NEET Test Series from KOTA - 10 Papers In MS WORD
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AC (NCERT)
274655
To increase the resonant frequency in series LCR circuit,
1 source frequency should be increased
2 another resistance should be added in series with the first resistance.
3 another capacitor should be added in series with the first capacitor.
4 the source frequency should be decreased.
Explanation:
(c) Resonant frequency, ${{f}_{r}}=\frac{1}{2\pi \sqrt{LC}}$
${{f}_{r}}\times \frac{1}{\sqrt{C}}$
When another capacitor is added in series, ${{\text{C}}_{\text{eq}}}$ decreases.
So, ${{f}_{r}}$ increases.
NCERT Page-247 / N-189
AC (NCERT)
274658
An ac voltage is applied to a resistance $R$ and an inductor $\text{L}$ in series. If $\text{R}$ and the inductive reactance are both equal to $3\text{ }\!\!\Omega\!\!\text{ }$, the phase difference between the applied voltage and the current in the circuit is
1 $\pi /6$.
2 $\pi /4$
3 $\pi /2$
4 zero
Explanation:
(b) The phase difference $\phi $ is given by
$\text{tan}\phi =\frac{{{\text{X}}_{\text{L}}}}{\text{R}}=\frac{3}{3}=1$
$\Rightarrow \phi =\frac{\pi }{4}$
NCERT Page-245 / N-187
AC (NCERT)
274659
An inductive circuit contains resistance of $10\text{ohms}$ and an inductance of 2 henry. If an A.C. voltage of 120 Volts and frequency $60\text{Hz}$ is applied to this circuit, the current would be nearly
1 $0.32\text{A}$
2 $0.16\text{A}$
3 $0.48\text{A}$
4 $0.80\text{A}$
Explanation:
(b)
NCERT Page-2 247 / N-187
AC (NCERT)
274660
In an LR circuit $f=50\text{Hz},\text{L}=2\text{H},\text{E}=5$ volts, $\text{R}=1\text{ }\!\!\Omega\!\!\text{ }$ then energy stored in inductor is
1 $50\text{J}$
2 $25\text{J}$
3 $100\text{J}$
4 None of these
Explanation:
(d) $\text{L}=2\text{H},\text{E}=5$ volts, $\text{R}=1\text{ }\!\!\Omega\!\!\text{ }$
Energy in inductor $=\frac{1}{2}\text{L}{{\text{I}}^{2}}\text{I}=\frac{\text{E}}{\text{Z}}$
$I=\frac{5}{\sqrt{{{R}^{2}}+{{(\omega L)}^{2}}}}=\frac{5}{\sqrt{1+4{{\pi }^{2}}\times {{50}^{2}}\times 4}}$$=\frac{5}{\sqrt{1+{{(200\pi )}^{2}}}}=\frac{5}{200\pi }$
Energy $=\frac{1}{2}\times 2\times \frac{5\times 5}{200\times 200{{\pi }^{2}}}=6.33\times {{10}^{-5}}$ joules
274655
To increase the resonant frequency in series LCR circuit,
1 source frequency should be increased
2 another resistance should be added in series with the first resistance.
3 another capacitor should be added in series with the first capacitor.
4 the source frequency should be decreased.
Explanation:
(c) Resonant frequency, ${{f}_{r}}=\frac{1}{2\pi \sqrt{LC}}$
${{f}_{r}}\times \frac{1}{\sqrt{C}}$
When another capacitor is added in series, ${{\text{C}}_{\text{eq}}}$ decreases.
So, ${{f}_{r}}$ increases.
NCERT Page-247 / N-189
AC (NCERT)
274658
An ac voltage is applied to a resistance $R$ and an inductor $\text{L}$ in series. If $\text{R}$ and the inductive reactance are both equal to $3\text{ }\!\!\Omega\!\!\text{ }$, the phase difference between the applied voltage and the current in the circuit is
1 $\pi /6$.
