155283
An LC resonant circuit contains a $400 \mathrm{pF}$ capacitor and an inductor of $400 \mu \mathrm{H}$. It is coupled to an antenna. Wavelength of radiated electromagnetic wave is
1 $754 \mathrm{~m}$
2 $377 \mathrm{~m}$
3 $377 \mathrm{~cm}$
4 $796 \mathrm{~m}$
Explanation:
A Given that, Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $\mathrm{C}=400 \mathrm{pF}=400 \times 10^{-12} \mathrm{~F}$ $\mathrm{L}=400 \mu \mathrm{H}=400 \times 10^{-6} \mathrm{~F}$ We know that, Frequency of the resonating LC circuit - $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{400 \times 10^{-12} \times 400 \times 10^{-6}}}$ $\mathrm{f}=397.7 \mathrm{kHz}$ We know that, Wavelength $(\lambda)=\frac{\text { Velocity }}{\text { Frequency }}$ $\lambda=\frac{3 \times 10^{8}}{397.7 \times 10^{3}}$ $\lambda=0.00754 \times 10^{5}$ $\lambda = 754 \mathrm{~m}$ Hence, An LC resonant circuit contains a $400 \mathrm{pF}$ capacitor and an inductor of $400 \mu \mathrm{H}$. It is coupled to an antenna. Wavelength of radiated electromagnetic wave is $754 \mathrm{~m}$.
AP EAMCET-25.08.2021
Alternating Current
155284
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 $\mathrm{f}_{0} / 4$
2 $2 \mathrm{f}_{0}$
3 $\mathrm{f}_{0}$
4 $\mathrm{f}_{0} / 2$
Explanation:
D We know that, Resonance frequency $\left(\mathrm{f}_{0}\right)=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ So, $\quad \mathrm{f}_{0} \propto \frac{1}{\sqrt{\mathrm{C}}}$ Therefore, $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=\sqrt{\frac{\mathrm{C}^{\prime}}{\mathrm{C}}}$ $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=\sqrt{\frac{4 \mathrm{C}}{\mathrm{C}}}$ $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=2$ $\mathrm{f}_{0}^{\prime}=\frac{\mathrm{f}_{0}}{2}$
AP EAMCET-19.08.2021
Alternating Current
155285
An oscillator circuit consists of an inductance of $0.5 \mathrm{mH}$ and a capacitor of $20 \mu \mathrm{F}$. The resonant frequency of the circuit is nearly
1 $15.92 \mathrm{~Hz}$
2 $159.2 \mathrm{~Hz}$
3 $1592 \mathrm{~Hz}$
4 $15910 \mathrm{~Hz}$
Explanation:
C Give data, $\mathrm{L}=0.5 \mathrm{mH}, \quad \mathrm{C}=20 \mu \mathrm{F}$ We know that, resonant frequency- $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{0.5 \times 10^{-3} \times 20 \times 10^{-6}}}$ $\mathrm{f}=1592.3 \mathrm{~Hz}$
AIIMS-25.05.2019(E) Shift-2]
Alternating Current
155286
LCR circuit, the resonance frequency of circuit increases two times of the initial circuit by changing $C$ and $C^{\prime}$ and $R$ from $100 \Omega$ to $400 \Omega$ while the inductance was kept the same. The ratio $\mathbf{C} / \mathbf{C}^{\prime}$, is
1 2
2 8
3 16
4 4
Explanation:
