356071
If \({i_1} = 3\sin \omega t\) and \({i_2} = 4\cos \omega t,\) then \({i_3}\) is
1 \(5\sin (\omega t + 53^\circ )\)
2 \(5\sin (\omega t + 45^\circ )\)
3 \( - 5\sin (\omega t + 53^\circ )\)
4 \(5\sin (\omega t + 37^\circ )\)
Explanation:
\({i_3} = {i_1} + {i_2} = 3\sin \omega t + 4\cos \omega t\) \( = 3\sin \omega t + 4\sin (\omega t + {90^o})\) \({i_{{3_{Max}}}} = \sqrt {{3^2} + {4^2}} = 5\) \(\tan \phi = \frac{4}{3} \Rightarrow \phi = {53^o}\) So \({i_3} = {i_{{3_{Max}}}}\sin (\omega t + \phi ) = 5\sin (\omega t + {53^o})\)
PHXII07:ALTERNATING CURRENT
356072
Find the intial phase angle for \(V = 2\sin \omega t + 3\cos \omega t\) is
1 \({\tan ^{ - 1}}\left( {\frac{3}{2}} \right)\)
2 \({\tan ^{ - 1}}\left( {\frac{4}{5}} \right)\)
3 \({\tan ^{ - 1}}\left( {\frac{1}{2}} \right)\)
4 \({\tan ^{ - 1}}\left( {\frac{3}{4}} \right)\)
Explanation:
Comparing the given equation \(2\sin \omega t + 3\cos \omega t\) with \({V_0}\sin \left( {\omega t + \phi } \right)\) \({V_0}\sin \omega t\cos \phi + {V_0}\cos \omega t\sin \phi \) \( \Rightarrow \,{V_0}\cos \phi = 2\) and \({V_0}\sin \phi = 3\) \( \Rightarrow \,\,\tan \phi = \frac{3}{2}\)
PHXII07:ALTERNATING CURRENT
356073
If a direct current of value a ampere is superimposed on an alternative current \(I=b \sin \omega t\) flowing through a wire, what is the effective value of the resulting current in the circuit?
As current at any instant in the circuit is \(\begin{gathered}I=I_{d c}+I_{a c}=a+b \sin \omega t \\I_{e f f}=\left[\dfrac{\int_{0}^{T} I^{2} d t}{\int_{0}^{T} d t}\right]^{1 / 2}=\left[\dfrac{1}{T} \int_{0}^{T}(a+b \sin \omega t)^{2} d t\right]^{1 / 2}\end{gathered}\) i.e., \(I_{e f f}=\left[\dfrac{1}{T} \int_{0}^{T}\left(a^{2}+2 a b \sin \omega t+b^{2} \sin ^{2} \omega t\right) d t\right]^{1 / 2}\) but as \(\dfrac{1}{T} \int_{0}^{T} \sin \omega t d t=0\) and \(\dfrac{1}{T} \int_{0}^{T} \sin ^{2} \omega t d t=\dfrac{1}{2}\) So, \(I_{e f f}=\left[a^{2}+\dfrac{1}{2} b^{2}\right]^{1 / 2}\)
PHXII07:ALTERNATING CURRENT
356074
The instantaneous current in an \({A C}\) circuit is \({i=5 \sqrt{2} \sin \left(100 t+\dfrac{\pi}{3}\right)}\). The rms value of current is
1 \(5\sqrt 2 \,A\)
2 \(5\,A\)
3 \(\sqrt 2 \,A\)
4 \(3.185\,A\)
Explanation:
