NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155231
The current in an $L-R$ circuit builds up to $(3 / 4)^{\text {th }}$ of its steady state value in 4 seconds. The time constant of this circuit is
1 $\frac{1}{\ln 2} \mathrm{sec}$
2 $\frac{2}{\ln 2} \mathrm{sec}$
3 $\frac{3}{\ln 2} \mathrm{sec}$
4 $\frac{4}{\ln 2} \mathrm{sec}$
Explanation:
B Given that, $\frac{3 \mathrm{I}_{0}}{4}=\mathrm{I}$ We know that, growth of current in LR circuit, $\mathrm{I}=\mathrm{I}_{0}\left[1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}\right]$ $\frac{3 \mathrm{I}_{0}}{4}=\mathrm{I}_{0}\left[1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}\right]$ $\frac{3}{4}=1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}$ $\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}=\frac{1}{4}$ $\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}=2^{-2}$ Taking natural $\log$ on both side, at $\mathrm{t}=4$ second $-\frac{4 \mathrm{R}}{\mathrm{L}}=-2 \ln 2$ $\mathrm{R}=\frac{2 \mathrm{~L} l \mathrm{n} 2}{4}=\frac{\mathrm{L} l \mathrm{n} 2}{2}$ The time constant for LR circuit, $\tau=\frac{\mathrm{L}}{\mathrm{R}}=\frac{\mathrm{L}}{\mathrm{L} \frac{\ln 2}{2}}$ $\tau=\frac{2}{\ln 2} \mathrm{sec}$
VITEEE-2016
Alternating Current
155232
The current graph for resonance in LC circuit is
1 a
2 b
3 c
4 d
Explanation:
C In LC circuit, if $X_{L}=X_{C}$ Then, $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ Then, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\left(\because \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\right)$ $\mathrm{Z}=\frac{\mathrm{E}_{0}}{\mathrm{I}_{0}}=0$ $\left(\mathrm{I}_{0}=\infty\right)$ This $\frac{1}{\sqrt{\mathrm{LC}}}$ is the natural frequency of LC circuit if the frequency of applied $\mathrm{AC}$ becomes equal to the natural frequency of $\mathrm{AC}$ circuit then the amplitude of current becomes infinite due to zero impedance.
VITEEE-2015
Alternating Current
155234
In an AC circuit, the potential across an inductance and resistance joined in series are respectively $16 \mathrm{~V}$ and $20 \mathrm{~V}$. The total potential difference across the circuit is
1 $20.0 \mathrm{~V}$
2 $25.6 \mathrm{~V}$
3 $31.9 \mathrm{~V}$
4 $33.6 \mathrm{~V}$
Explanation:
B Given that, $\mathrm{V}_{\mathrm{R}}=20$ Volt, $\mathrm{V}_{\mathrm{L}}=16$ Volt Then, total potential difference across the circuit is, $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}}$ $\mathrm{~V}=\sqrt{(20)^{2}+(16)^{2}}$ $\mathrm{~V}=\sqrt{656}$ $\mathrm{~V}=25.6 \text { Volt }$
VITEEE-2012
Alternating Current
155235
The following series L-C-R circuit, when driven by an emf source of angular frequency 70 kilo-radians per second, the circuit effectively behaves like
1 purely resistive circuit
2 series R-L circuit
3 series R-C circuit
4 series L-C circuit with $\mathrm{R}=0$
Explanation:
C Given that, $\omega=70 \mathrm{k}$ radians $/ \mathrm{sec}=70 \times 10^{3} \mathrm{rad} / \mathrm{sec}$ $\mathrm{L}=100 \times 10^{-6} \mathrm{H}, \mathrm{C}=1 \times 10^{-6} \mathrm{~F}, \mathrm{R}=10 \Omega$ We know that, If $X_{L}>X_{C}$ then circuit is called $R-L$ circuit. If $\mathrm{X}_{\mathrm{C}}>\mathrm{X}_{\mathrm{L}}$ then circuit is $\mathrm{R}-\mathrm{C}$ circuit. $X_{L}=\omega L$ $X_{L}=70 \times 10^{3} \times 100 \times 10^{-6}$ $X_{L}=7 \Omega$ Capacitive reactance, $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{70 \times 10^{3} \times 1 \times 10^{-6}}$ $\mathrm{X}_{\mathrm{C}}=\frac{1000}{70}$ $\mathrm{X}_{\mathrm{C}}=14.3 \Omega$ Since $X_{C}>X_{L}$ then circuit is called series $R-C$ circuit.