D Given, Resistance, $\mathrm{R}_{1}=100 \Omega, \mathrm{R}_{2}=400 \Omega$ Initial capacitance $\left(\mathrm{C}_{1}\right)=\mathrm{C}$ Final capacitance $\left(\mathrm{C}_{2}\right)=\mathrm{C}^{\prime}$ Inductance $(\mathrm{L})=\mathrm{L}_{1}=\mathrm{L}_{2}$ We know that, Resonance frequency $f_{1}=\frac{1}{2 \pi \sqrt{L C}}$ $f_{2}=\frac{1}{2 \pi \sqrt{L^{\prime}}}$ According to question, $\mathrm{f}_{2}=2 \mathrm{f}_{1}$ $\frac{1}{2 \pi \sqrt{\mathrm{LC} \mathrm{C}^{\prime}}}=\frac{2}{2 \pi \sqrt{\mathrm{LC}}}$ $\frac{1}{\sqrt{\mathrm{C}^{\prime}}}=\frac{2}{\sqrt{\mathrm{C}}}$ $\frac{\mathrm{C}}{\mathrm{C}^{\prime}}=4$
TS- EAMCET-04.05.2019
Alternating Current
155287
Find resonance frequency in the given circuit
1 $\frac{1}{\sqrt{\mathrm{LC}}}$
2 $\frac{2}{\sqrt{\mathrm{LC}}}$
3 $\frac{1}{2 \sqrt{\mathrm{LC}}}$
4 $\frac{4}{\sqrt{\mathrm{LC}}}$
Explanation:
A In parallel of inductance - $\mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ In parallel of capacitance - $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ At resonance, $\omega_{1}=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}} \& \omega_{2}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}=\frac{1}{\omega^{2}} \quad\left(\because \omega_{1}=\omega_{2}=\omega\right)$ Since, $\quad \omega_{\mathrm{eq}}=\frac{1}{\sqrt{\mathrm{L}_{\mathrm{eq}} \mathrm{C}_{\mathrm{eq}}}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{\mathrm{L}_{1} \mathrm{~L}_{2} \mathrm{C}_{1}+\mathrm{L}_{1} \mathrm{~L}_{2} \mathrm{C}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{1}{\omega^{2}}\left(\frac{\mathrm{L}_{2}+\mathrm{L}_{1}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}\right)}}$ $\omega_{\mathrm{eq}}=\omega=\frac{1}{\sqrt{\mathrm{LC}}}$
155283
An LC resonant circuit contains a $400 \mathrm{pF}$ capacitor and an inductor of $400 \mu \mathrm{H}$. It is coupled to an antenna. Wavelength of radiated electromagnetic wave is
1 $754 \mathrm{~m}$
2 $377 \mathrm{~m}$
3 $377 \mathrm{~cm}$
4 $796 \mathrm{~m}$
Explanation:
A Given that, Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $\mathrm{C}=400 \mathrm{pF}=400 \times 10^{-12} \mathrm{~F}$ $\mathrm{L}=400 \mu \mathrm{H}=400 \times 10^{-6} \mathrm{~F}$ We know that, Frequency of the resonating LC circuit - $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{400 \times 10^{-12} \times 400 \times 10^{-6}}}$ $\mathrm{f}=397.7 \mathrm{kHz}$ We know that, Wavelength $(\lambda)=\frac{\text { Velocity }}{\text { Frequency }}$ $\lambda=\frac{3 \times 10^{8}}{397.7 \times 10^{3}}$ $\lambda=0.00754 \times 10^{5}$ $\lambda = 754 \mathrm{~m}$ Hence, An LC resonant circuit contains a $400 \mathrm{pF}$ capacitor and an inductor of $400 \mu \mathrm{H}$. It is coupled to an antenna. Wavelength of radiated electromagnetic wave is $754 \mathrm{~m}$.
