274649
Resonance frequency of LCR series a.c. circuit is ${{f}_{0}}$. Now the capacitance is made 4 times, then the new resonance frequency will become
1 ${{\text{f}}_{0}}/4$
2 $2{{\text{f}}_{0}}$
3 ${{\text{f}}_{0}}$
4 ${{\text{f}}_{0}}/2$
Explanation:
(d) In LCR series circuit, resonance frequency ${{\text{f}}_{0}}$ is given by
$\begin{array}{*{35}{r}}
{} & \text{L}\omega =\frac{1}{\text{C}\omega }\Rightarrow {{\omega }^{2}}=\frac{1}{\text{LC}}\therefore \omega =\sqrt{\frac{1}{\text{LC}}}=2\pi {{\text{f}}_{0}} \\
{} & \therefore {{\text{f}}_{0}}=\frac{1}{2\pi \sqrt{\text{LC}}}\text{ }\!\!~\!\!\text{ or }\!\!~\!\!\text{ }{{\text{f}}_{0}}\alpha \frac{1}{\sqrt{\text{C}}} \\
\end{array}$
When the capacitance of the circuit is made 4 times, its resonant frequency become ${{\text{f}}_{0}}^{\text{ }\!\!'\!\!\text{ }}$
$\therefore \frac{\text{f}_{0}^{\text{ }\!\!'\!\!\text{ }}}{{{\text{f}}_{0}}}=\frac{\sqrt{\text{C}}}{\sqrt{4\text{C}}}\text{ }\!\!~\!\!\text{ or }\!\!~\!\!\text{ f}_{0}^{\text{ }\!\!'\!\!\text{ }}=\frac{{{\text{f}}_{0}}}{2}$
NCERT Page-248 / N-189
AC (NCERT)
274650
In an LCR series a.c. circuit, the voltage across each of the components, $\text{L},\text{C}$ and $\text{R}$ is $50\text{V}$. The voltage across the $\text{LC}$ combination will be
1 $100\text{V}$
2 $50\sqrt{2}\text{V}$
3 $50\text{V}$
4 $0\text{V}$
Explanation:
(d) Since the phase difference between $L\And C$ is $\pi $
$\therefore $ net voltage difference across $LC=50-50=0$
NCERT Page-245 / N- 187
AC (NCERT)
274651
If resistance of $100\text{ }\!\!\Omega\!\!\text{ }$, and inductance of 0.5 henry and capacitance of $10\times {{10}^{6}}$ farad are connected in series through $50\text{Hz}$ A.C. supply, then impedance is
274652
The current in resistance $R$ at resonance is
1 zero
2 minimum but finite
3 maximum but finite
4 infinite
Explanation:
(c) At resonance ${{X}_{L}}={{X}_{C}}\Rightarrow Z=R$& current is maximum but finite, which is ${{I}_{\text{max}}}=\frac{E}{R}$, where $E$ is applied voltage.
