274617
The voltage of an ac supply varies with time $\left( t \right)$ as $V=120\text{sin}100\pi \text{tcos}100\pi \text{t}$. The maximum voltage and frequency respectively are
1 120 volt,$100\text{Hz}$
2 $\frac{120}{\sqrt{2}}$ volt, $100\text{Hz}$
3 60 volt,$200\text{Hz}$
4 60 volt,$100\text{Hz}$
Explanation:
(d) $\text{V}=120\text{sin}100\pi \text{tcos}100\pi \text{t}\Rightarrow \text{V}=60\text{sin}200\pi \text{t}{{\text{V}}_{\text{max}}}=60\text{V}$ and $v=100\text{Hz}$
NCERT Page-236 / N-180
AC (NCERT)
274618
A resistance of $40\text{ }\!\!\Omega\!\!\text{ }$ is connected to a source of alternating current rated $220\text{V},50\text{Hz}$. Find the time taken by the current to change from its maximum value to $\text{rms}$ value :
1 $2.5\text{ms}$
2 $1.25\text{ms}$
3 $2.5\text{s}$
4 $0.25\text{s}$
Explanation:
(a)
NCERT Page-236 / N-180
AC (NCERT)
274619
A resistance of $20\text{ohm}$ is connected to a source of an alternating potential $\text{V}=200\text{cos}\left( 100\pi \text{t} \right)$. The time taken by the current to change from its peak value to rms value is
1 $2.5\times {{10}^{-3}}\text{s}$
2 $25\times {{10}^{-3}}\text{s}$
3 $0.25\text{s}$
4 $0.20\text{s}$
Explanation:
(a) The current and potential difference are in phase with the resistance. So, the time taken would be same as time for voltage to change from $\left( t=0 \right)$ that is peak value to rms value.
Time taken by voltage to achieve its rms value of $\frac{200}{\sqrt{2}}$.
$\begin{array}{*{35}{r}}
{} & \frac{200}{\sqrt{2}}=200\text{cos}\left( 100\pi t \right) \\
\Rightarrow & \text{cos}\left( 100\pi t \right)=\frac{1}{\sqrt{2}}=\text{cos}\left( \frac{\pi }{4} \right) \\
t= & \frac{1}{400}\text{ }\!\!~\!\!\text{ second }\!\!~\!\!\text{ }=2.5\times {{10}^{-3}}\text{ }\!\!~\!\!\text{ sec}\text{. }\!\!~\!\!\text{ } \\
\end{array}$
274617
The voltage of an ac supply varies with time $\left( t \right)$ as $V=120\text{sin}100\pi \text{tcos}100\pi \text{t}$. The maximum voltage and frequency respectively are
1 120 volt,$100\text{Hz}$
2 $\frac{120}{\sqrt{2}}$ volt, $100\text{Hz}$
3 60 volt,$200\text{Hz}$
4 60 volt,$100\text{Hz}$
Explanation:
(d) $\text{V}=120\text{sin}100\pi \text{tcos}100\pi \text{t}\Rightarrow \text{V}=60\text{sin}200\pi \text{t}{{\text{V}}_{\text{max}}}=60\text{V}$ and $v=100\text{Hz}$
NCERT Page-236 / N-180
AC (NCERT)
274618
A resistance of $40\text{ }\!\!\Omega\!\!\text{ }$ is connected to a source of alternating current rated $220\text{V},50\text{Hz}$. Find the time taken by the current to change from its maximum value to $\text{rms}$ value :
1 $2.5\text{ms}$
2 $1.25\text{ms}$
3 $2.5\text{s}$
4 $0.25\text{s}$
Explanation:
(a)
NCERT Page-236 / N-180
AC (NCERT)
274619
A resistance of $20\text{ohm}$ is connected to a source of an alternating potential $\text{V}=200\text{cos}\left( 100\pi \text{t} \right)$. The time taken by the current to change from its peak value to rms value is
1 $2.5\times {{10}^{-3}}\text{s}$
2 $25\times {{10}^{-3}}\text{s}$
3 $0.25\text{s}$
4 $0.20\text{s}$
Explanation:
(a) The current and potential difference are in phase with the resistance. So, the time taken would be same as time for voltage to change from $\left( t=0 \right)$ that is peak value to rms value.
Time taken by voltage to achieve its rms value of $\frac{200}{\sqrt{2}}$.