AP EAMCET-25.08.2021
Alternating Current
155284
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 $\mathrm{f}_{0} / 4$
2 $2 \mathrm{f}_{0}$
3 $\mathrm{f}_{0}$
4 $\mathrm{f}_{0} / 2$
Explanation:
D We know that, Resonance frequency $\left(\mathrm{f}_{0}\right)=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ So, $\quad \mathrm{f}_{0} \propto \frac{1}{\sqrt{\mathrm{C}}}$ Therefore, $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=\sqrt{\frac{\mathrm{C}^{\prime}}{\mathrm{C}}}$ $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=\sqrt{\frac{4 \mathrm{C}}{\mathrm{C}}}$ $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=2$ $\mathrm{f}_{0}^{\prime}=\frac{\mathrm{f}_{0}}{2}$
AP EAMCET-19.08.2021
Alternating Current
155285
An oscillator circuit consists of an inductance of $0.5 \mathrm{mH}$ and a capacitor of $20 \mu \mathrm{F}$. The resonant frequency of the circuit is nearly
1 $15.92 \mathrm{~Hz}$
2 $159.2 \mathrm{~Hz}$
3 $1592 \mathrm{~Hz}$
4 $15910 \mathrm{~Hz}$
Explanation:
C Give data, $\mathrm{L}=0.5 \mathrm{mH}, \quad \mathrm{C}=20 \mu \mathrm{F}$ We know that, resonant frequency- $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{0.5 \times 10^{-3} \times 20 \times 10^{-6}}}$ $\mathrm{f}=1592.3 \mathrm{~Hz}$
AIIMS-25.05.2019(E) Shift-2]
Alternating Current
155286
LCR circuit, the resonance frequency of circuit increases two times of the initial circuit by changing $C$ and $C^{\prime}$ and $R$ from $100 \Omega$ to $400 \Omega$ while the inductance was kept the same. The ratio $\mathbf{C} / \mathbf{C}^{\prime}$, is
1 2
2 8
3 16
4 4
Explanation:
D Given, Resistance, $\mathrm{R}_{1}=100 \Omega, \mathrm{R}_{2}=400 \Omega$ Initial capacitance $\left(\mathrm{C}_{1}\right)=\mathrm{C}$ Final capacitance $\left(\mathrm{C}_{2}\right)=\mathrm{C}^{\prime}$ Inductance $(\mathrm{L})=\mathrm{L}_{1}=\mathrm{L}_{2}$ We know that, Resonance frequency $f_{1}=\frac{1}{2 \pi \sqrt{L C}}$ $f_{2}=\frac{1}{2 \pi \sqrt{L^{\prime}}}$ According to question, $\mathrm{f}_{2}=2 \mathrm{f}_{1}$ $\frac{1}{2 \pi \sqrt{\mathrm{LC} \mathrm{C}^{\prime}}}=\frac{2}{2 \pi \sqrt{\mathrm{LC}}}$ $\frac{1}{\sqrt{\mathrm{C}^{\prime}}}=\frac{2}{\sqrt{\mathrm{C}}}$ $\frac{\mathrm{C}}{\mathrm{C}^{\prime}}=4$
TS- EAMCET-04.05.2019
Alternating Current
155287
Find resonance frequency in the given circuit
1 $\frac{1}{\sqrt{\mathrm{LC}}}$
2 $\frac{2}{\sqrt{\mathrm{LC}}}$
3 $\frac{1}{2 \sqrt{\mathrm{LC}}}$
4 $\frac{4}{\sqrt{\mathrm{LC}}}$
Explanation:
A In parallel of inductance - $\mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ In parallel of capacitance - $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ At resonance, $\omega_{1}=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}} \& \omega_{2}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}=\frac{1}{\omega^{2}} \quad\left(\because \omega_{1}=\omega_{2}=\omega\right)$ Since, $\quad \omega_{\mathrm{eq}}=\frac{1}{\sqrt{\mathrm{L}_{\mathrm{eq}} \mathrm{C}_{\mathrm{eq}}}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{\mathrm{L}_{1} \mathrm{~L}_{2} \mathrm{C}_{1}+\mathrm{L}_{1} \mathrm{~L}_{2} \mathrm{C}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{1}{\omega^{2}}\left(\frac{\mathrm{L}_{2}+\mathrm{L}_{1}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}\right)}}$ $\omega_{\mathrm{eq}}=\omega=\frac{1}{\sqrt{\mathrm{LC}}}$
155283
An LC resonant circuit contains a $400 \mathrm{pF}$ capacitor and an inductor of $400 \mu \mathrm{H}$. It is coupled to an antenna. Wavelength of radiated electromagnetic wave is
1 $754 \mathrm{~m}$
2 $377 \mathrm{~m}$
3 $377 \mathrm{~cm}$
4 $796 \mathrm{~m}$
Explanation:
A Given that, Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $\mathrm{C}=400 \mathrm{pF}=400 \times 10^{-12} \mathrm{~F}$ $\mathrm{L}=400 \mu \mathrm{H}=400 \times 10^{-6} \mathrm{~F}$ We know that, Frequency of the resonating LC circuit - $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{400 \times 10^{-12} \times 400 \times 10^{-6}}}$ $\mathrm{f}=397.7 \mathrm{kHz}$ We know that, Wavelength $(\lambda)=\frac{\text { Velocity }}{\text { Frequency }}$ $\lambda=\frac{3 \times 10^{8}}{397.7 \times 10^{3}}$ $\lambda=0.00754 \times 10^{5}$ $\lambda = 754 \mathrm{~m}$ Hence, An LC resonant circuit contains a $400 \mathrm{pF}$ capacitor and an inductor of $400 \mu \mathrm{H}$. It is coupled to an antenna. Wavelength of radiated electromagnetic wave is $754 \mathrm{~m}$.