155231
The current in an $L-R$ circuit builds up to $(3 / 4)^{\text {th }}$ of its steady state value in 4 seconds. The time constant of this circuit is
1 $\frac{1}{\ln 2} \mathrm{sec}$
2 $\frac{2}{\ln 2} \mathrm{sec}$
3 $\frac{3}{\ln 2} \mathrm{sec}$
4 $\frac{4}{\ln 2} \mathrm{sec}$
Explanation:
B Given that, $\frac{3 \mathrm{I}_{0}}{4}=\mathrm{I}$ We know that, growth of current in LR circuit, $\mathrm{I}=\mathrm{I}_{0}\left[1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}\right]$ $\frac{3 \mathrm{I}_{0}}{4}=\mathrm{I}_{0}\left[1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}\right]$ $\frac{3}{4}=1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}$ $\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}=\frac{1}{4}$ $\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}=2^{-2}$ Taking natural $\log$ on both side, at $\mathrm{t}=4$ second $-\frac{4 \mathrm{R}}{\mathrm{L}}=-2 \ln 2$ $\mathrm{R}=\frac{2 \mathrm{~L} l \mathrm{n} 2}{4}=\frac{\mathrm{L} l \mathrm{n} 2}{2}$ The time constant for LR circuit, $\tau=\frac{\mathrm{L}}{\mathrm{R}}=\frac{\mathrm{L}}{\mathrm{L} \frac{\ln 2}{2}}$ $\tau=\frac{2}{\ln 2} \mathrm{sec}$
VITEEE-2016
Alternating Current
155232
The current graph for resonance in LC circuit is
1 a
2 b
3 c
4 d
Explanation:
C In LC circuit, if $X_{L}=X_{C}$ Then, $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ Then, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\left(\because \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\right)$ $\mathrm{Z}=\frac{\mathrm{E}_{0}}{\mathrm{I}_{0}}=0$ $\left(\mathrm{I}_{0}=\infty\right)$ This $\frac{1}{\sqrt{\mathrm{LC}}}$ is the natural frequency of LC circuit if the frequency of applied $\mathrm{AC}$ becomes equal to the natural frequency of $\mathrm{AC}$ circuit then the amplitude of current becomes infinite due to zero impedance.
VITEEE-2015
Alternating Current
155234
In an AC circuit, the potential across an inductance and resistance joined in series are respectively $16 \mathrm{~V}$ and $20 \mathrm{~V}$. The total potential difference across the circuit is
1 $20.0 \mathrm{~V}$
2 $25.6 \mathrm{~V}$
3 $31.9 \mathrm{~V}$
4 $33.6 \mathrm{~V}$
Explanation:
B Given that, $\mathrm{V}_{\mathrm{R}}=20$ Volt, $\mathrm{V}_{\mathrm{L}}=16$ Volt Then, total potential difference across the circuit is, $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}}$ $\mathrm{~V}=\sqrt{(20)^{2}+(16)^{2}}$ $\mathrm{~V}=\sqrt{656}$ $\mathrm{~V}=25.6 \text { Volt }$
VITEEE-2012
Alternating Current
155235
The following series L-C-R circuit, when driven by an emf source of angular frequency 70 kilo-radians per second, the circuit effectively behaves like
1 purely resistive circuit
2 series R-L circuit
3 series R-C circuit
4 series L-C circuit with $\mathrm{R}=0$
Explanation:
C Given that, $\omega=70 \mathrm{k}$ radians $/ \mathrm{sec}=70 \times 10^{3} \mathrm{rad} / \mathrm{sec}$ $\mathrm{L}=100 \times 10^{-6} \mathrm{H}, \mathrm{C}=1 \times 10^{-6} \mathrm{~F}, \mathrm{R}=10 \Omega$ We know that, If $X_{L}>X_{C}$ then circuit is called $R-L$ circuit. If $\mathrm{X}_{\mathrm{C}}>\mathrm{X}_{\mathrm{L}}$ then circuit is $\mathrm{R}-\mathrm{C}$ circuit. $X_{L}=\omega L$ $X_{L}=70 \times 10^{3} \times 100 \times 10^{-6}$ $X_{L}=7 \Omega$ Capacitive reactance, $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{70 \times 10^{3} \times 1 \times 10^{-6}}$ $\mathrm{X}_{\mathrm{C}}=\frac{1000}{70}$ $\mathrm{X}_{\mathrm{C}}=14.3 \Omega$ Since $X_{C}>X_{L}$ then circuit is called series $R-C$ circuit.