2 $\pi /4$
3 $\pi /2$
4 zero
Explanation:
(b) The phase difference $\phi $ is given by
$\text{tan}\phi =\frac{{{\text{X}}_{\text{L}}}}{\text{R}}=\frac{3}{3}=1$
$\Rightarrow \phi =\frac{\pi }{4}$
NCERT Page-245 / N-187
AC (NCERT)
274659
An inductive circuit contains resistance of $10\text{ohms}$ and an inductance of 2 henry. If an A.C. voltage of 120 Volts and frequency $60\text{Hz}$ is applied to this circuit, the current would be nearly
1 $0.32\text{A}$
2 $0.16\text{A}$
3 $0.48\text{A}$
4 $0.80\text{A}$
Explanation:
(b)
NCERT Page-2 247 / N-187
AC (NCERT)
274660
In an LR circuit $f=50\text{Hz},\text{L}=2\text{H},\text{E}=5$ volts, $\text{R}=1\text{ }\!\!\Omega\!\!\text{ }$ then energy stored in inductor is
1 $50\text{J}$
2 $25\text{J}$
3 $100\text{J}$
4 None of these
Explanation:
(d) $\text{L}=2\text{H},\text{E}=5$ volts, $\text{R}=1\text{ }\!\!\Omega\!\!\text{ }$
Energy in inductor $=\frac{1}{2}\text{L}{{\text{I}}^{2}}\text{I}=\frac{\text{E}}{\text{Z}}$
$I=\frac{5}{\sqrt{{{R}^{2}}+{{(\omega L)}^{2}}}}=\frac{5}{\sqrt{1+4{{\pi }^{2}}\times {{50}^{2}}\times 4}}$$=\frac{5}{\sqrt{1+{{(200\pi )}^{2}}}}=\frac{5}{200\pi }$
Energy $=\frac{1}{2}\times 2\times \frac{5\times 5}{200\times 200{{\pi }^{2}}}=6.33\times {{10}^{-5}}$ joules
274655
To increase the resonant frequency in series LCR circuit,
1 source frequency should be increased
2 another resistance should be added in series with the first resistance.
3 another capacitor should be added in series with the first capacitor.
4 the source frequency should be decreased.
Explanation:
(c) Resonant frequency, ${{f}_{r}}=\frac{1}{2\pi \sqrt{LC}}$
${{f}_{r}}\times \frac{1}{\sqrt{C}}$
When another capacitor is added in series, ${{\text{C}}_{\text{eq}}}$ decreases.
So, ${{f}_{r}}$ increases.
NCERT Page-247 / N-189
AC (NCERT)
274658
An ac voltage is applied to a resistance $R$ and an inductor $\text{L}$ in series. If $\text{R}$ and the inductive reactance are both equal to $3\text{ }\!\!\Omega\!\!\text{ }$, the phase difference between the applied voltage and the current in the circuit is
1 $\pi /6$.
