356218
Assertion : In series \(L C R\) circuit resonance can take place. Reason : Resonance takes place if inductance and capacitive reactances are equal and opposite.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Resonance occurs when the inductive reactance \(\left(X_{L}\right)\) and capacitive reactance \(\left(X_{C}\right)\) are equal in magnitude and opposite in phase, resulting in the impedance being purely resistive. At this point, the circuit allows for maximum current flow and is in a state of resonance. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356219
For the series \(LCR\) circuit shown in the figure, the resonance frequency is
356220
To reduce the resonant frequency in an \(L-C-R\) series circuit with a generator
1 The iron core of the inductor should be removed
2 Dielectric in the capacitor should be removed
3 The generator frequency should be reduced
4 Another capacitor should be added in parallel to the first
Explanation:
Reasonant frequency in an \(L - C - R\) series circuit is \({\upsilon _r} = \frac{1}{{2\pi \sqrt {LC} }}\) To reduce \({\upsilon _r};C\) can be increased, by adding another capacitor in parallel to the first capacitor
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356221
Which of the following combinations should be selected for better tuning of an \(L-C-R\) circuit used for communication ?
1 \(R = 25\,\Omega ,L = 2.5\,H,C = 45\,\mu F\)
2 \(R = 20\Omega ,L = 1.5\,H,C = 35\,\mu F\)
3 \(R = 25\Omega ,L = 1.5\,H,C = 45\,\mu F\)
4 \(R = 15\Omega ,L = 3.5\,H,C = 30\,\mu F\)
Explanation:
For better tuning the resonance curve should be sharp. When \(Q\)-factor is large, the sharpness of resonance curve is more and vice-verse. \(Q = \frac{1}{R}\sqrt {\frac{L}{C}} \) For large value of \(Q\), \(R\) should be small, \(L\) should be high and \(C\) should be low. These conditions are best satisfied by the values in option (4)
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356222
An \(L C R\) circuit contains \(R = 50\,\Omega ,L = 1mH\) and \(C = 0.1\mu F\). The impedance of the circuit will be minimum for a frequency of
1 \(\frac{{{{10}^5}}}{{2\pi }}Hz\)
2 \(\frac{{{{10}^6}}}{{2\pi }}Hz\)
3 \(2\pi \times {10^5}\;Hz\)
4 \(2\pi \times {10^6}\;Hz\)
Explanation:
The impedance of the \(L C R\) circuit will be minimum for a resonant frequency \(v_{r}\). \(v_{r}=\dfrac{1}{2 \pi \sqrt{L C}}\) Here, \(L = 1mH = 1 \times {10^{ - 3}}H\) \(C = 0.1\mu F = 0.1 \times {10^{ - 6}}\;F\) \(\therefore {v_r} = \frac{1}{{2\pi \sqrt {\left( {1 \times {{10}^{ - 3}}H} \right)\left( {0.1 \times {{10}^{ - 6}}F} \right)} }}\) \( = \frac{{{{10}^5}}}{{2\pi }}Hz\)
356218
Assertion : In series \(L C R\) circuit resonance can take place. Reason : Resonance takes place if inductance and capacitive reactances are equal and opposite.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Resonance occurs when the inductive reactance \(\left(X_{L}\right)\) and capacitive reactance \(\left(X_{C}\right)\) are equal in magnitude and opposite in phase, resulting in the impedance being purely resistive. At this point, the circuit allows for maximum current flow and is in a state of resonance. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356219
For the series \(LCR\) circuit shown in the figure, the resonance frequency is
356220
To reduce the resonant frequency in an \(L-C-R\) series circuit with a generator
1 The iron core of the inductor should be removed
2 Dielectric in the capacitor should be removed
3 The generator frequency should be reduced
4 Another capacitor should be added in parallel to the first
Explanation:
Reasonant frequency in an \(L - C - R\) series circuit is \({\upsilon _r} = \frac{1}{{2\pi \sqrt {LC} }}\) To reduce \({\upsilon _r};C\) can be increased, by adding another capacitor in parallel to the first capacitor
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356221
Which of the following combinations should be selected for better tuning of an \(L-C-R\) circuit used for communication ?