\({i_{r m s}=\dfrac{i_{0}}{\sqrt{2}}=\dfrac{5 \sqrt{2}}{\sqrt{2}}=5 {~A}}\)
PHXII07:ALTERNATING CURRENT
356075
A direct current of \(2\;A\) is superimposed on an alternating current \(I=5 \sin \omega t\) flowing through a wire. The rms value of the resulting current will be
1 4.5
2 13.5
3 16.5
4 12.5
Explanation:
Given \(I=2+5 \sin \omega t\) \(\begin{aligned}& I_{r m s}=\left[\dfrac{\int_{0}^{T} I^{2} d t}{\int_{0}^{T} d t}\right] \\& =\left[\dfrac{1}{T} \int_{0}^{T}(2+5 \sin \omega t)^{2} d t\right]^{1 / 2} \\& =\left[\dfrac{1}{T} \int_{0}^{T}\left(4+25 \sin ^{2} \omega t+20 \sin \omega t\right) d t\right]^{1 / 2}\end{aligned}\) But \(\int_{0}^{T} \sin \omega t d t=0\) \(\begin{aligned}& I_{r m s}=\dfrac{1}{T} 4 T+\dfrac{25}{T} \int_{0}^{T} \sin ^{2} \omega t d t \\& =4+\dfrac{25}{T}\left(\dfrac{T}{2}\right)=16.5\end{aligned}\)
356071
If \({i_1} = 3\sin \omega t\) and \({i_2} = 4\cos \omega t,\) then \({i_3}\) is
1 \(5\sin (\omega t + 53^\circ )\)
2 \(5\sin (\omega t + 45^\circ )\)
3 \( - 5\sin (\omega t + 53^\circ )\)
4 \(5\sin (\omega t + 37^\circ )\)
Explanation:
\({i_3} = {i_1} + {i_2} = 3\sin \omega t + 4\cos \omega t\) \( = 3\sin \omega t + 4\sin (\omega t + {90^o})\) \({i_{{3_{Max}}}} = \sqrt {{3^2} + {4^2}} = 5\) \(\tan \phi = \frac{4}{3} \Rightarrow \phi = {53^o}\) So \({i_3} = {i_{{3_{Max}}}}\sin (\omega t + \phi ) = 5\sin (\omega t + {53^o})\)
PHXII07:ALTERNATING CURRENT
356072
Find the intial phase angle for \(V = 2\sin \omega t + 3\cos \omega t\) is
1 \({\tan ^{ - 1}}\left( {\frac{3}{2}} \right)\)
2 \({\tan ^{ - 1}}\left( {\frac{4}{5}} \right)\)
3 \({\tan ^{ - 1}}\left( {\frac{1}{2}} \right)\)
4 \({\tan ^{ - 1}}\left( {\frac{3}{4}} \right)\)
Explanation:
Comparing the given equation \(2\sin \omega t + 3\cos \omega t\) with \({V_0}\sin \left( {\omega t + \phi } \right)\) \({V_0}\sin \omega t\cos \phi + {V_0}\cos \omega t\sin \phi \) \( \Rightarrow \,{V_0}\cos \phi = 2\) and \({V_0}\sin \phi = 3\) \( \Rightarrow \,\,\tan \phi = \frac{3}{2}\)
PHXII07:ALTERNATING CURRENT
356073
If a direct current of value a ampere is superimposed on an alternative current \(I=b \sin \omega t\) flowing through a wire, what is the effective value of the resulting current in the circuit?