AP EAMCET-25.08.2021
Alternating Current
155284
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 $\mathrm{f}_{0} / 4$
2 $2 \mathrm{f}_{0}$
3 $\mathrm{f}_{0}$
4 $\mathrm{f}_{0} / 2$
Explanation:
D We know that, Resonance frequency $\left(\mathrm{f}_{0}\right)=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ So, $\quad \mathrm{f}_{0} \propto \frac{1}{\sqrt{\mathrm{C}}}$ Therefore, $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=\sqrt{\frac{\mathrm{C}^{\prime}}{\mathrm{C}}}$ $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=\sqrt{\frac{4 \mathrm{C}}{\mathrm{C}}}$ $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=2$ $\mathrm{f}_{0}^{\prime}=\frac{\mathrm{f}_{0}}{2}$
AP EAMCET-19.08.2021
Alternating Current
155285
An oscillator circuit consists of an inductance of $0.5 \mathrm{mH}$ and a capacitor of $20 \mu \mathrm{F}$. The resonant frequency of the circuit is nearly
1 $15.92 \mathrm{~Hz}$
2 $159.2 \mathrm{~Hz}$
3 $1592 \mathrm{~Hz}$
4 $15910 \mathrm{~Hz}$
Explanation:
C Give data, $\mathrm{L}=0.5 \mathrm{mH}, \quad \mathrm{C}=20 \mu \mathrm{F}$ We know that, resonant frequency- $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{0.5 \times 10^{-3} \times 20 \times 10^{-6}}}$ $\mathrm{f}=1592.3 \mathrm{~Hz}$
AIIMS-25.05.2019(E) Shift-2]
Alternating Current
155286
LCR circuit, the resonance frequency of circuit increases two times of the initial circuit by changing $C$ and $C^{\prime}$ and $R$ from $100 \Omega$ to $400 \Omega$ while the inductance was kept the same. The ratio $\mathbf{C} / \mathbf{C}^{\prime}$, is
1 2
2 8
3 16
4 4
Explanation:
D Given, Resistance, $\mathrm{R}_{1}=100 \Omega, \mathrm{R}_{2}=400 \Omega$ Initial capacitance $\left(\mathrm{C}_{1}\right)=\mathrm{C}$ Final capacitance $\left(\mathrm{C}_{2}\right)=\mathrm{C}^{\prime}$ Inductance $(\mathrm{L})=\mathrm{L}_{1}=\mathrm{L}_{2}$ We know that, Resonance frequency $f_{1}=\frac{1}{2 \pi \sqrt{L C}}$ $f_{2}=\frac{1}{2 \pi \sqrt{L^{\prime}}}$ According to question, $\mathrm{f}_{2}=2 \mathrm{f}_{1}$ $\frac{1}{2 \pi \sqrt{\mathrm{LC} \mathrm{C}^{\prime}}}=\frac{2}{2 \pi \sqrt{\mathrm{LC}}}$ $\frac{1}{\sqrt{\mathrm{C}^{\prime}}}=\frac{2}{\sqrt{\mathrm{C}}}$ $\frac{\mathrm{C}}{\mathrm{C}^{\prime}}=4$
TS- EAMCET-04.05.2019
Alternating Current
155287
Find resonance frequency in the given circuit
1 $\frac{1}{\sqrt{\mathrm{LC}}}$
2 $\frac{2}{\sqrt{\mathrm{LC}}}$
3 $\frac{1}{2 \sqrt{\mathrm{LC}}}$
4 $\frac{4}{\sqrt{\mathrm{LC}}}$
Explanation:
A In parallel of inductance - $\mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ In parallel of capacitance - $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ At resonance, $\omega_{1}=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}} \& \omega_{2}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}=\frac{1}{\omega^{2}} \quad\left(\because \omega_{1}=\omega_{2}=\omega\right)$ Since, $\quad \omega_{\mathrm{eq}}=\frac{1}{\sqrt{\mathrm{L}_{\mathrm{eq}} \mathrm{C}_{\mathrm{eq}}}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{\mathrm{L}_{1} \mathrm{~L}_{2} \mathrm{C}_{1}+\mathrm{L}_{1} \mathrm{~L}_{2} \mathrm{C}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{1}{\omega^{2}}\left(\frac{\mathrm{L}_{2}+\mathrm{L}_{1}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}\right)}}$ $\omega_{\mathrm{eq}}=\omega=\frac{1}{\sqrt{\mathrm{LC}}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