1 \(R = 25\,\Omega ,L = 2.5\,H,C = 45\,\mu F\)
2 \(R = 20\Omega ,L = 1.5\,H,C = 35\,\mu F\)
3 \(R = 25\Omega ,L = 1.5\,H,C = 45\,\mu F\)
4 \(R = 15\Omega ,L = 3.5\,H,C = 30\,\mu F\)
Explanation:
For better tuning the resonance curve should be sharp. When \(Q\)-factor is large, the sharpness of resonance curve is more and vice-verse. \(Q = \frac{1}{R}\sqrt {\frac{L}{C}} \) For large value of \(Q\), \(R\) should be small, \(L\) should be high and \(C\) should be low. These conditions are best satisfied by the values in option (4)
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356222
An \(L C R\) circuit contains \(R = 50\,\Omega ,L = 1mH\) and \(C = 0.1\mu F\). The impedance of the circuit will be minimum for a frequency of
1 \(\frac{{{{10}^5}}}{{2\pi }}Hz\)
2 \(\frac{{{{10}^6}}}{{2\pi }}Hz\)
3 \(2\pi \times {10^5}\;Hz\)
4 \(2\pi \times {10^6}\;Hz\)
Explanation:
The impedance of the \(L C R\) circuit will be minimum for a resonant frequency \(v_{r}\). \(v_{r}=\dfrac{1}{2 \pi \sqrt{L C}}\) Here, \(L = 1mH = 1 \times {10^{ - 3}}H\) \(C = 0.1\mu F = 0.1 \times {10^{ - 6}}\;F\) \(\therefore {v_r} = \frac{1}{{2\pi \sqrt {\left( {1 \times {{10}^{ - 3}}H} \right)\left( {0.1 \times {{10}^{ - 6}}F} \right)} }}\) \( = \frac{{{{10}^5}}}{{2\pi }}Hz\)
356218
Assertion : In series \(L C R\) circuit resonance can take place. Reason : Resonance takes place if inductance and capacitive reactances are equal and opposite.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Resonance occurs when the inductive reactance \(\left(X_{L}\right)\) and capacitive reactance \(\left(X_{C}\right)\) are equal in magnitude and opposite in phase, resulting in the impedance being purely resistive. At this point, the circuit allows for maximum current flow and is in a state of resonance. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356219
For the series \(LCR\) circuit shown in the figure, the resonance frequency is
356220
To reduce the resonant frequency in an \(L-C-R\) series circuit with a generator
1 The iron core of the inductor should be removed
2 Dielectric in the capacitor should be removed
3 The generator frequency should be reduced
4 Another capacitor should be added in parallel to the first
Explanation:
Reasonant frequency in an \(L - C - R\) series circuit is \({\upsilon _r} = \frac{1}{{2\pi \sqrt {LC} }}\) To reduce \({\upsilon _r};C\) can be increased, by adding another capacitor in parallel to the first capacitor
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356221
Which of the following combinations should be selected for better tuning of an \(L-C-R\) circuit used for communication ?
1 \(R = 25\,\Omega ,L = 2.5\,H,C = 45\,\mu F\)
2 \(R = 20\Omega ,L = 1.5\,H,C = 35\,\mu F\)
3 \(R = 25\Omega ,L = 1.5\,H,C = 45\,\mu F\)
4 \(R = 15\Omega ,L = 3.5\,H,C = 30\,\mu F\)
Explanation:
For better tuning the resonance curve should be sharp. When \(Q\)-factor is large, the sharpness of resonance curve is more and vice-verse. \(Q = \frac{1}{R}\sqrt {\frac{L}{C}} \) For large value of \(Q\), \(R\) should be small, \(L\) should be high and \(C\) should be low. These conditions are best satisfied by the values in option (4)
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356222
An \(L C R\) circuit contains \(R = 50\,\Omega ,L = 1mH\) and \(C = 0.1\mu F\). The impedance of the circuit will be minimum for a frequency of
1 \(\frac{{{{10}^5}}}{{2\pi }}Hz\)
2 \(\frac{{{{10}^6}}}{{2\pi }}Hz\)
3 \(2\pi \times {10^5}\;Hz\)
4 \(2\pi \times {10^6}\;Hz\)
Explanation:
The impedance of the \(L C R\) circuit will be minimum for a resonant frequency \(v_{r}\). \(v_{r}=\dfrac{1}{2 \pi \sqrt{L C}}\) Here, \(L = 1mH = 1 \times {10^{ - 3}}H\) \(C = 0.1\mu F = 0.1 \times {10^{ - 6}}\;F\) \(\therefore {v_r} = \frac{1}{{2\pi \sqrt {\left( {1 \times {{10}^{ - 3}}H} \right)\left( {0.1 \times {{10}^{ - 6}}F} \right)} }}\) \( = \frac{{{{10}^5}}}{{2\pi }}Hz\)
356218
Assertion : In series \(L C R\) circuit resonance can take place. Reason : Resonance takes place if inductance and capacitive reactances are equal and opposite.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Resonance occurs when the inductive reactance \(\left(X_{L}\right)\) and capacitive reactance \(\left(X_{C}\right)\) are equal in magnitude and opposite in phase, resulting in the impedance being purely resistive. At this point, the circuit allows for maximum current flow and is in a state of resonance. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356219
For the series \(LCR\) circuit shown in the figure, the resonance frequency is
356220
To reduce the resonant frequency in an \(L-C-R\) series circuit with a generator
1 The iron core of the inductor should be removed
2 Dielectric in the capacitor should be removed
3 The generator frequency should be reduced
4 Another capacitor should be added in parallel to the first
Explanation:
Reasonant frequency in an \(L - C - R\) series circuit is \({\upsilon _r} = \frac{1}{{2\pi \sqrt {LC} }}\) To reduce \({\upsilon _r};C\) can be increased, by adding another capacitor in parallel to the first capacitor
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356221
Which of the following combinations should be selected for better tuning of an \(L-C-R\) circuit used for communication ?
