NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155088
The maximum current in a coil of $0.50 \mathrm{H}$ having resistance of $100 \Omega$, when connected to a supply of $240 \mathrm{~V}, 50 \mathrm{~Hz}$ ac is
1 $0.3 \mathrm{~A}$
2 $0.1 \mathrm{~A}$
3 $1.3 \mathrm{~A}$
4 $0.6 \mathrm{~mA}$
Explanation:
C $\mathrm{L}=0.50 \mathrm{H}, \mathrm{R}=100 \Omega, \mathrm{V}=240 \mathrm{~V}, \mathrm{f}=50 \mathrm{~Hz}$ Angular frequency $(\omega)=2 \pi \mathrm{f}$ $=2 \times \pi \times 50=100 \pi \mathrm{rad} / \mathrm{sec}$ Peak voltage $\left(\mathrm{V}_{0}\right)=\sqrt{2} \mathrm{~V}=\sqrt{2} \times 240=339.41 \mathrm{~V}$ Maximum current in the circuit, $\mathrm{I}_{0}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}}}=\frac{339.41}{\sqrt{(100)^{2}+(100 \pi)^{2} \times(0.50)^{2}}}$ $=1.82 \mathrm{Amp} \approx 1.3 \mathrm{~A}$
J and K CET- 1998
Alternating Current
155090
In an $L-R$ circuit, time constant is that time in which current grows from zero to the value (Where $I_{\mathbf{0}}$ is steady state current)
1 $0.63 \mathrm{I}_{0}$
2 $0.50 \mathrm{I}_{0}$
3 $0.37 \mathrm{I}_{0}$
4 $\mathrm{I}_{0}$
Explanation:
A Current at any instant of time $t$ after closing an LR circuit is given by, $\mathrm{I}=\mathrm{I}_{0}\left[\mathrm{I}-\mathrm{e}^{-\frac{\mathrm{R}}{\mathrm{L}} \mathrm{t}}\right] \text {. }$ According to question, Time $(\mathrm{t})=$ Time constant So, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}}$ $I=I_{0}\left[1-e^{-\frac{R}{L} \times \frac{L}{R}}\right]=I_{0}\left[1-e^{-1}\right]$ $=I_{0}\left[1-\frac{1}{e}\right]=I_{0}\left[1-\frac{1}{2.718}\right]=0.63 \mathrm{I}_{0}$
UP CPMT-2006
Alternating Current
155091
In an oscillating $\mathrm{L}-\mathrm{C}$ circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor, when the energy is stored equally between the electric and magnetic field is
1 $\frac{Q}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{2}}$
3 $\frac{\mathrm{Q}}{\sqrt{3}}$
4 $\frac{\mathrm{Q}}{3}$
Explanation:
B The energy stored in the capacitor when fully charged in any $\mathrm{L}-\mathrm{C}$ circuit is, $\mathrm{E}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ Let the charge on the capacitor when energy is half that can be stored in the capacitor be Q'. $\frac{\mathrm{E}}{2}=\frac{1}{2} \frac{\mathrm{Q}^{\prime 2}}{\mathrm{C}}$ Put value $\mathrm{E}$ from eq ${ }^{\mathrm{n}}$ (i) in $\mathrm{eq}^{\mathrm{n}}$ (ii) - $\frac{1}{2} \times \frac{1}{2} \times \frac{\mathrm{Q}^{2}}{\mathrm{C}}=\frac{1}{2} \frac{\mathrm{Q}^{\prime 2}}{\mathrm{C}}$ $\mathrm{Q}^{\prime}=\frac{\mathrm{Q}}{\sqrt{2}}$
UP CPMT-2005
Alternating Current
155093
A coil of inductance $300 \mathrm{mH}$ and resistance $2 \Omega$ is connected to a source of voltage $2 \mathrm{~V}$. The current reaches half of its steady state value in