155283
An LC resonant circuit contains a $400 \mathrm{pF}$ capacitor and an inductor of $400 \mu \mathrm{H}$. It is coupled to an antenna. Wavelength of radiated electromagnetic wave is
1 $754 \mathrm{~m}$
2 $377 \mathrm{~m}$
3 $377 \mathrm{~cm}$
4 $796 \mathrm{~m}$
Explanation:
A Given that, Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $\mathrm{C}=400 \mathrm{pF}=400 \times 10^{-12} \mathrm{~F}$ $\mathrm{L}=400 \mu \mathrm{H}=400 \times 10^{-6} \mathrm{~F}$ We know that, Frequency of the resonating LC circuit - $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{400 \times 10^{-12} \times 400 \times 10^{-6}}}$ $\mathrm{f}=397.7 \mathrm{kHz}$ We know that, Wavelength $(\lambda)=\frac{\text { Velocity }}{\text { Frequency }}$ $\lambda=\frac{3 \times 10^{8}}{397.7 \times 10^{3}}$ $\lambda=0.00754 \times 10^{5}$ $\lambda = 754 \mathrm{~m}$ Hence, An LC resonant circuit contains a $400 \mathrm{pF}$ capacitor and an inductor of $400 \mu \mathrm{H}$. It is coupled to an antenna. Wavelength of radiated electromagnetic wave is $754 \mathrm{~m}$.
AP EAMCET-25.08.2021
Alternating Current
155284
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 $\mathrm{f}_{0} / 4$
2 $2 \mathrm{f}_{0}$
3 $\mathrm{f}_{0}$
4 $\mathrm{f}_{0} / 2$
Explanation:
D We know that, Resonance frequency $\left(\mathrm{f}_{0}\right)=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ So, $\quad \mathrm{f}_{0} \propto \frac{1}{\sqrt{\mathrm{C}}}$ Therefore, $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=\sqrt{\frac{\mathrm{C}^{\prime}}{\mathrm{C}}}$ $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=\sqrt{\frac{4 \mathrm{C}}{\mathrm{C}}}$ $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=2$ $\mathrm{f}_{0}^{\prime}=\frac{\mathrm{f}_{0}}{2}$
AP EAMCET-19.08.2021
Alternating Current
155285
An oscillator circuit consists of an inductance of $0.5 \mathrm{mH}$ and a capacitor of $20 \mu \mathrm{F}$. The resonant frequency of the circuit is nearly
1 $15.92 \mathrm{~Hz}$
2 $159.2 \mathrm{~Hz}$
3 $1592 \mathrm{~Hz}$
4 $15910 \mathrm{~Hz}$
Explanation:
C Give data, $\mathrm{L}=0.5 \mathrm{mH}, \quad \mathrm{C}=20 \mu \mathrm{F}$ We know that, resonant frequency- $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{0.5 \times 10^{-3} \times 20 \times 10^{-6}}}$ $\mathrm{f}=1592.3 \mathrm{~Hz}$
AIIMS-25.05.2019(E) Shift-2]
Alternating Current
155286
LCR circuit, the resonance frequency of circuit increases two times of the initial circuit by changing $C$ and $C^{\prime}$ and $R$ from $100 \Omega$ to $400 \Omega$ while the inductance was kept the same. The ratio $\mathbf{C} / \mathbf{C}^{\prime}$, is
