155288
In the given circuit the reading of voltmeter $V_{1}$ and $V_{2}$ are 300 volt each. The reading of the voltmeter $V_{3}$ and ammeter $A$ are respectively
1 $150 \mathrm{~V}$ and $2.2 \mathrm{~A}$
2 $220 \mathrm{~V}$ and $2.2 \mathrm{~A}$
3 $220 \mathrm{~V}$ and $2.0 \mathrm{~A}$
4 $100 \mathrm{~V}$ and $2.0 \mathrm{~A}$
Explanation:
B Given that, $\mathrm{V}_{1}=\mathrm{V}_{2}=300 \mathrm{~V}$ Here, $V_{1}, V_{2}$ are same but opposite in phase with each other So, cancel each other. Thus, $\mathrm{V}_{3}$ is equal to voltage source $220 \mathrm{~V}$. And Ammeter current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{220}{100}$ $\mathrm{I}_{\mathrm{A}}=2.2 \mathrm{~A}$
VITEEE-2018
Alternating Current
155289
Consider two series resonant circuits with components $L_{1}, C_{1}$ and $L_{2}, C_{2}$ with same resonant frequency, $\omega$. When connected in series, the resonant frequency of the combination is
1 $2 \omega$
2 $\frac{\omega}{2}$
3 $3 \omega$
4 $\omega$
Explanation:
D In series of inductance - $\mathrm{L}_{\mathrm{eq}}=\mathrm{L}_{1}+\mathrm{L}_{2}$ In series of capacitance, $\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ Resonant frequency is - $\omega_{1}=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}} \text { and } \omega_{2}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\omega_{1}=\omega_{2}=\omega$ $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ Then, $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\mathrm{L}_{\mathrm{eq}} \mathrm{C}_{\mathrm{eq}}}}=\frac{1}{\sqrt{\left(\mathrm{L}_{1}+\mathrm{L}_{2}\right)\left(\frac{\mathrm{C}_{1} \times \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right)}}$ $\because \quad \mathrm{L}_{1} \mathrm{C}_{1}=\frac{1}{\omega^{2}}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{1}{\omega^{2}}\left(\frac{\mathrm{C}_{2}+\mathrm{C}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right)}}$ $\omega_{\mathrm{eq}}=\omega$ $\omega_{\mathrm{eq}}=\omega$
TS EAMCET(Medical)-2017
Alternating Current
155290
In non-resonant circuit, what will be the nature of the circuit for frequencies higher than the resonant frequency?
1 Resistive
2 Capacitive
3 Inductive
4 None of these
Explanation:
C Let, $\omega_{\mathrm{R}}$ is the resonant frequency Then, $0 \lt \omega \lt \omega_{\mathrm{R}}$ At resonant frequency $X_{L}=X_{C}$ $\omega L=\frac{1}{\omega C}$ For frequencies higher than the resonant frequency- $\mathrm{X}_{\mathrm{L}}>\mathrm{X}_{\mathrm{C}}$ So, nature of circuit is inductive.
Manipal UGET-2012
Alternating Current
155291
The values of $L, C$ and $R$ for a circuit are $1 H$, $9 F$ and $3 \Omega$. What is the quality factor for the circuit at resonance ?