155088
The maximum current in a coil of $0.50 \mathrm{H}$ having resistance of $100 \Omega$, when connected to a supply of $240 \mathrm{~V}, 50 \mathrm{~Hz}$ ac is
1 $0.3 \mathrm{~A}$
2 $0.1 \mathrm{~A}$
3 $1.3 \mathrm{~A}$
4 $0.6 \mathrm{~mA}$
Explanation:
C $\mathrm{L}=0.50 \mathrm{H}, \mathrm{R}=100 \Omega, \mathrm{V}=240 \mathrm{~V}, \mathrm{f}=50 \mathrm{~Hz}$ Angular frequency $(\omega)=2 \pi \mathrm{f}$ $=2 \times \pi \times 50=100 \pi \mathrm{rad} / \mathrm{sec}$ Peak voltage $\left(\mathrm{V}_{0}\right)=\sqrt{2} \mathrm{~V}=\sqrt{2} \times 240=339.41 \mathrm{~V}$ Maximum current in the circuit, $\mathrm{I}_{0}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}}}=\frac{339.41}{\sqrt{(100)^{2}+(100 \pi)^{2} \times(0.50)^{2}}}$ $=1.82 \mathrm{Amp} \approx 1.3 \mathrm{~A}$
J and K CET- 1998
Alternating Current
155090
In an $L-R$ circuit, time constant is that time in which current grows from zero to the value (Where $I_{\mathbf{0}}$ is steady state current)
1 $0.63 \mathrm{I}_{0}$
2 $0.50 \mathrm{I}_{0}$
3 $0.37 \mathrm{I}_{0}$
4 $\mathrm{I}_{0}$
Explanation:
A Current at any instant of time $t$ after closing an LR circuit is given by, $\mathrm{I}=\mathrm{I}_{0}\left[\mathrm{I}-\mathrm{e}^{-\frac{\mathrm{R}}{\mathrm{L}} \mathrm{t}}\right] \text {. }$ According to question, Time $(\mathrm{t})=$ Time constant So, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}}$ $I=I_{0}\left[1-e^{-\frac{R}{L} \times \frac{L}{R}}\right]=I_{0}\left[1-e^{-1}\right]$ $=I_{0}\left[1-\frac{1}{e}\right]=I_{0}\left[1-\frac{1}{2.718}\right]=0.63 \mathrm{I}_{0}$
UP CPMT-2006
Alternating Current
155091
In an oscillating $\mathrm{L}-\mathrm{C}$ circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor, when the energy is stored equally between the electric and magnetic field is
1 $\frac{Q}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{2}}$
3 $\frac{\mathrm{Q}}{\sqrt{3}}$
4 $\frac{\mathrm{Q}}{3}$
Explanation:
B The energy stored in the capacitor when fully charged in any $\mathrm{L}-\mathrm{C}$ circuit is, $\mathrm{E}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ Let the charge on the capacitor when energy is half that can be stored in the capacitor be Q'. $\frac{\mathrm{E}}{2}=\frac{1}{2} \frac{\mathrm{Q}^{\prime 2}}{\mathrm{C}}$ Put value $\mathrm{E}$ from eq ${ }^{\mathrm{n}}$ (i) in $\mathrm{eq}^{\mathrm{n}}$ (ii) - $\frac{1}{2} \times \frac{1}{2} \times \frac{\mathrm{Q}^{2}}{\mathrm{C}}=\frac{1}{2} \frac{\mathrm{Q}^{\prime 2}}{\mathrm{C}}$ $\mathrm{Q}^{\prime}=\frac{\mathrm{Q}}{\sqrt{2}}$
UP CPMT-2005
Alternating Current
155093
A coil of inductance $300 \mathrm{mH}$ and resistance $2 \Omega$ is connected to a source of voltage $2 \mathrm{~V}$. The current reaches half of its steady state value in