155231
The current in an $L-R$ circuit builds up to $(3 / 4)^{\text {th }}$ of its steady state value in 4 seconds. The time constant of this circuit is
1 $\frac{1}{\ln 2} \mathrm{sec}$
2 $\frac{2}{\ln 2} \mathrm{sec}$
3 $\frac{3}{\ln 2} \mathrm{sec}$
4 $\frac{4}{\ln 2} \mathrm{sec}$
Explanation:
B Given that, $\frac{3 \mathrm{I}_{0}}{4}=\mathrm{I}$ We know that, growth of current in LR circuit, $\mathrm{I}=\mathrm{I}_{0}\left[1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}\right]$ $\frac{3 \mathrm{I}_{0}}{4}=\mathrm{I}_{0}\left[1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}\right]$ $\frac{3}{4}=1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}$ $\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}=\frac{1}{4}$ $\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}=2^{-2}$ Taking natural $\log$ on both side, at $\mathrm{t}=4$ second $-\frac{4 \mathrm{R}}{\mathrm{L}}=-2 \ln 2$ $\mathrm{R}=\frac{2 \mathrm{~L} l \mathrm{n} 2}{4}=\frac{\mathrm{L} l \mathrm{n} 2}{2}$ The time constant for LR circuit, $\tau=\frac{\mathrm{L}}{\mathrm{R}}=\frac{\mathrm{L}}{\mathrm{L} \frac{\ln 2}{2}}$ $\tau=\frac{2}{\ln 2} \mathrm{sec}$
VITEEE-2016
Alternating Current
155232
The current graph for resonance in LC circuit is
1 a
2 b
3 c
4 d
Explanation:
C In LC circuit, if $X_{L}=X_{C}$ Then, $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ Then, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\left(\because \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\right)$ $\mathrm{Z}=\frac{\mathrm{E}_{0}}{\mathrm{I}_{0}}=0$ $\left(\mathrm{I}_{0}=\infty\right)$ This $\frac{1}{\sqrt{\mathrm{LC}}}$ is the natural frequency of LC circuit if the frequency of applied $\mathrm{AC}$ becomes equal to the natural frequency of $\mathrm{AC}$ circuit then the amplitude of current becomes infinite due to zero impedance.
VITEEE-2015
Alternating Current
155234
In an AC circuit, the potential across an inductance and resistance joined in series are respectively $16 \mathrm{~V}$ and $20 \mathrm{~V}$. The total potential difference across the circuit is
1 $20.0 \mathrm{~V}$
2 $25.6 \mathrm{~V}$
3 $31.9 \mathrm{~V}$
4 $33.6 \mathrm{~V}$
Explanation:
B Given that, $\mathrm{V}_{\mathrm{R}}=20$ Volt, $\mathrm{V}_{\mathrm{L}}=16$ Volt Then, total potential difference across the circuit is, $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}}$ $\mathrm{~V}=\sqrt{(20)^{2}+(16)^{2}}$ $\mathrm{~V}=\sqrt{656}$ $\mathrm{~V}=25.6 \text { Volt }$
VITEEE-2012
Alternating Current
155235
The following series L-C-R circuit, when driven by an emf source of angular frequency 70 kilo-radians per second, the circuit effectively behaves like
1 purely resistive circuit
2 series R-L circuit
3 series R-C circuit
4 series L-C circuit with $\mathrm{R}=0$
Explanation:
C Given that, $\omega=70 \mathrm{k}$ radians $/ \mathrm{sec}=70 \times 10^{3} \mathrm{rad} / \mathrm{sec}$ $\mathrm{L}=100 \times 10^{-6} \mathrm{H}, \mathrm{C}=1 \times 10^{-6} \mathrm{~F}, \mathrm{R}=10 \Omega$ We know that, If $X_{L}>X_{C}$ then circuit is called $R-L$ circuit. If $\mathrm{X}_{\mathrm{C}}>\mathrm{X}_{\mathrm{L}}$ then circuit is $\mathrm{R}-\mathrm{C}$ circuit. $X_{L}=\omega L$ $X_{L}=70 \times 10^{3} \times 100 \times 10^{-6}$ $X_{L}=7 \Omega$ Capacitive reactance, $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{70 \times 10^{3} \times 1 \times 10^{-6}}$ $\mathrm{X}_{\mathrm{C}}=\frac{1000}{70}$ $\mathrm{X}_{\mathrm{C}}=14.3 \Omega$ Since $X_{C}>X_{L}$ then circuit is called series $R-C$ circuit.