As current at any instant in the circuit is \(\begin{gathered}I=I_{d c}+I_{a c}=a+b \sin \omega t \\I_{e f f}=\left[\dfrac{\int_{0}^{T} I^{2} d t}{\int_{0}^{T} d t}\right]^{1 / 2}=\left[\dfrac{1}{T} \int_{0}^{T}(a+b \sin \omega t)^{2} d t\right]^{1 / 2}\end{gathered}\) i.e., \(I_{e f f}=\left[\dfrac{1}{T} \int_{0}^{T}\left(a^{2}+2 a b \sin \omega t+b^{2} \sin ^{2} \omega t\right) d t\right]^{1 / 2}\) but as \(\dfrac{1}{T} \int_{0}^{T} \sin \omega t d t=0\) and \(\dfrac{1}{T} \int_{0}^{T} \sin ^{2} \omega t d t=\dfrac{1}{2}\) So, \(I_{e f f}=\left[a^{2}+\dfrac{1}{2} b^{2}\right]^{1 / 2}\)
PHXII07:ALTERNATING CURRENT
356074
The instantaneous current in an \({A C}\) circuit is \({i=5 \sqrt{2} \sin \left(100 t+\dfrac{\pi}{3}\right)}\). The rms value of current is
1 \(5\sqrt 2 \,A\)
2 \(5\,A\)
3 \(\sqrt 2 \,A\)
4 \(3.185\,A\)
Explanation:
\({i_{r m s}=\dfrac{i_{0}}{\sqrt{2}}=\dfrac{5 \sqrt{2}}{\sqrt{2}}=5 {~A}}\)
PHXII07:ALTERNATING CURRENT
356075
A direct current of \(2\;A\) is superimposed on an alternating current \(I=5 \sin \omega t\) flowing through a wire. The rms value of the resulting current will be
1 4.5
2 13.5
3 16.5
4 12.5
Explanation:
Given \(I=2+5 \sin \omega t\) \(\begin{aligned}& I_{r m s}=\left[\dfrac{\int_{0}^{T} I^{2} d t}{\int_{0}^{T} d t}\right] \\& =\left[\dfrac{1}{T} \int_{0}^{T}(2+5 \sin \omega t)^{2} d t\right]^{1 / 2} \\& =\left[\dfrac{1}{T} \int_{0}^{T}\left(4+25 \sin ^{2} \omega t+20 \sin \omega t\right) d t\right]^{1 / 2}\end{aligned}\) But \(\int_{0}^{T} \sin \omega t d t=0\) \(\begin{aligned}& I_{r m s}=\dfrac{1}{T} 4 T+\dfrac{25}{T} \int_{0}^{T} \sin ^{2} \omega t d t \\& =4+\dfrac{25}{T}\left(\dfrac{T}{2}\right)=16.5\end{aligned}\)
356071
If \({i_1} = 3\sin \omega t\) and \({i_2} = 4\cos \omega t,\) then \({i_3}\) is
1 \(5\sin (\omega t + 53^\circ )\)
2 \(5\sin (\omega t + 45^\circ )\)
3 \( - 5\sin (\omega t + 53^\circ )\)
4 \(5\sin (\omega t + 37^\circ )\)
Explanation:
\({i_3} = {i_1} + {i_2} = 3\sin \omega t + 4\cos \omega t\) \( = 3\sin \omega t + 4\sin (\omega t + {90^o})\) \({i_{{3_{Max}}}} = \sqrt {{3^2} + {4^2}} = 5\) \(\tan \phi = \frac{4}{3} \Rightarrow \phi = {53^o}\) So \({i_3} = {i_{{3_{Max}}}}\sin (\omega t + \phi ) = 5\sin (\omega t + {53^o})\)
PHXII07:ALTERNATING CURRENT
356072
Find the intial phase angle for \(V = 2\sin \omega t + 3\cos \omega t\) is
1 \({\tan ^{ - 1}}\left( {\frac{3}{2}} \right)\)
2 \({\tan ^{ - 1}}\left( {\frac{4}{5}} \right)\)
3 \({\tan ^{ - 1}}\left( {\frac{1}{2}} \right)\)
4 \({\tan ^{ - 1}}\left( {\frac{3}{4}} \right)\)
Explanation:
Comparing the given equation \(2\sin \omega t + 3\cos \omega t\) with \({V_0}\sin \left( {\omega t + \phi } \right)\) \({V_0}\sin \omega t\cos \phi + {V_0}\cos \omega t\sin \phi \) \( \Rightarrow \,{V_0}\cos \phi = 2\) and \({V_0}\sin \phi = 3\) \( \Rightarrow \,\,\tan \phi = \frac{3}{2}\)
PHXII07:ALTERNATING CURRENT
356073
If a direct current of value a ampere is superimposed on an alternative current \(I=b \sin \omega t\) flowing through a wire, what is the effective value of the resulting current in the circuit?