1 1
2 9
3 $\frac{1}{9}$
4 $\frac{1}{3}$
Explanation:
C Given that, $\mathrm{L}=1 \mathrm{H}, \mathrm{C}=9 \mathrm{~F}, \mathrm{R}=3 \Omega$ We know that, $\text { Quality factor }(\mathrm{Q}) =\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}} \quad\left(\because \omega=\frac{1}{\sqrt{\mathrm{LC}}}\right)$ $\mathrm{Q} =\frac{1}{3} \sqrt{\frac{1}{9}}$ $\mathrm{Q} =\frac{1}{9}$
155288
In the given circuit the reading of voltmeter $V_{1}$ and $V_{2}$ are 300 volt each. The reading of the voltmeter $V_{3}$ and ammeter $A$ are respectively
1 $150 \mathrm{~V}$ and $2.2 \mathrm{~A}$
2 $220 \mathrm{~V}$ and $2.2 \mathrm{~A}$
3 $220 \mathrm{~V}$ and $2.0 \mathrm{~A}$
4 $100 \mathrm{~V}$ and $2.0 \mathrm{~A}$
Explanation:
B Given that, $\mathrm{V}_{1}=\mathrm{V}_{2}=300 \mathrm{~V}$ Here, $V_{1}, V_{2}$ are same but opposite in phase with each other So, cancel each other. Thus, $\mathrm{V}_{3}$ is equal to voltage source $220 \mathrm{~V}$. And Ammeter current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{220}{100}$ $\mathrm{I}_{\mathrm{A}}=2.2 \mathrm{~A}$
VITEEE-2018
Alternating Current
155289
Consider two series resonant circuits with components $L_{1}, C_{1}$ and $L_{2}, C_{2}$ with same resonant frequency, $\omega$. When connected in series, the resonant frequency of the combination is
1 $2 \omega$
2 $\frac{\omega}{2}$
3 $3 \omega$
4 $\omega$
Explanation:
D In series of inductance - $\mathrm{L}_{\mathrm{eq}}=\mathrm{L}_{1}+\mathrm{L}_{2}$ In series of capacitance, $\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ Resonant frequency is - $\omega_{1}=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}} \text { and } \omega_{2}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\omega_{1}=\omega_{2}=\omega$ $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ Then, $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\mathrm{L}_{\mathrm{eq}} \mathrm{C}_{\mathrm{eq}}}}=\frac{1}{\sqrt{\left(\mathrm{L}_{1}+\mathrm{L}_{2}\right)\left(\frac{\mathrm{C}_{1} \times \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right)}}$ $\because \quad \mathrm{L}_{1} \mathrm{C}_{1}=\frac{1}{\omega^{2}}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{1}{\omega^{2}}\left(\frac{\mathrm{C}_{2}+\mathrm{C}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right)}}$ $\omega_{\mathrm{eq}}=\omega$ $\omega_{\mathrm{eq}}=\omega$
TS EAMCET(Medical)-2017
Alternating Current
155290
In non-resonant circuit, what will be the nature of the circuit for frequencies higher than the resonant frequency?
1 Resistive
2 Capacitive
3 Inductive
4 None of these
Explanation:
C Let, $\omega_{\mathrm{R}}$ is the resonant frequency Then, $0 \lt \omega \lt \omega_{\mathrm{R}}$ At resonant frequency $X_{L}=X_{C}$ $\omega L=\frac{1}{\omega C}$ For frequencies higher than the resonant frequency- $\mathrm{X}_{\mathrm{L}}>\mathrm{X}_{\mathrm{C}}$ So, nature of circuit is inductive.
Manipal UGET-2012
Alternating Current
155291
The values of $L, C$ and $R$ for a circuit are $1 H$, $9 F$ and $3 \Omega$. What is the quality factor for the circuit at resonance ?