1 2
2 8
3 16
4 4
Explanation:
D Given, Resistance, $\mathrm{R}_{1}=100 \Omega, \mathrm{R}_{2}=400 \Omega$ Initial capacitance $\left(\mathrm{C}_{1}\right)=\mathrm{C}$ Final capacitance $\left(\mathrm{C}_{2}\right)=\mathrm{C}^{\prime}$ Inductance $(\mathrm{L})=\mathrm{L}_{1}=\mathrm{L}_{2}$ We know that, Resonance frequency $f_{1}=\frac{1}{2 \pi \sqrt{L C}}$ $f_{2}=\frac{1}{2 \pi \sqrt{L^{\prime}}}$ According to question, $\mathrm{f}_{2}=2 \mathrm{f}_{1}$ $\frac{1}{2 \pi \sqrt{\mathrm{LC} \mathrm{C}^{\prime}}}=\frac{2}{2 \pi \sqrt{\mathrm{LC}}}$ $\frac{1}{\sqrt{\mathrm{C}^{\prime}}}=\frac{2}{\sqrt{\mathrm{C}}}$ $\frac{\mathrm{C}}{\mathrm{C}^{\prime}}=4$
TS- EAMCET-04.05.2019
Alternating Current
155287
Find resonance frequency in the given circuit
1 $\frac{1}{\sqrt{\mathrm{LC}}}$
2 $\frac{2}{\sqrt{\mathrm{LC}}}$
3 $\frac{1}{2 \sqrt{\mathrm{LC}}}$
4 $\frac{4}{\sqrt{\mathrm{LC}}}$
Explanation:
A In parallel of inductance - $\mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ In parallel of capacitance - $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ At resonance, $\omega_{1}=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}} \& \omega_{2}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}=\frac{1}{\omega^{2}} \quad\left(\because \omega_{1}=\omega_{2}=\omega\right)$ Since, $\quad \omega_{\mathrm{eq}}=\frac{1}{\sqrt{\mathrm{L}_{\mathrm{eq}} \mathrm{C}_{\mathrm{eq}}}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{\mathrm{L}_{1} \mathrm{~L}_{2} \mathrm{C}_{1}+\mathrm{L}_{1} \mathrm{~L}_{2} \mathrm{C}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{1}{\omega^{2}}\left(\frac{\mathrm{L}_{2}+\mathrm{L}_{1}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}\right)}}$ $\omega_{\mathrm{eq}}=\omega=\frac{1}{\sqrt{\mathrm{LC}}}$
155283
An LC resonant circuit contains a $400 \mathrm{pF}$ capacitor and an inductor of $400 \mu \mathrm{H}$. It is coupled to an antenna. Wavelength of radiated electromagnetic wave is
1 $754 \mathrm{~m}$
2 $377 \mathrm{~m}$
3 $377 \mathrm{~cm}$
4 $796 \mathrm{~m}$
Explanation:
A Given that, Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $\mathrm{C}=400 \mathrm{pF}=400 \times 10^{-12} \mathrm{~F}$ $\mathrm{L}=400 \mu \mathrm{H}=400 \times 10^{-6} \mathrm{~F}$ We know that, Frequency of the resonating LC circuit - $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{400 \times 10^{-12} \times 400 \times 10^{-6}}}$ $\mathrm{f}=397.7 \mathrm{kHz}$ We know that, Wavelength $(\lambda)=\frac{\text { Velocity }}{\text { Frequency }}$ $\lambda=\frac{3 \times 10^{8}}{397.7 \times 10^{3}}$ $\lambda=0.00754 \times 10^{5}$ $\lambda = 754 \mathrm{~m}$ Hence, An LC resonant circuit contains a $400 \mathrm{pF}$ capacitor and an inductor of $400 \mu \mathrm{H}$. It is coupled to an antenna. Wavelength of radiated electromagnetic wave is $754 \mathrm{~m}$.