155088
The maximum current in a coil of $0.50 \mathrm{H}$ having resistance of $100 \Omega$, when connected to a supply of $240 \mathrm{~V}, 50 \mathrm{~Hz}$ ac is
1 $0.3 \mathrm{~A}$
2 $0.1 \mathrm{~A}$
3 $1.3 \mathrm{~A}$
4 $0.6 \mathrm{~mA}$
Explanation:
C $\mathrm{L}=0.50 \mathrm{H}, \mathrm{R}=100 \Omega, \mathrm{V}=240 \mathrm{~V}, \mathrm{f}=50 \mathrm{~Hz}$ Angular frequency $(\omega)=2 \pi \mathrm{f}$ $=2 \times \pi \times 50=100 \pi \mathrm{rad} / \mathrm{sec}$ Peak voltage $\left(\mathrm{V}_{0}\right)=\sqrt{2} \mathrm{~V}=\sqrt{2} \times 240=339.41 \mathrm{~V}$ Maximum current in the circuit, $\mathrm{I}_{0}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}}}=\frac{339.41}{\sqrt{(100)^{2}+(100 \pi)^{2} \times(0.50)^{2}}}$ $=1.82 \mathrm{Amp} \approx 1.3 \mathrm{~A}$
J and K CET- 1998
Alternating Current
155090
In an $L-R$ circuit, time constant is that time in which current grows from zero to the value (Where $I_{\mathbf{0}}$ is steady state current)
1 $0.63 \mathrm{I}_{0}$
2 $0.50 \mathrm{I}_{0}$
3 $0.37 \mathrm{I}_{0}$
4 $\mathrm{I}_{0}$
Explanation:
A Current at any instant of time $t$ after closing an LR circuit is given by, $\mathrm{I}=\mathrm{I}_{0}\left[\mathrm{I}-\mathrm{e}^{-\frac{\mathrm{R}}{\mathrm{L}} \mathrm{t}}\right] \text {. }$ According to question, Time $(\mathrm{t})=$ Time constant So, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}}$ $I=I_{0}\left[1-e^{-\frac{R}{L} \times \frac{L}{R}}\right]=I_{0}\left[1-e^{-1}\right]$ $=I_{0}\left[1-\frac{1}{e}\right]=I_{0}\left[1-\frac{1}{2.718}\right]=0.63 \mathrm{I}_{0}$
UP CPMT-2006
Alternating Current
155091
In an oscillating $\mathrm{L}-\mathrm{C}$ circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor, when the energy is stored equally between the electric and magnetic field is
1 $\frac{Q}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{2}}$
3 $\frac{\mathrm{Q}}{\sqrt{3}}$
4 $\frac{\mathrm{Q}}{3}$
Explanation:
B The energy stored in the capacitor when fully charged in any $\mathrm{L}-\mathrm{C}$ circuit is, $\mathrm{E}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ Let the charge on the capacitor when energy is half that can be stored in the capacitor be Q'. $\frac{\mathrm{E}}{2}=\frac{1}{2} \frac{\mathrm{Q}^{\prime 2}}{\mathrm{C}}$ Put value $\mathrm{E}$ from eq ${ }^{\mathrm{n}}$ (i) in $\mathrm{eq}^{\mathrm{n}}$ (ii) - $\frac{1}{2} \times \frac{1}{2} \times \frac{\mathrm{Q}^{2}}{\mathrm{C}}=\frac{1}{2} \frac{\mathrm{Q}^{\prime 2}}{\mathrm{C}}$ $\mathrm{Q}^{\prime}=\frac{\mathrm{Q}}{\sqrt{2}}$
UP CPMT-2005
Alternating Current
155093
A coil of inductance $300 \mathrm{mH}$ and resistance $2 \Omega$ is connected to a source of voltage $2 \mathrm{~V}$. The current reaches half of its steady state value in
