Answer
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Hint: The output of the circuit is maximum at one particular frequency, that frequency is known as the resonant frequency of that electrical circuit. At resonant frequency, the capacitive reactance and inductive reactance are equal.
Formula used:
The resonant frequency of an L-C-R circuit is given by $\omega = \dfrac{1}{{\sqrt {LC} }}$, where L= inductance of the circuit, and C=capacitance of the circuit
Complete step by step solution:
First, let’s assume initial resonant frequency, inductance, and capacitance as ${\omega _o},{L_o},{C_o}$respectively.
Now after the operation resonant frequency, inductance, and capacitance changes to${\omega _{req}},{L_{req}},{C_{req}}$.
We know that
${L_{req}} = 2{L_o}$and${C_{req}} = 2{C_o}$
We also know that
${\omega _{req}} = \dfrac{1}{{\sqrt {{L_{req}}{C_{req}}} }}$
So, now by substituting these values in the formula of the required
resonant frequency(${\omega _{req}}$). We will get
${\omega _{req}} = \dfrac{1}{{\sqrt {\left( {2{L_o}} \right)\left( {2{C_o}} \right)} }}$
$ \Rightarrow {\omega _{req}} = \dfrac{1}{{2\sqrt {{L_o}{C_o}} }}$
$ \Rightarrow {\omega _{req}} = \dfrac{1}{2}{\omega _o}$ as ${\omega _o} = \dfrac{1}{{\sqrt {{L_o}{C_o}} }}$.
Hence we can say that resonant frequency is reduced to half of the initial value.
Additional information: In a parallel L-C-R circuit, the potential difference across the capacitor and the inductor are the same in magnitude but it is opposite in direction at the resonant frequency. As no power is consumed by the conductor and the capacitor and potential difference across the resistance is maximum, it results in the maximum power generated by the circuit.
Hence, the correct option is (A), Decrease to one-half of the original value.
Note:
1. The Resonant frequency of the circuit is independent of the resistance of the electrical circuit.
2. At the resonant frequency amplitude of the current is maximum in the electrical circuit.
3. Resonant frequency of the electrical circuit is inversely proportional to the square root of inductance and capacitance.
Formula used:
The resonant frequency of an L-C-R circuit is given by $\omega = \dfrac{1}{{\sqrt {LC} }}$, where L= inductance of the circuit, and C=capacitance of the circuit
Complete step by step solution:
First, let’s assume initial resonant frequency, inductance, and capacitance as ${\omega _o},{L_o},{C_o}$respectively.
Now after the operation resonant frequency, inductance, and capacitance changes to${\omega _{req}},{L_{req}},{C_{req}}$.
We know that
${L_{req}} = 2{L_o}$and${C_{req}} = 2{C_o}$
We also know that
${\omega _{req}} = \dfrac{1}{{\sqrt {{L_{req}}{C_{req}}} }}$
So, now by substituting these values in the formula of the required
resonant frequency(${\omega _{req}}$). We will get
${\omega _{req}} = \dfrac{1}{{\sqrt {\left( {2{L_o}} \right)\left( {2{C_o}} \right)} }}$
$ \Rightarrow {\omega _{req}} = \dfrac{1}{{2\sqrt {{L_o}{C_o}} }}$
$ \Rightarrow {\omega _{req}} = \dfrac{1}{2}{\omega _o}$ as ${\omega _o} = \dfrac{1}{{\sqrt {{L_o}{C_o}} }}$.
Hence we can say that resonant frequency is reduced to half of the initial value.
Additional information: In a parallel L-C-R circuit, the potential difference across the capacitor and the inductor are the same in magnitude but it is opposite in direction at the resonant frequency. As no power is consumed by the conductor and the capacitor and potential difference across the resistance is maximum, it results in the maximum power generated by the circuit.
Hence, the correct option is (A), Decrease to one-half of the original value.
Note:
1. The Resonant frequency of the circuit is independent of the resistance of the electrical circuit.
2. At the resonant frequency amplitude of the current is maximum in the electrical circuit.
3. Resonant frequency of the electrical circuit is inversely proportional to the square root of inductance and capacitance.
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