NCERT Page-244 / N-186
AC (NCERT)
274654
For a series $RLC$ circuit $R={{X}_{L}}=2{{X}_{C}}$. The impedance of the circuit and phase difference between $V$ and $I$ respectively will be
274649
Resonance frequency of LCR series a.c. circuit is ${{f}_{0}}$. Now the capacitance is made 4 times, then the new resonance frequency will become
1 ${{\text{f}}_{0}}/4$
2 $2{{\text{f}}_{0}}$
3 ${{\text{f}}_{0}}$
4 ${{\text{f}}_{0}}/2$
Explanation:
(d) In LCR series circuit, resonance frequency ${{\text{f}}_{0}}$ is given by
$\begin{array}{*{35}{r}}
{} & \text{L}\omega =\frac{1}{\text{C}\omega }\Rightarrow {{\omega }^{2}}=\frac{1}{\text{LC}}\therefore \omega =\sqrt{\frac{1}{\text{LC}}}=2\pi {{\text{f}}_{0}} \\
{} & \therefore {{\text{f}}_{0}}=\frac{1}{2\pi \sqrt{\text{LC}}}\text{ }\!\!~\!\!\text{ or }\!\!~\!\!\text{ }{{\text{f}}_{0}}\alpha \frac{1}{\sqrt{\text{C}}} \\
\end{array}$
When the capacitance of the circuit is made 4 times, its resonant frequency become ${{\text{f}}_{0}}^{\text{ }\!\!'\!\!\text{ }}$
$\therefore \frac{\text{f}_{0}^{\text{ }\!\!'\!\!\text{ }}}{{{\text{f}}_{0}}}=\frac{\sqrt{\text{C}}}{\sqrt{4\text{C}}}\text{ }\!\!~\!\!\text{ or }\!\!~\!\!\text{ f}_{0}^{\text{ }\!\!'\!\!\text{ }}=\frac{{{\text{f}}_{0}}}{2}$
NCERT Page-248 / N-189
AC (NCERT)
274650
In an LCR series a.c. circuit, the voltage across each of the components, $\text{L},\text{C}$ and $\text{R}$ is $50\text{V}$. The voltage across the $\text{LC}$ combination will be
1 $100\text{V}$
2 $50\sqrt{2}\text{V}$
3 $50\text{V}$
4 $0\text{V}$
Explanation:
(d) Since the phase difference between $L\And C$ is $\pi $
$\therefore $ net voltage difference across $LC=50-50=0$
NCERT Page-245 / N- 187
AC (NCERT)
274651
If resistance of $100\text{ }\!\!\Omega\!\!\text{ }$, and inductance of 0.5 henry and capacitance of $10\times {{10}^{6}}$ farad are connected in series through $50\text{Hz}$ A.C. supply, then impedance is
274652
The current in resistance $R$ at resonance is
1 zero
2 minimum but finite
3 maximum but finite
4 infinite
Explanation:
(c) At resonance ${{X}_{L}}={{X}_{C}}\Rightarrow Z=R$& current is maximum but finite, which is ${{I}_{\text{max}}}=\frac{E}{R}$, where $E$ is applied voltage.
NCERT Page-244 / N-186
AC (NCERT)
274654
For a series $RLC$ circuit $R={{X}_{L}}=2{{X}_{C}}$. The impedance of the circuit and phase difference between $V$ and $I$ respectively will be
NEET Test Series from KOTA - 10 Papers In MS WORD
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AC (NCERT)
274649
Resonance frequency of LCR series a.c. circuit is ${{f}_{0}}$. Now the capacitance is made 4 times, then the new resonance frequency will become
1 ${{\text{f}}_{0}}/4$
2 $2{{\text{f}}_{0}}$
3 ${{\text{f}}_{0}}$
4 ${{\text{f}}_{0}}/2$
Explanation:
(d) In LCR series circuit, resonance frequency ${{\text{f}}_{0}}$ is given by
$\begin{array}{*{35}{r}}
{} & \text{L}\omega =\frac{1}{\text{C}\omega }\Rightarrow {{\omega }^{2}}=\frac{1}{\text{LC}}\therefore \omega =\sqrt{\frac{1}{\text{LC}}}=2\pi {{\text{f}}_{0}} \\
{} & \therefore {{\text{f}}_{0}}=\frac{1}{2\pi \sqrt{\text{LC}}}\text{ }\!\!~\!\!\text{ or }\!\!~\!\!\text{ }{{\text{f}}_{0}}\alpha \frac{1}{\sqrt{\text{C}}} \\
\end{array}$
When the capacitance of the circuit is made 4 times, its resonant frequency become ${{\text{f}}_{0}}^{\text{ }\!\!'\!\!\text{ }}$