155088
The maximum current in a coil of $0.50 \mathrm{H}$ having resistance of $100 \Omega$, when connected to a supply of $240 \mathrm{~V}, 50 \mathrm{~Hz}$ ac is
1 $0.3 \mathrm{~A}$
2 $0.1 \mathrm{~A}$
3 $1.3 \mathrm{~A}$
4 $0.6 \mathrm{~mA}$
Explanation:
C $\mathrm{L}=0.50 \mathrm{H}, \mathrm{R}=100 \Omega, \mathrm{V}=240 \mathrm{~V}, \mathrm{f}=50 \mathrm{~Hz}$ Angular frequency $(\omega)=2 \pi \mathrm{f}$ $=2 \times \pi \times 50=100 \pi \mathrm{rad} / \mathrm{sec}$ Peak voltage $\left(\mathrm{V}_{0}\right)=\sqrt{2} \mathrm{~V}=\sqrt{2} \times 240=339.41 \mathrm{~V}$ Maximum current in the circuit, $\mathrm{I}_{0}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}}}=\frac{339.41}{\sqrt{(100)^{2}+(100 \pi)^{2} \times(0.50)^{2}}}$ $=1.82 \mathrm{Amp} \approx 1.3 \mathrm{~A}$
J and K CET- 1998
Alternating Current
155090
In an $L-R$ circuit, time constant is that time in which current grows from zero to the value (Where $I_{\mathbf{0}}$ is steady state current)
1 $0.63 \mathrm{I}_{0}$
2 $0.50 \mathrm{I}_{0}$
3 $0.37 \mathrm{I}_{0}$
4 $\mathrm{I}_{0}$
Explanation:
A Current at any instant of time $t$ after closing an LR circuit is given by, $\mathrm{I}=\mathrm{I}_{0}\left[\mathrm{I}-\mathrm{e}^{-\frac{\mathrm{R}}{\mathrm{L}} \mathrm{t}}\right] \text {. }$ According to question, Time $(\mathrm{t})=$ Time constant So, $\mathrm{t}=\frac{\mathrm{L}}{\mathrm{R}}$ $I=I_{0}\left[1-e^{-\frac{R}{L} \times \frac{L}{R}}\right]=I_{0}\left[1-e^{-1}\right]$ $=I_{0}\left[1-\frac{1}{e}\right]=I_{0}\left[1-\frac{1}{2.718}\right]=0.63 \mathrm{I}_{0}$
UP CPMT-2006
Alternating Current
155091
In an oscillating $\mathrm{L}-\mathrm{C}$ circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor, when the energy is stored equally between the electric and magnetic field is
1 $\frac{Q}{2}$
2 $\frac{\mathrm{Q}}{\sqrt{2}}$
3 $\frac{\mathrm{Q}}{\sqrt{3}}$
4 $\frac{\mathrm{Q}}{3}$
Explanation:
B The energy stored in the capacitor when fully charged in any $\mathrm{L}-\mathrm{C}$ circuit is, $\mathrm{E}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}}$ Let the charge on the capacitor when energy is half that can be stored in the capacitor be Q'. $\frac{\mathrm{E}}{2}=\frac{1}{2} \frac{\mathrm{Q}^{\prime 2}}{\mathrm{C}}$ Put value $\mathrm{E}$ from eq ${ }^{\mathrm{n}}$ (i) in $\mathrm{eq}^{\mathrm{n}}$ (ii) - $\frac{1}{2} \times \frac{1}{2} \times \frac{\mathrm{Q}^{2}}{\mathrm{C}}=\frac{1}{2} \frac{\mathrm{Q}^{\prime 2}}{\mathrm{C}}$ $\mathrm{Q}^{\prime}=\frac{\mathrm{Q}}{\sqrt{2}}$
UP CPMT-2005
Alternating Current
155093
A coil of inductance $300 \mathrm{mH}$ and resistance $2 \Omega$ is connected to a source of voltage $2 \mathrm{~V}$. The current reaches half of its steady state value in
