356114
A capacitor of capacitance \(100\,\mu F\) is charged to a potential of \(12\,V\) and connected to a \(6.4\,mH\) inductor to produce oscillations. The maximum current in the circuit would be
Total energy is conserved. max energy in \(C=\) max energy in \(L\) \(\Rightarrow \dfrac{1}{2} C V^{2}=\dfrac{1}{2} L i^{2}\) \(\Rightarrow i=\sqrt{\dfrac{C}{L}} V\) \( \Rightarrow i = \sqrt {\frac{{{{10}^{ - 4}}}}{{6.4 \times {{10}^{ - 3}}}}} \times 12 = \frac{{12}}{8} = 1.5\;A\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356115
In the given circuit, the resonant frequency is
356116
The frequency of the output signal becomes \(x\) times by doubling of the value of the capacitance in the \(LC\) oscillator circuit. The value of \(x\) is
356117
An \(L-C\) circuit is in the state of resonance. If \(C = 0.1\mu F\) and \(L = 0.25H\), neglecting ohmic resistance of circuit, then what is the frequency of oscillations?
1 \(1007\;Hz\)
2 \(100\;Hz\)
3 \(109\;Hz\)
4 \(500\;Hz\)
Explanation:
From the formula, the frequency of oscillation is \(f = \frac{1}{{2\pi \sqrt {LC} }} = \frac{1}{{2\pi \sqrt {0.25 \times 0.1 \times {{10}^{ - 6}}} }}\) \(\,\,\,\, \approx 1007\;Hz\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII07:ALTERNATING CURRENT
356114
A capacitor of capacitance \(100\,\mu F\) is charged to a potential of \(12\,V\) and connected to a \(6.4\,mH\) inductor to produce oscillations. The maximum current in the circuit would be
Total energy is conserved. max energy in \(C=\) max energy in \(L\) \(\Rightarrow \dfrac{1}{2} C V^{2}=\dfrac{1}{2} L i^{2}\) \(\Rightarrow i=\sqrt{\dfrac{C}{L}} V\) \( \Rightarrow i = \sqrt {\frac{{{{10}^{ - 4}}}}{{6.4 \times {{10}^{ - 3}}}}} \times 12 = \frac{{12}}{8} = 1.5\;A\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356115
In the given circuit, the resonant frequency is
356116
The frequency of the output signal becomes \(x\) times by doubling of the value of the capacitance in the \(LC\) oscillator circuit. The value of \(x\) is
356117
An \(L-C\) circuit is in the state of resonance. If \(C = 0.1\mu F\) and \(L = 0.25H\), neglecting ohmic resistance of circuit, then what is the frequency of oscillations?
1 \(1007\;Hz\)
2 \(100\;Hz\)
3 \(109\;Hz\)
4 \(500\;Hz\)
Explanation:
From the formula, the frequency of oscillation is \(f = \frac{1}{{2\pi \sqrt {LC} }} = \frac{1}{{2\pi \sqrt {0.25 \times 0.1 \times {{10}^{ - 6}}} }}\) \(\,\,\,\, \approx 1007\;Hz\)
356114
A capacitor of capacitance \(100\,\mu F\) is charged to a potential of \(12\,V\) and connected to a \(6.4\,mH\) inductor to produce oscillations. The maximum current in the circuit would be
Total energy is conserved. max energy in \(C=\) max energy in \(L\) \(\Rightarrow \dfrac{1}{2} C V^{2}=\dfrac{1}{2} L i^{2}\) \(\Rightarrow i=\sqrt{\dfrac{C}{L}} V\) \( \Rightarrow i = \sqrt {\frac{{{{10}^{ - 4}}}}{{6.4 \times {{10}^{ - 3}}}}} \times 12 = \frac{{12}}{8} = 1.5\;A\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356115
In the given circuit, the resonant frequency is
356116
The frequency of the output signal becomes \(x\) times by doubling of the value of the capacitance in the \(LC\) oscillator circuit. The value of \(x\) is
356117
An \(L-C\) circuit is in the state of resonance. If \(C = 0.1\mu F\) and \(L = 0.25H\), neglecting ohmic resistance of circuit, then what is the frequency of oscillations?
1 \(1007\;Hz\)
2 \(100\;Hz\)
3 \(109\;Hz\)
4 \(500\;Hz\)
Explanation:
From the formula, the frequency of oscillation is \(f = \frac{1}{{2\pi \sqrt {LC} }} = \frac{1}{{2\pi \sqrt {0.25 \times 0.1 \times {{10}^{ - 6}}} }}\) \(\,\,\,\, \approx 1007\;Hz\)
356114
A capacitor of capacitance \(100\,\mu F\) is charged to a potential of \(12\,V\) and connected to a \(6.4\,mH\) inductor to produce oscillations. The maximum current in the circuit would be
Total energy is conserved. max energy in \(C=\) max energy in \(L\) \(\Rightarrow \dfrac{1}{2} C V^{2}=\dfrac{1}{2} L i^{2}\) \(\Rightarrow i=\sqrt{\dfrac{C}{L}} V\) \( \Rightarrow i = \sqrt {\frac{{{{10}^{ - 4}}}}{{6.4 \times {{10}^{ - 3}}}}} \times 12 = \frac{{12}}{8} = 1.5\;A\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356115
In the given circuit, the resonant frequency is
356116
The frequency of the output signal becomes \(x\) times by doubling of the value of the capacitance in the \(LC\) oscillator circuit. The value of \(x\) is
356117
An \(L-C\) circuit is in the state of resonance. If \(C = 0.1\mu F\) and \(L = 0.25H\), neglecting ohmic resistance of circuit, then what is the frequency of oscillations?
1 \(1007\;Hz\)
2 \(100\;Hz\)
3 \(109\;Hz\)
4 \(500\;Hz\)
Explanation:
From the formula, the frequency of oscillation is \(f = \frac{1}{{2\pi \sqrt {LC} }} = \frac{1}{{2\pi \sqrt {0.25 \times 0.1 \times {{10}^{ - 6}}} }}\) \(\,\,\,\, \approx 1007\;Hz\)
