERNATING CURRENT
356096
A direct current of 5\(A\) is superimposed on an alternating current \(I = 10\sin \omega t\) flowing through a wire. The effective value of the resulting current will be
356093
In an \(A C\) circuit, the \(r m s\) value of the current, \(I_{r m s}\), is related to the peak current \(I_{0}\) as
1 \(I_{r m s}=\dfrac{1}{\pi} I_{0}\)
2 \(I_{r m s}=\dfrac{1}{\sqrt{2}} I_{0}\)
3 \(I_{r m s}=\sqrt{2} I_{0}\)
4 \(I_{r m s}=\pi I_{0}\)
Explanation:
\(I_{r m s}=\dfrac{I_{0}}{\sqrt{2}}\) So, correct option is (2).
PHXII07:ALTERNATING CURRENT
356094
The voltage time \((V - t)\) graph for triangular wave having peak value \(V_{o}\) is as shown in figure. The rms value of \(V\) in time interval from \(t = 0\) to \(T/4\) is
1 \(\dfrac{V_{o}}{\sqrt{3}}\)
2 \(\dfrac{V_{o}}{2}\)
3 \(\dfrac{V_{o}}{\sqrt{2}}\)
4 None of these
Explanation:
The equation of the line from 0 to \(\dfrac{T}{4}\) is \(\begin{gathered}V=\dfrac{V_{o}}{T / 4} t \Rightarrow V=\dfrac{4 V_{o}}{T} t \\V_{r m s}=\sqrt{ < V^{2}>} \\ < V^{2}>=\left(\dfrac{4 V_{o}}{T}\right)^{2}\left\{\dfrac{\int_{0}^{T / 4} t^{2} d t}{\int_{0}^{T / 4} d t}\right\} \Rightarrow V_{r m s}=\dfrac{V_{o}}{\sqrt{3}}\end{gathered}\)
PHXII07:ALTERNATING CURRENT
356095
A periodic voltage \({V}\) varies with time \({t}\), as shown in the figure. \({T}\) is the time period. The rms value of the voltage is
1 \({\dfrac{V_{0}}{8}}\)
2 \({\dfrac{V_{0}}{2}}\)
3 \({V_{0}}\)
4 \({\dfrac{V_{0}}{4}}\)
Explanation:
Root mean square value \({\langle V\rangle=\sqrt{\dfrac{\int_{0}^{T / 4} V_{0}^{2} d t}{\int_{0}^{T} d t}}=\sqrt{\dfrac{V_{0}^{2}\left(\dfrac{T}{4}\right)}{T}}}\) \({=\sqrt{\dfrac{V_{0}^{2}}{4}}=\dfrac{V_{0}}{2}}\)
PHXII07:ALTERNATING CURRENT
356096
A direct current of 5\(A\) is superimposed on an alternating current \(I = 10\sin \omega t\) flowing through a wire. The effective value of the resulting current will be
356093
In an \(A C\) circuit, the \(r m s\) value of the current, \(I_{r m s}\), is related to the peak current \(I_{0}\) as
1 \(I_{r m s}=\dfrac{1}{\pi} I_{0}\)
2 \(I_{r m s}=\dfrac{1}{\sqrt{2}} I_{0}\)
3 \(I_{r m s}=\sqrt{2} I_{0}\)
4 \(I_{r m s}=\pi I_{0}\)
Explanation:
\(I_{r m s}=\dfrac{I_{0}}{\sqrt{2}}\) So, correct option is (2).
PHXII07:ALTERNATING CURRENT
356094
The voltage time \((V - t)\) graph for triangular wave having peak value \(V_{o}\) is as shown in figure. The rms value of \(V\) in time interval from \(t = 0\) to \(T/4\) is
1 \(\dfrac{V_{o}}{\sqrt{3}}\)
2 \(\dfrac{V_{o}}{2}\)
3 \(\dfrac{V_{o}}{\sqrt{2}}\)
4 None of these
Explanation:
The equation of the line from 0 to \(\dfrac{T}{4}\) is \(\begin{gathered}V=\dfrac{V_{o}}{T / 4} t \Rightarrow V=\dfrac{4 V_{o}}{T} t \\V_{r m s}=\sqrt{ < V^{2}>} \\ < V^{2}>=\left(\dfrac{4 V_{o}}{T}\right)^{2}\left\{\dfrac{\int_{0}^{T / 4} t^{2} d t}{\int_{0}^{T / 4} d t}\right\} \Rightarrow V_{r m s}=\dfrac{V_{o}}{\sqrt{3}}\end{gathered}\)
PHXII07:ALTERNATING CURRENT
356095
A periodic voltage \({V}\) varies with time \({t}\), as shown in the figure. \({T}\) is the time period. The rms value of the voltage is
1 \({\dfrac{V_{0}}{8}}\)
2 \({\dfrac{V_{0}}{2}}\)
3 \({V_{0}}\)
4 \({\dfrac{V_{0}}{4}}\)
Explanation:
Root mean square value \({\langle V\rangle=\sqrt{\dfrac{\int_{0}^{T / 4} V_{0}^{2} d t}{\int_{0}^{T} d t}}=\sqrt{\dfrac{V_{0}^{2}\left(\dfrac{T}{4}\right)}{T}}}\) \({=\sqrt{\dfrac{V_{0}^{2}}{4}}=\dfrac{V_{0}}{2}}\)
PHXII07:ALTERNATING CURRENT
356096
A direct current of 5\(A\) is superimposed on an alternating current \(I = 10\sin \omega t\) flowing through a wire. The effective value of the resulting current will be
