274611
Determine the rms value of the emf given by $\text{E}($ in volt $)=8\text{sin}\left( \omega \text{t} \right)+6\text{sin}\left( 2\omega \text{t} \right)$
274612
The alternating current of equivalent value of $\frac{{{I}_{0}}}{\sqrt{2}}$ is
1 peak current
2 r.m.s. current
3 D.C. current
4 all of these
Explanation:
(b) $\frac{{{I}_{0}}}{\sqrt{2}}=$ RMS current
NCERT Page-236 / N-179
AC (NCERT)
274613
The peak value of the a.c. current flowing through a resistor is given by
1 ${{\text{I}}_{0}}={{\text{e}}_{0}}/\text{R}$
2 $I=e/R$
3 ${{\text{I}}_{0}}={{\text{e}}_{0}}$
4 ${{I}_{0}}=R/{{e}_{0}}$
Explanation:
(a) Peak value, ${{I}_{0}}=\frac{{{e}_{0}}}{R}$
NCERT Page-234 / N-179
AC (NCERT)
274614
The sinusoidal A.C. current flows through a resistor of resistance $R$. If the peak current is ${{I}_{p}}$, then power dissipated is
1 $\text{I}_{\text{p}}^{2}\text{Rcos}\theta $
2 $\frac{1}{2}{{I}_{p}}^{2}R$
3 $\frac{4}{\pi }I_{p}^{2}R$
4 $\frac{1}{{{\pi }^{2}}}{{I}_{p}}^{2}R$
Explanation:
(b) The value of r.m.s current is ${{I}_{\text{rms}}}=\frac{{{I}_{P}}}{\sqrt{2}}$
so power dissipated is $\text{P}=\text{I}_{\text{rms}}^{2}\text{R}=\frac{1}{2}\text{I}_{\text{P}}^{2}\text{R}$
NCERT Page-235 / N-179
AC (NCERT)
274615
The r.m.s value of an a.c. of $50\text{Hz}$ is $10\text{amp}$. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be
1 $2\times {{10}^{-2}}\text{sec}$ and $14.14\text{amp}$
2 $1\times {{10}^{-2}}\text{sec}$ and $7.07\text{amp}$
3 $5\times {{10}^{-3}}\text{sec}$ and $7.07\text{amp}$
4 $5\times {{10}^{-3}}\text{sec}$ and $14.14\text{amp}$
274611
Determine the rms value of the emf given by $\text{E}($ in volt $)=8\text{sin}\left( \omega \text{t} \right)+6\text{sin}\left( 2\omega \text{t} \right)$
274612
The alternating current of equivalent value of $\frac{{{I}_{0}}}{\sqrt{2}}$ is
1 peak current
2 r.m.s. current
3 D.C. current
4 all of these
Explanation:
(b) $\frac{{{I}_{0}}}{\sqrt{2}}=$ RMS current
NCERT Page-236 / N-179
AC (NCERT)
274613
The peak value of the a.c. current flowing through a resistor is given by
1 ${{\text{I}}_{0}}={{\text{e}}_{0}}/\text{R}$
2 $I=e/R$
3 ${{\text{I}}_{0}}={{\text{e}}_{0}}$
4 ${{I}_{0}}=R/{{e}_{0}}$
Explanation:
(a) Peak value, ${{I}_{0}}=\frac{{{e}_{0}}}{R}$
NCERT Page-234 / N-179
AC (NCERT)
274614
The sinusoidal A.C. current flows through a resistor of resistance $R$. If the peak current is ${{I}_{p}}$, then power dissipated is
1 $\text{I}_{\text{p}}^{2}\text{Rcos}\theta $
2 $\frac{1}{2}{{I}_{p}}^{2}R$
3 $\frac{4}{\pi }I_{p}^{2}R$
4 $\frac{1}{{{\pi }^{2}}}{{I}_{p}}^{2}R$
Explanation:
(b) The value of r.m.s current is ${{I}_{\text{rms}}}=\frac{{{I}_{P}}}{\sqrt{2}}$