As current at any instant in the circuit is \(\begin{gathered}I=I_{d c}+I_{a c}=a+b \sin \omega t \\I_{e f f}=\left[\dfrac{\int_{0}^{T} I^{2} d t}{\int_{0}^{T} d t}\right]^{1 / 2}=\left[\dfrac{1}{T} \int_{0}^{T}(a+b \sin \omega t)^{2} d t\right]^{1 / 2}\end{gathered}\) i.e., \(I_{e f f}=\left[\dfrac{1}{T} \int_{0}^{T}\left(a^{2}+2 a b \sin \omega t+b^{2} \sin ^{2} \omega t\right) d t\right]^{1 / 2}\) but as \(\dfrac{1}{T} \int_{0}^{T} \sin \omega t d t=0\) and \(\dfrac{1}{T} \int_{0}^{T} \sin ^{2} \omega t d t=\dfrac{1}{2}\) So, \(I_{e f f}=\left[a^{2}+\dfrac{1}{2} b^{2}\right]^{1 / 2}\)
PHXII07:ALTERNATING CURRENT
356074
The instantaneous current in an \({A C}\) circuit is \({i=5 \sqrt{2} \sin \left(100 t+\dfrac{\pi}{3}\right)}\). The rms value of current is
1 \(5\sqrt 2 \,A\)
2 \(5\,A\)
3 \(\sqrt 2 \,A\)
4 \(3.185\,A\)
Explanation:
\({i_{r m s}=\dfrac{i_{0}}{\sqrt{2}}=\dfrac{5 \sqrt{2}}{\sqrt{2}}=5 {~A}}\)
PHXII07:ALTERNATING CURRENT
356075
A direct current of \(2\;A\) is superimposed on an alternating current \(I=5 \sin \omega t\) flowing through a wire. The rms value of the resulting current will be
1 4.5
2 13.5
3 16.5
4 12.5
Explanation:
Given \(I=2+5 \sin \omega t\) \(\begin{aligned}& I_{r m s}=\left[\dfrac{\int_{0}^{T} I^{2} d t}{\int_{0}^{T} d t}\right] \\& =\left[\dfrac{1}{T} \int_{0}^{T}(2+5 \sin \omega t)^{2} d t\right]^{1 / 2} \\& =\left[\dfrac{1}{T} \int_{0}^{T}\left(4+25 \sin ^{2} \omega t+20 \sin \omega t\right) d t\right]^{1 / 2}\end{aligned}\) But \(\int_{0}^{T} \sin \omega t d t=0\) \(\begin{aligned}& I_{r m s}=\dfrac{1}{T} 4 T+\dfrac{25}{T} \int_{0}^{T} \sin ^{2} \omega t d t \\& =4+\dfrac{25}{T}\left(\dfrac{T}{2}\right)=16.5\end{aligned}\)
356071
If \({i_1} = 3\sin \omega t\) and \({i_2} = 4\cos \omega t,\) then \({i_3}\) is
1 \(5\sin (\omega t + 53^\circ )\)
2 \(5\sin (\omega t + 45^\circ )\)
3 \( - 5\sin (\omega t + 53^\circ )\)
4 \(5\sin (\omega t + 37^\circ )\)
Explanation:
\({i_3} = {i_1} + {i_2} = 3\sin \omega t + 4\cos \omega t\) \( = 3\sin \omega t + 4\sin (\omega t + {90^o})\) \({i_{{3_{Max}}}} = \sqrt {{3^2} + {4^2}} = 5\) \(\tan \phi = \frac{4}{3} \Rightarrow \phi = {53^o}\) So \({i_3} = {i_{{3_{Max}}}}\sin (\omega t + \phi ) = 5\sin (\omega t + {53^o})\)
PHXII07:ALTERNATING CURRENT
356072
Find the intial phase angle for \(V = 2\sin \omega t + 3\cos \omega t\) is
1 \({\tan ^{ - 1}}\left( {\frac{3}{2}} \right)\)
2 \({\tan ^{ - 1}}\left( {\frac{4}{5}} \right)\)
3 \({\tan ^{ - 1}}\left( {\frac{1}{2}} \right)\)
4 \({\tan ^{ - 1}}\left( {\frac{3}{4}} \right)\)
Explanation:
Comparing the given equation \(2\sin \omega t + 3\cos \omega t\) with \({V_0}\sin \left( {\omega t + \phi } \right)\) \({V_0}\sin \omega t\cos \phi + {V_0}\cos \omega t\sin \phi \) \( \Rightarrow \,{V_0}\cos \phi = 2\) and \({V_0}\sin \phi = 3\) \( \Rightarrow \,\,\tan \phi = \frac{3}{2}\)
PHXII07:ALTERNATING CURRENT
356073
If a direct current of value a ampere is superimposed on an alternative current \(I=b \sin \omega t\) flowing through a wire, what is the effective value of the resulting current in the circuit?