$\therefore \frac{\text{f}_{0}^{\text{ }\!\!'\!\!\text{ }}}{{{\text{f}}_{0}}}=\frac{\sqrt{\text{C}}}{\sqrt{4\text{C}}}\text{ }\!\!~\!\!\text{ or }\!\!~\!\!\text{ f}_{0}^{\text{ }\!\!'\!\!\text{ }}=\frac{{{\text{f}}_{0}}}{2}$
NCERT Page-248 / N-189
AC (NCERT)
274650
In an LCR series a.c. circuit, the voltage across each of the components, $\text{L},\text{C}$ and $\text{R}$ is $50\text{V}$. The voltage across the $\text{LC}$ combination will be
1 $100\text{V}$
2 $50\sqrt{2}\text{V}$
3 $50\text{V}$
4 $0\text{V}$
Explanation:
(d) Since the phase difference between $L\And C$ is $\pi $
$\therefore $ net voltage difference across $LC=50-50=0$
NCERT Page-245 / N- 187
AC (NCERT)
274651
If resistance of $100\text{ }\!\!\Omega\!\!\text{ }$, and inductance of 0.5 henry and capacitance of $10\times {{10}^{6}}$ farad are connected in series through $50\text{Hz}$ A.C. supply, then impedance is
274652
The current in resistance $R$ at resonance is
1 zero
2 minimum but finite
3 maximum but finite
4 infinite
Explanation:
(c) At resonance ${{X}_{L}}={{X}_{C}}\Rightarrow Z=R$& current is maximum but finite, which is ${{I}_{\text{max}}}=\frac{E}{R}$, where $E$ is applied voltage.
NCERT Page-244 / N-186
AC (NCERT)
274654
For a series $RLC$ circuit $R={{X}_{L}}=2{{X}_{C}}$. The impedance of the circuit and phase difference between $V$ and $I$ respectively will be
274649
Resonance frequency of LCR series a.c. circuit is ${{f}_{0}}$. Now the capacitance is made 4 times, then the new resonance frequency will become
1 ${{\text{f}}_{0}}/4$
2 $2{{\text{f}}_{0}}$
3 ${{\text{f}}_{0}}$
4 ${{\text{f}}_{0}}/2$
Explanation:
(d) In LCR series circuit, resonance frequency ${{\text{f}}_{0}}$ is given by
$\begin{array}{*{35}{r}}
{} & \text{L}\omega =\frac{1}{\text{C}\omega }\Rightarrow {{\omega }^{2}}=\frac{1}{\text{LC}}\therefore \omega =\sqrt{\frac{1}{\text{LC}}}=2\pi {{\text{f}}_{0}} \\
{} & \therefore {{\text{f}}_{0}}=\frac{1}{2\pi \sqrt{\text{LC}}}\text{ }\!\!~\!\!\text{ or }\!\!~\!\!\text{ }{{\text{f}}_{0}}\alpha \frac{1}{\sqrt{\text{C}}} \\
\end{array}$
When the capacitance of the circuit is made 4 times, its resonant frequency become ${{\text{f}}_{0}}^{\text{ }\!\!'\!\!\text{ }}$
$\therefore \frac{\text{f}_{0}^{\text{ }\!\!'\!\!\text{ }}}{{{\text{f}}_{0}}}=\frac{\sqrt{\text{C}}}{\sqrt{4\text{C}}}\text{ }\!\!~\!\!\text{ or }\!\!~\!\!\text{ f}_{0}^{\text{ }\!\!'\!\!\text{ }}=\frac{{{\text{f}}_{0}}}{2}$
NCERT Page-248 / N-189
AC (NCERT)
274650
In an LCR series a.c. circuit, the voltage across each of the components, $\text{L},\text{C}$ and $\text{R}$ is $50\text{V}$. The voltage across the $\text{LC}$ combination will be
1 $100\text{V}$
2 $50\sqrt{2}\text{V}$
3 $50\text{V}$
4 $0\text{V}$
Explanation:
(d) Since the phase difference between $L\And C$ is $\pi $
$\therefore $ net voltage difference across $LC=50-50=0$
NCERT Page-245 / N- 187
AC (NCERT)
274651
If resistance of $100\text{ }\!\!\Omega\!\!\text{ }$, and inductance of 0.5 henry and capacitance of $10\times {{10}^{6}}$ farad are connected in series through $50\text{Hz}$ A.C. supply, then impedance is
274652
The current in resistance $R$ at resonance is
1 zero
2 minimum but finite
3 maximum but finite
4 infinite
Explanation:
(c) At resonance ${{X}_{L}}={{X}_{C}}\Rightarrow Z=R$& current is maximum but finite, which is ${{I}_{\text{max}}}=\frac{E}{R}$, where $E$ is applied voltage.