so power dissipated is $\text{P}=\text{I}_{\text{rms}}^{2}\text{R}=\frac{1}{2}\text{I}_{\text{P}}^{2}\text{R}$
NCERT Page-235 / N-179
AC (NCERT)
274615
The r.m.s value of an a.c. of $50\text{Hz}$ is $10\text{amp}$. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be
1 $2\times {{10}^{-2}}\text{sec}$ and $14.14\text{amp}$
2 $1\times {{10}^{-2}}\text{sec}$ and $7.07\text{amp}$
3 $5\times {{10}^{-3}}\text{sec}$ and $7.07\text{amp}$
4 $5\times {{10}^{-3}}\text{sec}$ and $14.14\text{amp}$
274611
Determine the rms value of the emf given by $\text{E}($ in volt $)=8\text{sin}\left( \omega \text{t} \right)+6\text{sin}\left( 2\omega \text{t} \right)$
274612
The alternating current of equivalent value of $\frac{{{I}_{0}}}{\sqrt{2}}$ is
1 peak current
2 r.m.s. current
3 D.C. current
4 all of these
Explanation:
(b) $\frac{{{I}_{0}}}{\sqrt{2}}=$ RMS current
NCERT Page-236 / N-179
AC (NCERT)
274613
The peak value of the a.c. current flowing through a resistor is given by
1 ${{\text{I}}_{0}}={{\text{e}}_{0}}/\text{R}$
2 $I=e/R$
3 ${{\text{I}}_{0}}={{\text{e}}_{0}}$
4 ${{I}_{0}}=R/{{e}_{0}}$
Explanation:
(a) Peak value, ${{I}_{0}}=\frac{{{e}_{0}}}{R}$
NCERT Page-234 / N-179
AC (NCERT)
274614
The sinusoidal A.C. current flows through a resistor of resistance $R$. If the peak current is ${{I}_{p}}$, then power dissipated is
1 $\text{I}_{\text{p}}^{2}\text{Rcos}\theta $
2 $\frac{1}{2}{{I}_{p}}^{2}R$
3 $\frac{4}{\pi }I_{p}^{2}R$
4 $\frac{1}{{{\pi }^{2}}}{{I}_{p}}^{2}R$
Explanation:
(b) The value of r.m.s current is ${{I}_{\text{rms}}}=\frac{{{I}_{P}}}{\sqrt{2}}$
so power dissipated is $\text{P}=\text{I}_{\text{rms}}^{2}\text{R}=\frac{1}{2}\text{I}_{\text{P}}^{2}\text{R}$
NCERT Page-235 / N-179
AC (NCERT)
274615
The r.m.s value of an a.c. of $50\text{Hz}$ is $10\text{amp}$. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be
1 $2\times {{10}^{-2}}\text{sec}$ and $14.14\text{amp}$
2 $1\times {{10}^{-2}}\text{sec}$ and $7.07\text{amp}$
3 $5\times {{10}^{-3}}\text{sec}$ and $7.07\text{amp}$
4 $5\times {{10}^{-3}}\text{sec}$ and $14.14\text{amp}$
274611
Determine the rms value of the emf given by $\text{E}($ in volt $)=8\text{sin}\left( \omega \text{t} \right)+6\text{sin}\left( 2\omega \text{t} \right)$
274612
The alternating current of equivalent value of $\frac{{{I}_{0}}}{\sqrt{2}}$ is
1 peak current
2 r.m.s. current
3 D.C. current
4 all of these
Explanation:
(b) $\frac{{{I}_{0}}}{\sqrt{2}}=$ RMS current
NCERT Page-236 / N-179
AC (NCERT)
274613
The peak value of the a.c. current flowing through a resistor is given by
1 ${{\text{I}}_{0}}={{\text{e}}_{0}}/\text{R}$