1 1
2 9
3 $\frac{1}{9}$
4 $\frac{1}{3}$
Explanation:
C Given that, $\mathrm{L}=1 \mathrm{H}, \mathrm{C}=9 \mathrm{~F}, \mathrm{R}=3 \Omega$ We know that, $\text { Quality factor }(\mathrm{Q}) =\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}} \quad\left(\because \omega=\frac{1}{\sqrt{\mathrm{LC}}}\right)$ $\mathrm{Q} =\frac{1}{3} \sqrt{\frac{1}{9}}$ $\mathrm{Q} =\frac{1}{9}$
155288
In the given circuit the reading of voltmeter $V_{1}$ and $V_{2}$ are 300 volt each. The reading of the voltmeter $V_{3}$ and ammeter $A$ are respectively
1 $150 \mathrm{~V}$ and $2.2 \mathrm{~A}$
2 $220 \mathrm{~V}$ and $2.2 \mathrm{~A}$
3 $220 \mathrm{~V}$ and $2.0 \mathrm{~A}$
4 $100 \mathrm{~V}$ and $2.0 \mathrm{~A}$
Explanation:
B Given that, $\mathrm{V}_{1}=\mathrm{V}_{2}=300 \mathrm{~V}$ Here, $V_{1}, V_{2}$ are same but opposite in phase with each other So, cancel each other. Thus, $\mathrm{V}_{3}$ is equal to voltage source $220 \mathrm{~V}$. And Ammeter current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{220}{100}$ $\mathrm{I}_{\mathrm{A}}=2.2 \mathrm{~A}$
VITEEE-2018
Alternating Current
155289
Consider two series resonant circuits with components $L_{1}, C_{1}$ and $L_{2}, C_{2}$ with same resonant frequency, $\omega$. When connected in series, the resonant frequency of the combination is
1 $2 \omega$
2 $\frac{\omega}{2}$
3 $3 \omega$
4 $\omega$
Explanation:
D In series of inductance - $\mathrm{L}_{\mathrm{eq}}=\mathrm{L}_{1}+\mathrm{L}_{2}$ In series of capacitance, $\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ Resonant frequency is - $\omega_{1}=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}} \text { and } \omega_{2}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\omega_{1}=\omega_{2}=\omega$ $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ Then, $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\mathrm{L}_{\mathrm{eq}} \mathrm{C}_{\mathrm{eq}}}}=\frac{1}{\sqrt{\left(\mathrm{L}_{1}+\mathrm{L}_{2}\right)\left(\frac{\mathrm{C}_{1} \times \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right)}}$ $\because \quad \mathrm{L}_{1} \mathrm{C}_{1}=\frac{1}{\omega^{2}}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{1}{\omega^{2}}\left(\frac{\mathrm{C}_{2}+\mathrm{C}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right)}}$ $\omega_{\mathrm{eq}}=\omega$ $\omega_{\mathrm{eq}}=\omega$
TS EAMCET(Medical)-2017
Alternating Current
155290
In non-resonant circuit, what will be the nature of the circuit for frequencies higher than the resonant frequency?
1 Resistive
2 Capacitive
3 Inductive
4 None of these
Explanation:
C Let, $\omega_{\mathrm{R}}$ is the resonant frequency Then, $0 \lt \omega \lt \omega_{\mathrm{R}}$ At resonant frequency $X_{L}=X_{C}$ $\omega L=\frac{1}{\omega C}$ For frequencies higher than the resonant frequency- $\mathrm{X}_{\mathrm{L}}>\mathrm{X}_{\mathrm{C}}$ So, nature of circuit is inductive.
Manipal UGET-2012
Alternating Current
155291
The values of $L, C$ and $R$ for a circuit are $1 H$, $9 F$ and $3 \Omega$. What is the quality factor for the circuit at resonance ?