155231
The current in an $L-R$ circuit builds up to $(3 / 4)^{\text {th }}$ of its steady state value in 4 seconds. The time constant of this circuit is
1 $\frac{1}{\ln 2} \mathrm{sec}$
2 $\frac{2}{\ln 2} \mathrm{sec}$
3 $\frac{3}{\ln 2} \mathrm{sec}$
4 $\frac{4}{\ln 2} \mathrm{sec}$
Explanation:
B Given that, $\frac{3 \mathrm{I}_{0}}{4}=\mathrm{I}$ We know that, growth of current in LR circuit, $\mathrm{I}=\mathrm{I}_{0}\left[1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}\right]$ $\frac{3 \mathrm{I}_{0}}{4}=\mathrm{I}_{0}\left[1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}\right]$ $\frac{3}{4}=1-\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}$ $\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}=\frac{1}{4}$ $\mathrm{e}^{(-\mathrm{Rt} / \mathrm{L})}=2^{-2}$ Taking natural $\log$ on both side, at $\mathrm{t}=4$ second $-\frac{4 \mathrm{R}}{\mathrm{L}}=-2 \ln 2$ $\mathrm{R}=\frac{2 \mathrm{~L} l \mathrm{n} 2}{4}=\frac{\mathrm{L} l \mathrm{n} 2}{2}$ The time constant for LR circuit, $\tau=\frac{\mathrm{L}}{\mathrm{R}}=\frac{\mathrm{L}}{\mathrm{L} \frac{\ln 2}{2}}$ $\tau=\frac{2}{\ln 2} \mathrm{sec}$
VITEEE-2016
Alternating Current
155232
The current graph for resonance in LC circuit is
1 a
2 b
3 c
4 d
Explanation:
C In LC circuit, if $X_{L}=X_{C}$ Then, $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$ Then, $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\left(\because \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\right)$ $\mathrm{Z}=\frac{\mathrm{E}_{0}}{\mathrm{I}_{0}}=0$ $\left(\mathrm{I}_{0}=\infty\right)$ This $\frac{1}{\sqrt{\mathrm{LC}}}$ is the natural frequency of LC circuit if the frequency of applied $\mathrm{AC}$ becomes equal to the natural frequency of $\mathrm{AC}$ circuit then the amplitude of current becomes infinite due to zero impedance.
VITEEE-2015
Alternating Current
155234
In an AC circuit, the potential across an inductance and resistance joined in series are respectively $16 \mathrm{~V}$ and $20 \mathrm{~V}$. The total potential difference across the circuit is
1 $20.0 \mathrm{~V}$
2 $25.6 \mathrm{~V}$
3 $31.9 \mathrm{~V}$
4 $33.6 \mathrm{~V}$
Explanation:
B Given that, $\mathrm{V}_{\mathrm{R}}=20$ Volt, $\mathrm{V}_{\mathrm{L}}=16$ Volt Then, total potential difference across the circuit is, $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{L}}^{2}}$ $\mathrm{~V}=\sqrt{(20)^{2}+(16)^{2}}$ $\mathrm{~V}=\sqrt{656}$ $\mathrm{~V}=25.6 \text { Volt }$
VITEEE-2012
Alternating Current
155235
The following series L-C-R circuit, when driven by an emf source of angular frequency 70 kilo-radians per second, the circuit effectively behaves like
1 purely resistive circuit
2 series R-L circuit
3 series R-C circuit
4 series L-C circuit with $\mathrm{R}=0$
Explanation:
C Given that, $\omega=70 \mathrm{k}$ radians $/ \mathrm{sec}=70 \times 10^{3} \mathrm{rad} / \mathrm{sec}$ $\mathrm{L}=100 \times 10^{-6} \mathrm{H}, \mathrm{C}=1 \times 10^{-6} \mathrm{~F}, \mathrm{R}=10 \Omega$ We know that, If $X_{L}>X_{C}$ then circuit is called $R-L$ circuit. If $\mathrm{X}_{\mathrm{C}}>\mathrm{X}_{\mathrm{L}}$ then circuit is $\mathrm{R}-\mathrm{C}$ circuit. $X_{L}=\omega L$ $X_{L}=70 \times 10^{3} \times 100 \times 10^{-6}$ $X_{L}=7 \Omega$ Capacitive reactance, $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{70 \times 10^{3} \times 1 \times 10^{-6}}$ $\mathrm{X}_{\mathrm{C}}=\frac{1000}{70}$ $\mathrm{X}_{\mathrm{C}}=14.3 \Omega$ Since $X_{C}>X_{L}$ then circuit is called series $R-C$ circuit.