NCERT Page-244 / N-186
AC (NCERT)
274654
For a series $RLC$ circuit $R={{X}_{L}}=2{{X}_{C}}$. The impedance of the circuit and phase difference between $V$ and $I$ respectively will be
274649
Resonance frequency of LCR series a.c. circuit is ${{f}_{0}}$. Now the capacitance is made 4 times, then the new resonance frequency will become
1 ${{\text{f}}_{0}}/4$
2 $2{{\text{f}}_{0}}$
3 ${{\text{f}}_{0}}$
4 ${{\text{f}}_{0}}/2$
Explanation:
(d) In LCR series circuit, resonance frequency ${{\text{f}}_{0}}$ is given by
$\begin{array}{*{35}{r}}
{} & \text{L}\omega =\frac{1}{\text{C}\omega }\Rightarrow {{\omega }^{2}}=\frac{1}{\text{LC}}\therefore \omega =\sqrt{\frac{1}{\text{LC}}}=2\pi {{\text{f}}_{0}} \\
{} & \therefore {{\text{f}}_{0}}=\frac{1}{2\pi \sqrt{\text{LC}}}\text{ }\!\!~\!\!\text{ or }\!\!~\!\!\text{ }{{\text{f}}_{0}}\alpha \frac{1}{\sqrt{\text{C}}} \\
\end{array}$
When the capacitance of the circuit is made 4 times, its resonant frequency become ${{\text{f}}_{0}}^{\text{ }\!\!'\!\!\text{ }}$
$\therefore \frac{\text{f}_{0}^{\text{ }\!\!'\!\!\text{ }}}{{{\text{f}}_{0}}}=\frac{\sqrt{\text{C}}}{\sqrt{4\text{C}}}\text{ }\!\!~\!\!\text{ or }\!\!~\!\!\text{ f}_{0}^{\text{ }\!\!'\!\!\text{ }}=\frac{{{\text{f}}_{0}}}{2}$
NCERT Page-248 / N-189
AC (NCERT)
274650
In an LCR series a.c. circuit, the voltage across each of the components, $\text{L},\text{C}$ and $\text{R}$ is $50\text{V}$. The voltage across the $\text{LC}$ combination will be
1 $100\text{V}$
2 $50\sqrt{2}\text{V}$
3 $50\text{V}$
4 $0\text{V}$
Explanation:
(d) Since the phase difference between $L\And C$ is $\pi $
$\therefore $ net voltage difference across $LC=50-50=0$
NCERT Page-245 / N- 187
AC (NCERT)
274651
If resistance of $100\text{ }\!\!\Omega\!\!\text{ }$, and inductance of 0.5 henry and capacitance of $10\times {{10}^{6}}$ farad are connected in series through $50\text{Hz}$ A.C. supply, then impedance is
274652
The current in resistance $R$ at resonance is
1 zero
2 minimum but finite
3 maximum but finite
4 infinite
Explanation:
(c) At resonance ${{X}_{L}}={{X}_{C}}\Rightarrow Z=R$& current is maximum but finite, which is ${{I}_{\text{max}}}=\frac{E}{R}$, where $E$ is applied voltage.
NCERT Page-244 / N-186
AC (NCERT)
274654
For a series $RLC$ circuit $R={{X}_{L}}=2{{X}_{C}}$. The impedance of the circuit and phase difference between $V$ and $I$ respectively will be