1 1
2 9
3 $\frac{1}{9}$
4 $\frac{1}{3}$
Explanation:
C Given that, $\mathrm{L}=1 \mathrm{H}, \mathrm{C}=9 \mathrm{~F}, \mathrm{R}=3 \Omega$ We know that, $\text { Quality factor }(\mathrm{Q}) =\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}} \quad\left(\because \omega=\frac{1}{\sqrt{\mathrm{LC}}}\right)$ $\mathrm{Q} =\frac{1}{3} \sqrt{\frac{1}{9}}$ $\mathrm{Q} =\frac{1}{9}$
155288
In the given circuit the reading of voltmeter $V_{1}$ and $V_{2}$ are 300 volt each. The reading of the voltmeter $V_{3}$ and ammeter $A$ are respectively
1 $150 \mathrm{~V}$ and $2.2 \mathrm{~A}$
2 $220 \mathrm{~V}$ and $2.2 \mathrm{~A}$
3 $220 \mathrm{~V}$ and $2.0 \mathrm{~A}$
4 $100 \mathrm{~V}$ and $2.0 \mathrm{~A}$
Explanation:
B Given that, $\mathrm{V}_{1}=\mathrm{V}_{2}=300 \mathrm{~V}$ Here, $V_{1}, V_{2}$ are same but opposite in phase with each other So, cancel each other. Thus, $\mathrm{V}_{3}$ is equal to voltage source $220 \mathrm{~V}$. And Ammeter current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{220}{100}$ $\mathrm{I}_{\mathrm{A}}=2.2 \mathrm{~A}$
VITEEE-2018
Alternating Current
155289
Consider two series resonant circuits with components $L_{1}, C_{1}$ and $L_{2}, C_{2}$ with same resonant frequency, $\omega$. When connected in series, the resonant frequency of the combination is
1 $2 \omega$
2 $\frac{\omega}{2}$
3 $3 \omega$
4 $\omega$
Explanation:
D In series of inductance - $\mathrm{L}_{\mathrm{eq}}=\mathrm{L}_{1}+\mathrm{L}_{2}$ In series of capacitance, $\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$ Resonant frequency is - $\omega_{1}=\frac{1}{\sqrt{\mathrm{L}_{1} \mathrm{C}_{1}}} \text { and } \omega_{2}=\frac{1}{\sqrt{\mathrm{L}_{2} \mathrm{C}_{2}}}$ $\omega_{1}=\omega_{2}=\omega$ $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ Then, $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\mathrm{L}_{\mathrm{eq}} \mathrm{C}_{\mathrm{eq}}}}=\frac{1}{\sqrt{\left(\mathrm{L}_{1}+\mathrm{L}_{2}\right)\left(\frac{\mathrm{C}_{1} \times \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right)}}$ $\because \quad \mathrm{L}_{1} \mathrm{C}_{1}=\frac{1}{\omega^{2}}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\omega_{\mathrm{eq}}=\frac{1}{\sqrt{\frac{1}{\omega^{2}}\left(\frac{\mathrm{C}_{2}+\mathrm{C}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right)}}$ $\omega_{\mathrm{eq}}=\omega$ $\omega_{\mathrm{eq}}=\omega$
TS EAMCET(Medical)-2017
Alternating Current
155290
In non-resonant circuit, what will be the nature of the circuit for frequencies higher than the resonant frequency?
1 Resistive
2 Capacitive
3 Inductive
4 None of these
Explanation:
C Let, $\omega_{\mathrm{R}}$ is the resonant frequency Then, $0 \lt \omega \lt \omega_{\mathrm{R}}$ At resonant frequency $X_{L}=X_{C}$ $\omega L=\frac{1}{\omega C}$ For frequencies higher than the resonant frequency- $\mathrm{X}_{\mathrm{L}}>\mathrm{X}_{\mathrm{C}}$ So, nature of circuit is inductive.
Manipal UGET-2012
Alternating Current
155291
The values of $L, C$ and $R$ for a circuit are $1 H$, $9 F$ and $3 \Omega$. What is the quality factor for the circuit at resonance ?
1 1
2 9
3 $\frac{1}{9}$
4 $\frac{1}{3}$
Explanation:
C Given that, $\mathrm{L}=1 \mathrm{H}, \mathrm{C}=9 \mathrm{~F}, \mathrm{R}=3 \Omega$ We know that, $\text { Quality factor }(\mathrm{Q}) =\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}} \quad\left(\because \omega=\frac{1}{\sqrt{\mathrm{LC}}}\right)$ $\mathrm{Q} =\frac{1}{3} \sqrt{\frac{1}{9}}$ $\mathrm{Q} =\frac{1}{9}$