1 \(R = 25\,\Omega ,L = 2.5\,H,C = 45\,\mu F\)
2 \(R = 20\Omega ,L = 1.5\,H,C = 35\,\mu F\)
3 \(R = 25\Omega ,L = 1.5\,H,C = 45\,\mu F\)
4 \(R = 15\Omega ,L = 3.5\,H,C = 30\,\mu F\)
Explanation:
For better tuning the resonance curve should be sharp. When \(Q\)-factor is large, the sharpness of resonance curve is more and vice-verse. \(Q = \frac{1}{R}\sqrt {\frac{L}{C}} \) For large value of \(Q\), \(R\) should be small, \(L\) should be high and \(C\) should be low. These conditions are best satisfied by the values in option (4)
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356222
An \(L C R\) circuit contains \(R = 50\,\Omega ,L = 1mH\) and \(C = 0.1\mu F\). The impedance of the circuit will be minimum for a frequency of
1 \(\frac{{{{10}^5}}}{{2\pi }}Hz\)
2 \(\frac{{{{10}^6}}}{{2\pi }}Hz\)
3 \(2\pi \times {10^5}\;Hz\)
4 \(2\pi \times {10^6}\;Hz\)
Explanation:
The impedance of the \(L C R\) circuit will be minimum for a resonant frequency \(v_{r}\). \(v_{r}=\dfrac{1}{2 \pi \sqrt{L C}}\) Here, \(L = 1mH = 1 \times {10^{ - 3}}H\) \(C = 0.1\mu F = 0.1 \times {10^{ - 6}}\;F\) \(\therefore {v_r} = \frac{1}{{2\pi \sqrt {\left( {1 \times {{10}^{ - 3}}H} \right)\left( {0.1 \times {{10}^{ - 6}}F} \right)} }}\) \( = \frac{{{{10}^5}}}{{2\pi }}Hz\)
356218
Assertion : In series \(L C R\) circuit resonance can take place. Reason : Resonance takes place if inductance and capacitive reactances are equal and opposite.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Resonance occurs when the inductive reactance \(\left(X_{L}\right)\) and capacitive reactance \(\left(X_{C}\right)\) are equal in magnitude and opposite in phase, resulting in the impedance being purely resistive. At this point, the circuit allows for maximum current flow and is in a state of resonance. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356219
For the series \(LCR\) circuit shown in the figure, the resonance frequency is
356220
To reduce the resonant frequency in an \(L-C-R\) series circuit with a generator
1 The iron core of the inductor should be removed
2 Dielectric in the capacitor should be removed
3 The generator frequency should be reduced
4 Another capacitor should be added in parallel to the first
Explanation:
Reasonant frequency in an \(L - C - R\) series circuit is \({\upsilon _r} = \frac{1}{{2\pi \sqrt {LC} }}\) To reduce \({\upsilon _r};C\) can be increased, by adding another capacitor in parallel to the first capacitor
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356221
Which of the following combinations should be selected for better tuning of an \(L-C-R\) circuit used for communication ?
1 \(R = 25\,\Omega ,L = 2.5\,H,C = 45\,\mu F\)
2 \(R = 20\Omega ,L = 1.5\,H,C = 35\,\mu F\)
3 \(R = 25\Omega ,L = 1.5\,H,C = 45\,\mu F\)
4 \(R = 15\Omega ,L = 3.5\,H,C = 30\,\mu F\)
Explanation:
For better tuning the resonance curve should be sharp. When \(Q\)-factor is large, the sharpness of resonance curve is more and vice-verse. \(Q = \frac{1}{R}\sqrt {\frac{L}{C}} \) For large value of \(Q\), \(R\) should be small, \(L\) should be high and \(C\) should be low. These conditions are best satisfied by the values in option (4)
NCERT Exemplar
PHXII07:ALTERNATING CURRENT
356222
An \(L C R\) circuit contains \(R = 50\,\Omega ,L = 1mH\) and \(C = 0.1\mu F\). The impedance of the circuit will be minimum for a frequency of
1 \(\frac{{{{10}^5}}}{{2\pi }}Hz\)
2 \(\frac{{{{10}^6}}}{{2\pi }}Hz\)
3 \(2\pi \times {10^5}\;Hz\)
4 \(2\pi \times {10^6}\;Hz\)
Explanation:
The impedance of the \(L C R\) circuit will be minimum for a resonant frequency \(v_{r}\). \(v_{r}=\dfrac{1}{2 \pi \sqrt{L C}}\) Here, \(L = 1mH = 1 \times {10^{ - 3}}H\) \(C = 0.1\mu F = 0.1 \times {10^{ - 6}}\;F\) \(\therefore {v_r} = \frac{1}{{2\pi \sqrt {\left( {1 \times {{10}^{ - 3}}H} \right)\left( {0.1 \times {{10}^{ - 6}}F} \right)} }}\) \( = \frac{{{{10}^5}}}{{2\pi }}Hz\)