$\begin{array}{*{35}{r}}
{} & \frac{200}{\sqrt{2}}=200\text{cos}\left( 100\pi t \right) \\
\Rightarrow & \text{cos}\left( 100\pi t \right)=\frac{1}{\sqrt{2}}=\text{cos}\left( \frac{\pi }{4} \right) \\
t= & \frac{1}{400}\text{ }\!\!~\!\!\text{ second }\!\!~\!\!\text{ }=2.5\times {{10}^{-3}}\text{ }\!\!~\!\!\text{ sec}\text{. }\!\!~\!\!\text{ } \\
\end{array}$
274617
The voltage of an ac supply varies with time $\left( t \right)$ as $V=120\text{sin}100\pi \text{tcos}100\pi \text{t}$. The maximum voltage and frequency respectively are
1 120 volt,$100\text{Hz}$
2 $\frac{120}{\sqrt{2}}$ volt, $100\text{Hz}$
3 60 volt,$200\text{Hz}$
4 60 volt,$100\text{Hz}$
Explanation:
(d) $\text{V}=120\text{sin}100\pi \text{tcos}100\pi \text{t}\Rightarrow \text{V}=60\text{sin}200\pi \text{t}{{\text{V}}_{\text{max}}}=60\text{V}$ and $v=100\text{Hz}$
NCERT Page-236 / N-180
AC (NCERT)
274618
A resistance of $40\text{ }\!\!\Omega\!\!\text{ }$ is connected to a source of alternating current rated $220\text{V},50\text{Hz}$. Find the time taken by the current to change from its maximum value to $\text{rms}$ value :
1 $2.5\text{ms}$
2 $1.25\text{ms}$
3 $2.5\text{s}$
4 $0.25\text{s}$
Explanation:
(a)
NCERT Page-236 / N-180
AC (NCERT)
274619
A resistance of $20\text{ohm}$ is connected to a source of an alternating potential $\text{V}=200\text{cos}\left( 100\pi \text{t} \right)$. The time taken by the current to change from its peak value to rms value is
1 $2.5\times {{10}^{-3}}\text{s}$
2 $25\times {{10}^{-3}}\text{s}$
3 $0.25\text{s}$
4 $0.20\text{s}$
Explanation:
(a) The current and potential difference are in phase with the resistance. So, the time taken would be same as time for voltage to change from $\left( t=0 \right)$ that is peak value to rms value.
Time taken by voltage to achieve its rms value of $\frac{200}{\sqrt{2}}$.
$\begin{array}{*{35}{r}}
{} & \frac{200}{\sqrt{2}}=200\text{cos}\left( 100\pi t \right) \\
\Rightarrow & \text{cos}\left( 100\pi t \right)=\frac{1}{\sqrt{2}}=\text{cos}\left( \frac{\pi }{4} \right) \\
t= & \frac{1}{400}\text{ }\!\!~\!\!\text{ second }\!\!~\!\!\text{ }=2.5\times {{10}^{-3}}\text{ }\!\!~\!\!\text{ sec}\text{. }\!\!~\!\!\text{ } \\
\end{array}$
274617
The voltage of an ac supply varies with time $\left( t \right)$ as $V=120\text{sin}100\pi \text{tcos}100\pi \text{t}$. The maximum voltage and frequency respectively are
1 120 volt,$100\text{Hz}$
2 $\frac{120}{\sqrt{2}}$ volt, $100\text{Hz}$
3 60 volt,$200\text{Hz}$
4 60 volt,$100\text{Hz}$
Explanation:
(d) $\text{V}=120\text{sin}100\pi \text{tcos}100\pi \text{t}\Rightarrow \text{V}=60\text{sin}200\pi \text{t}{{\text{V}}_{\text{max}}}=60\text{V}$ and $v=100\text{Hz}$
NCERT Page-236 / N-180
AC (NCERT)
274618
A resistance of $40\text{ }\!\!\Omega\!\!\text{ }$ is connected to a source of alternating current rated $220\text{V},50\text{Hz}$. Find the time taken by the current to change from its maximum value to $\text{rms}$ value :
1 $2.5\text{ms}$
2 $1.25\text{ms}$
3 $2.5\text{s}$
4 $0.25\text{s}$
Explanation:
(a)
NCERT Page-236 / N-180
AC (NCERT)
274619
A resistance of $20\text{ohm}$ is connected to a source of an alternating potential $\text{V}=200\text{cos}\left( 100\pi \text{t} \right)$. The time taken by the current to change from its peak value to rms value is
1 $2.5\times {{10}^{-3}}\text{s}$
2 $25\times {{10}^{-3}}\text{s}$
3 $0.25\text{s}$
4 $0.20\text{s}$
Explanation:
(a) The current and potential difference are in phase with the resistance. So, the time taken would be same as time for voltage to change from $\left( t=0 \right)$ that is peak value to rms value.
Time taken by voltage to achieve its rms value of $\frac{200}{\sqrt{2}}$.
$\begin{array}{*{35}{r}}
{} & \frac{200}{\sqrt{2}}=200\text{cos}\left( 100\pi t \right) \\
\Rightarrow & \text{cos}\left( 100\pi t \right)=\frac{1}{\sqrt{2}}=\text{cos}\left( \frac{\pi }{4} \right) \\
t= & \frac{1}{400}\text{ }\!\!~\!\!\text{ second }\!\!~\!\!\text{ }=2.5\times {{10}^{-3}}\text{ }\!\!~\!\!\text{ sec}\text{. }\!\!~\!\!\text{ } \\
\end{array}$