As current at any instant in the circuit is \(\begin{gathered}I=I_{d c}+I_{a c}=a+b \sin \omega t \\I_{e f f}=\left[\dfrac{\int_{0}^{T} I^{2} d t}{\int_{0}^{T} d t}\right]^{1 / 2}=\left[\dfrac{1}{T} \int_{0}^{T}(a+b \sin \omega t)^{2} d t\right]^{1 / 2}\end{gathered}\) i.e., \(I_{e f f}=\left[\dfrac{1}{T} \int_{0}^{T}\left(a^{2}+2 a b \sin \omega t+b^{2} \sin ^{2} \omega t\right) d t\right]^{1 / 2}\) but as \(\dfrac{1}{T} \int_{0}^{T} \sin \omega t d t=0\) and \(\dfrac{1}{T} \int_{0}^{T} \sin ^{2} \omega t d t=\dfrac{1}{2}\) So, \(I_{e f f}=\left[a^{2}+\dfrac{1}{2} b^{2}\right]^{1 / 2}\)
PHXII07:ALTERNATING CURRENT
356074
The instantaneous current in an \({A C}\) circuit is \({i=5 \sqrt{2} \sin \left(100 t+\dfrac{\pi}{3}\right)}\). The rms value of current is
1 \(5\sqrt 2 \,A\)
2 \(5\,A\)
3 \(\sqrt 2 \,A\)
4 \(3.185\,A\)
Explanation:
\({i_{r m s}=\dfrac{i_{0}}{\sqrt{2}}=\dfrac{5 \sqrt{2}}{\sqrt{2}}=5 {~A}}\)
PHXII07:ALTERNATING CURRENT
356075
A direct current of \(2\;A\) is superimposed on an alternating current \(I=5 \sin \omega t\) flowing through a wire. The rms value of the resulting current will be
1 4.5
2 13.5
3 16.5
4 12.5
Explanation:
Given \(I=2+5 \sin \omega t\) \(\begin{aligned}& I_{r m s}=\left[\dfrac{\int_{0}^{T} I^{2} d t}{\int_{0}^{T} d t}\right] \\& =\left[\dfrac{1}{T} \int_{0}^{T}(2+5 \sin \omega t)^{2} d t\right]^{1 / 2} \\& =\left[\dfrac{1}{T} \int_{0}^{T}\left(4+25 \sin ^{2} \omega t+20 \sin \omega t\right) d t\right]^{1 / 2}\end{aligned}\) But \(\int_{0}^{T} \sin \omega t d t=0\) \(\begin{aligned}& I_{r m s}=\dfrac{1}{T} 4 T+\dfrac{25}{T} \int_{0}^{T} \sin ^{2} \omega t d t \\& =4+\dfrac{25}{T}\left(\dfrac{T}{2}\right)=16.5\end{aligned}\)
356071
If \({i_1} = 3\sin \omega t\) and \({i_2} = 4\cos \omega t,\) then \({i_3}\) is
1 \(5\sin (\omega t + 53^\circ )\)
2 \(5\sin (\omega t + 45^\circ )\)
3 \( - 5\sin (\omega t + 53^\circ )\)
4 \(5\sin (\omega t + 37^\circ )\)
Explanation:
\({i_3} = {i_1} + {i_2} = 3\sin \omega t + 4\cos \omega t\) \( = 3\sin \omega t + 4\sin (\omega t + {90^o})\) \({i_{{3_{Max}}}} = \sqrt {{3^2} + {4^2}} = 5\) \(\tan \phi = \frac{4}{3} \Rightarrow \phi = {53^o}\) So \({i_3} = {i_{{3_{Max}}}}\sin (\omega t + \phi ) = 5\sin (\omega t + {53^o})\)
PHXII07:ALTERNATING CURRENT
356072
Find the intial phase angle for \(V = 2\sin \omega t + 3\cos \omega t\) is
1 \({\tan ^{ - 1}}\left( {\frac{3}{2}} \right)\)