AP EAMCET-25.08.2021
Alternating Current
155284
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 $\mathrm{f}_{0} / 4$
2 $2 \mathrm{f}_{0}$
3 $\mathrm{f}_{0}$
4 $\mathrm{f}_{0} / 2$
Explanation:
D We know that, Resonance frequency $\left(\mathrm{f}_{0}\right)=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ So, $\quad \mathrm{f}_{0} \propto \frac{1}{\sqrt{\mathrm{C}}}$ Therefore, $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=\sqrt{\frac{\mathrm{C}^{\prime}}{\mathrm{C}}}$ $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=\sqrt{\frac{4 \mathrm{C}}{\mathrm{C}}}$ $\frac{\mathrm{f}_{0}}{\mathrm{f}_{0}^{\prime}}=2$ $\mathrm{f}_{0}^{\prime}=\frac{\mathrm{f}_{0}}{2}$
AP EAMCET-19.08.2021
Alternating Current
155285
An oscillator circuit consists of an inductance of $0.5 \mathrm{mH}$ and a capacitor of $20 \mu \mathrm{F}$. The resonant frequency of the circuit is nearly
1 $15.92 \mathrm{~Hz}$
2 $159.2 \mathrm{~Hz}$
3 $1592 \mathrm{~Hz}$
4 $15910 \mathrm{~Hz}$
Explanation:
C Give data, $\mathrm{L}=0.5 \mathrm{mH}, \quad \mathrm{C}=20 \mu \mathrm{F}$ We know that, resonant frequency- $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{0.5 \times 10^{-3} \times 20 \times 10^{-6}}}$ $\mathrm{f}=1592.3 \mathrm{~Hz}$
AIIMS-25.05.2019(E) Shift-2]
Alternating Current
155286
LCR circuit, the resonance frequency of circuit increases two times of the initial circuit by changing $C$ and $C^{\prime}$ and $R$ from $100 \Omega$ to $400 \Omega$ while the inductance was kept the same. The ratio $\mathbf{C} / \mathbf{C}^{\prime}$, is
1 2
2 8
3 16
4 4
Explanation:
D Given, Resistance, $\mathrm{R}_{1}=100 \Omega, \mathrm{R}_{2}=400 \Omega$ Initial capacitance $\left(\mathrm{C}_{1}\right)=\mathrm{C}$ Final capacitance $\left(\mathrm{C}_{2}\right)=\mathrm{C}^{\prime}$ Inductance $(\mathrm{L})=\mathrm{L}_{1}=\mathrm{L}_{2}$ We know that, Resonance frequency $f_{1}=\frac{1}{2 \pi \sqrt{L C}}$ $f_{2}=\frac{1}{2 \pi \sqrt{L^{\prime}}}$ According to question, $\mathrm{f}_{2}=2 \mathrm{f}_{1}$ $\frac{1}{2 \pi \sqrt{\mathrm{LC} \mathrm{C}^{\prime}}}=\frac{2}{2 \pi \sqrt{\mathrm{LC}}}$ $\frac{1}{\sqrt{\mathrm{C}^{\prime}}}=\frac{2}{\sqrt{\mathrm{C}}}$ $\frac{\mathrm{C}}{\mathrm{C}^{\prime}}=4$
TS- EAMCET-04.05.2019
Alternating Current
155287
Find resonance frequency in the given circuit
1 $\frac{1}{\sqrt{\mathrm{LC}}}$
2 $\frac{2}{\sqrt{\mathrm{LC}}}$
3 $\frac{1}{2 \sqrt{\mathrm{LC}}}$
4 $\frac{4}{\sqrt{\mathrm{LC}}}$
Explanation:
A In parallel of inductance - $\mathrm{L}_{\mathrm{eq}}=\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$ In parallel of capacitance - $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ At resonance, $\omega_{1}=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}} \& \omega_{2}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}=\frac{1}{\omega^{2}} \quad\left(\because \omega_{1}=\omega_{2}=\omega\right)$ Since, $\quad \omega_{\mathrm{eq}}=\frac{1}{\sqrt{\mathrm{L}_{\mathrm{eq}} \mathrm{C}_{\mathrm{eq}}}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{\mathrm{L}_{1} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{\mathrm{L}_{1} \mathrm{~L}_{2} \mathrm{C}_{1}+\mathrm{L}_{1} \mathrm{~L}_{2} \mathrm{C}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}}}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{1}{\omega^{2}}\left(\frac{\mathrm{L}_{2}+\mathrm{L}_{1}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}\right)}}$ $\omega_{\mathrm{eq}}=\omega=\frac{1}{\sqrt{\mathrm{LC}}}$