2 $I=e/R$
3 ${{\text{I}}_{0}}={{\text{e}}_{0}}$
4 ${{I}_{0}}=R/{{e}_{0}}$
Explanation:
(a) Peak value, ${{I}_{0}}=\frac{{{e}_{0}}}{R}$
NCERT Page-234 / N-179
AC (NCERT)
274614
The sinusoidal A.C. current flows through a resistor of resistance $R$. If the peak current is ${{I}_{p}}$, then power dissipated is
1 $\text{I}_{\text{p}}^{2}\text{Rcos}\theta $
2 $\frac{1}{2}{{I}_{p}}^{2}R$
3 $\frac{4}{\pi }I_{p}^{2}R$
4 $\frac{1}{{{\pi }^{2}}}{{I}_{p}}^{2}R$
Explanation:
(b) The value of r.m.s current is ${{I}_{\text{rms}}}=\frac{{{I}_{P}}}{\sqrt{2}}$
so power dissipated is $\text{P}=\text{I}_{\text{rms}}^{2}\text{R}=\frac{1}{2}\text{I}_{\text{P}}^{2}\text{R}$
NCERT Page-235 / N-179
AC (NCERT)
274615
The r.m.s value of an a.c. of $50\text{Hz}$ is $10\text{amp}$. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be
1 $2\times {{10}^{-2}}\text{sec}$ and $14.14\text{amp}$
2 $1\times {{10}^{-2}}\text{sec}$ and $7.07\text{amp}$
3 $5\times {{10}^{-3}}\text{sec}$ and $7.07\text{amp}$
4 $5\times {{10}^{-3}}\text{sec}$ and $14.14\text{amp}$
274611
Determine the rms value of the emf given by $\text{E}($ in volt $)=8\text{sin}\left( \omega \text{t} \right)+6\text{sin}\left( 2\omega \text{t} \right)$
274612
The alternating current of equivalent value of $\frac{{{I}_{0}}}{\sqrt{2}}$ is
1 peak current
2 r.m.s. current
3 D.C. current
4 all of these
Explanation:
(b) $\frac{{{I}_{0}}}{\sqrt{2}}=$ RMS current
NCERT Page-236 / N-179
AC (NCERT)
274613
The peak value of the a.c. current flowing through a resistor is given by
1 ${{\text{I}}_{0}}={{\text{e}}_{0}}/\text{R}$
2 $I=e/R$
3 ${{\text{I}}_{0}}={{\text{e}}_{0}}$
4 ${{I}_{0}}=R/{{e}_{0}}$
Explanation:
(a) Peak value, ${{I}_{0}}=\frac{{{e}_{0}}}{R}$
NCERT Page-234 / N-179
AC (NCERT)
274614
The sinusoidal A.C. current flows through a resistor of resistance $R$. If the peak current is ${{I}_{p}}$, then power dissipated is
1 $\text{I}_{\text{p}}^{2}\text{Rcos}\theta $
2 $\frac{1}{2}{{I}_{p}}^{2}R$
3 $\frac{4}{\pi }I_{p}^{2}R$
4 $\frac{1}{{{\pi }^{2}}}{{I}_{p}}^{2}R$
Explanation:
(b) The value of r.m.s current is ${{I}_{\text{rms}}}=\frac{{{I}_{P}}}{\sqrt{2}}$
so power dissipated is $\text{P}=\text{I}_{\text{rms}}^{2}\text{R}=\frac{1}{2}\text{I}_{\text{P}}^{2}\text{R}$
NCERT Page-235 / N-179
AC (NCERT)
274615
The r.m.s value of an a.c. of $50\text{Hz}$ is $10\text{amp}$. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be
1 $2\times {{10}^{-2}}\text{sec}$ and $14.14\text{amp}$
2 $1\times {{10}^{-2}}\text{sec}$ and $7.07\text{amp}$
3 $5\times {{10}^{-3}}\text{sec}$ and $7.07\text{amp}$
4 $5\times {{10}^{-3}}\text{sec}$ and $14.14\text{amp}$