2 \({\tan ^{ - 1}}\left( {\frac{4}{5}} \right)\)
3 \({\tan ^{ - 1}}\left( {\frac{1}{2}} \right)\)
4 \({\tan ^{ - 1}}\left( {\frac{3}{4}} \right)\)
Explanation:
Comparing the given equation \(2\sin \omega t + 3\cos \omega t\) with \({V_0}\sin \left( {\omega t + \phi } \right)\) \({V_0}\sin \omega t\cos \phi + {V_0}\cos \omega t\sin \phi \) \( \Rightarrow \,{V_0}\cos \phi = 2\) and \({V_0}\sin \phi = 3\) \( \Rightarrow \,\,\tan \phi = \frac{3}{2}\)
PHXII07:ALTERNATING CURRENT
356073
If a direct current of value a ampere is superimposed on an alternative current \(I=b \sin \omega t\) flowing through a wire, what is the effective value of the resulting current in the circuit?
As current at any instant in the circuit is \(\begin{gathered}I=I_{d c}+I_{a c}=a+b \sin \omega t \\I_{e f f}=\left[\dfrac{\int_{0}^{T} I^{2} d t}{\int_{0}^{T} d t}\right]^{1 / 2}=\left[\dfrac{1}{T} \int_{0}^{T}(a+b \sin \omega t)^{2} d t\right]^{1 / 2}\end{gathered}\) i.e., \(I_{e f f}=\left[\dfrac{1}{T} \int_{0}^{T}\left(a^{2}+2 a b \sin \omega t+b^{2} \sin ^{2} \omega t\right) d t\right]^{1 / 2}\) but as \(\dfrac{1}{T} \int_{0}^{T} \sin \omega t d t=0\) and \(\dfrac{1}{T} \int_{0}^{T} \sin ^{2} \omega t d t=\dfrac{1}{2}\) So, \(I_{e f f}=\left[a^{2}+\dfrac{1}{2} b^{2}\right]^{1 / 2}\)
PHXII07:ALTERNATING CURRENT
356074
The instantaneous current in an \({A C}\) circuit is \({i=5 \sqrt{2} \sin \left(100 t+\dfrac{\pi}{3}\right)}\). The rms value of current is
1 \(5\sqrt 2 \,A\)
2 \(5\,A\)
3 \(\sqrt 2 \,A\)
4 \(3.185\,A\)
Explanation:
\({i_{r m s}=\dfrac{i_{0}}{\sqrt{2}}=\dfrac{5 \sqrt{2}}{\sqrt{2}}=5 {~A}}\)
PHXII07:ALTERNATING CURRENT
356075
A direct current of \(2\;A\) is superimposed on an alternating current \(I=5 \sin \omega t\) flowing through a wire. The rms value of the resulting current will be
1 4.5
2 13.5
3 16.5
4 12.5
Explanation:
Given \(I=2+5 \sin \omega t\) \(\begin{aligned}& I_{r m s}=\left[\dfrac{\int_{0}^{T} I^{2} d t}{\int_{0}^{T} d t}\right] \\& =\left[\dfrac{1}{T} \int_{0}^{T}(2+5 \sin \omega t)^{2} d t\right]^{1 / 2} \\& =\left[\dfrac{1}{T} \int_{0}^{T}\left(4+25 \sin ^{2} \omega t+20 \sin \omega t\right) d t\right]^{1 / 2}\end{aligned}\) But \(\int_{0}^{T} \sin \omega t d t=0\) \(\begin{aligned}& I_{r m s}=\dfrac{1}{T} 4 T+\dfrac{25}{T} \int_{0}^{T} \sin ^{2} \omega t d t \\& =4+\dfrac{25}{T}\left(\dfrac{T}{2}\right)=16.5\end{aligned}\)