2 $\pi /4$
3 $\pi /2$
4 zero
Explanation:
(b) The phase difference $\phi $ is given by
$\text{tan}\phi =\frac{{{\text{X}}_{\text{L}}}}{\text{R}}=\frac{3}{3}=1$
$\Rightarrow \phi =\frac{\pi }{4}$
NCERT Page-245 / N-187
AC (NCERT)
274659
An inductive circuit contains resistance of $10\text{ohms}$ and an inductance of 2 henry. If an A.C. voltage of 120 Volts and frequency $60\text{Hz}$ is applied to this circuit, the current would be nearly
1 $0.32\text{A}$
2 $0.16\text{A}$
3 $0.48\text{A}$
4 $0.80\text{A}$
Explanation:
(b)
NCERT Page-2 247 / N-187
AC (NCERT)
274660
In an LR circuit $f=50\text{Hz},\text{L}=2\text{H},\text{E}=5$ volts, $\text{R}=1\text{ }\!\!\Omega\!\!\text{ }$ then energy stored in inductor is
1 $50\text{J}$
2 $25\text{J}$
3 $100\text{J}$
4 None of these
Explanation:
(d) $\text{L}=2\text{H},\text{E}=5$ volts, $\text{R}=1\text{ }\!\!\Omega\!\!\text{ }$
Energy in inductor $=\frac{1}{2}\text{L}{{\text{I}}^{2}}\text{I}=\frac{\text{E}}{\text{Z}}$
$I=\frac{5}{\sqrt{{{R}^{2}}+{{(\omega L)}^{2}}}}=\frac{5}{\sqrt{1+4{{\pi }^{2}}\times {{50}^{2}}\times 4}}$$=\frac{5}{\sqrt{1+{{(200\pi )}^{2}}}}=\frac{5}{200\pi }$
Energy $=\frac{1}{2}\times 2\times \frac{5\times 5}{200\times 200{{\pi }^{2}}}=6.33\times {{10}^{-5}}$ joules
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
AC (NCERT)
274655
To increase the resonant frequency in series LCR circuit,
1 source frequency should be increased
2 another resistance should be added in series with the first resistance.
3 another capacitor should be added in series with the first capacitor.
4 the source frequency should be decreased.
Explanation:
(c) Resonant frequency, ${{f}_{r}}=\frac{1}{2\pi \sqrt{LC}}$
${{f}_{r}}\times \frac{1}{\sqrt{C}}$
When another capacitor is added in series, ${{\text{C}}_{\text{eq}}}$ decreases.
So, ${{f}_{r}}$ increases.
NCERT Page-247 / N-189
AC (NCERT)
274658
An ac voltage is applied to a resistance $R$ and an inductor $\text{L}$ in series. If $\text{R}$ and the inductive reactance are both equal to $3\text{ }\!\!\Omega\!\!\text{ }$, the phase difference between the applied voltage and the current in the circuit is
1 $\pi /6$.
2 $\pi /4$
3 $\pi /2$
4 zero
Explanation:
(b) The phase difference $\phi $ is given by
$\text{tan}\phi =\frac{{{\text{X}}_{\text{L}}}}{\text{R}}=\frac{3}{3}=1$
$\Rightarrow \phi =\frac{\pi }{4}$
NCERT Page-245 / N-187
AC (NCERT)
274659
An inductive circuit contains resistance of $10\text{ohms}$ and an inductance of 2 henry. If an A.C. voltage of 120 Volts and frequency $60\text{Hz}$ is applied to this circuit, the current would be nearly
1 $0.32\text{A}$
2 $0.16\text{A}$
3 $0.48\text{A}$
4 $0.80\text{A}$
Explanation:
(b)
NCERT Page-2 247 / N-187
AC (NCERT)
274660
In an LR circuit $f=50\text{Hz},\text{L}=2\text{H},\text{E}=5$ volts, $\text{R}=1\text{ }\!\!\Omega\!\!\text{ }$ then energy stored in inductor is
1 $50\text{J}$
2 $25\text{J}$
3 $100\text{J}$
4 None of these
Explanation:
(d) $\text{L}=2\text{H},\text{E}=5$ volts, $\text{R}=1\text{ }\!\!\Omega\!\!\text{ }$
Energy in inductor $=\frac{1}{2}\text{L}{{\text{I}}^{2}}\text{I}=\frac{\text{E}}{\text{Z}}$
$I=\frac{5}{\sqrt{{{R}^{2}}+{{(\omega L)}^{2}}}}=\frac{5}{\sqrt{1+4{{\pi }^{2}}\times {{50}^{2}}\times 4}}$$=\frac{5}{\sqrt{1+{{(200\pi )}^{2}}}}=\frac{5}{200\pi }$
Energy $=\frac{1}{2}\times 2\times \frac{5\times 5}{200\times 200{{\pi }^{2}}}=6.33\times {{10}^{-5}}$ joules
