
If the value of $C$ in a series RLC circuit is decreased, the resonant frequency
(A) is not affected
(B) increases
(C) is reduced to zero
(D) decreases
Answer
568.5k+ views
Hint: The solution of this question can be determined by using the formula of the resonant frequency in the series RLC circuit. The equation of the series RLC circuit shows the relation between the resonant frequency ${v_r}$, inductance $L$, and the capacitance $C$.
Formula used:
The resonant frequency in the series RLC circuit is,
${v_r} = \dfrac{1}{{2\pi \sqrt {LC} }}$
Where, ${v_r}$ is the resonant frequency in the circuit, $L$ is the inductance of the circuit and $C$ is the capacitance of the circuit.
Complete step by step answer:
Given that,
The value of $C$ in a series RLC circuit is decreased.
The resonant frequency in the series RLC circuit is,
${v_r} = \dfrac{1}{{2\pi \sqrt {LC} }}\,....................\left( 1 \right)$
Assume that the inductance of the circuit is to be constant, then the above equation is written as,
\[\Rightarrow {v_r} \propto \dfrac{1}{{\sqrt C }}\]
From the above equation, the resonant frequency of the series RLC circuit is inversely proportional to the square root of the capacitance.
By the above equation, the value of the capacitance $C$ in a series RLC circuit is decreased, then the resonant frequency in a series RLC circuit is increased.
Hence, option (B) is the exact and correct answer for the given condition in the question.
Note:
By equation (1), the resonant frequency in the RLC circuit is inversely proportional to the inductance in a series RLC circuit and also inversely proportional to the capacitance in a series RLC circuit. Resonant frequency increases as both the inductance and capacitance decreases.
Formula used:
The resonant frequency in the series RLC circuit is,
${v_r} = \dfrac{1}{{2\pi \sqrt {LC} }}$
Where, ${v_r}$ is the resonant frequency in the circuit, $L$ is the inductance of the circuit and $C$ is the capacitance of the circuit.
Complete step by step answer:
Given that,
The value of $C$ in a series RLC circuit is decreased.
The resonant frequency in the series RLC circuit is,
${v_r} = \dfrac{1}{{2\pi \sqrt {LC} }}\,....................\left( 1 \right)$
Assume that the inductance of the circuit is to be constant, then the above equation is written as,
\[\Rightarrow {v_r} \propto \dfrac{1}{{\sqrt C }}\]
From the above equation, the resonant frequency of the series RLC circuit is inversely proportional to the square root of the capacitance.
By the above equation, the value of the capacitance $C$ in a series RLC circuit is decreased, then the resonant frequency in a series RLC circuit is increased.
Hence, option (B) is the exact and correct answer for the given condition in the question.
Note:
By equation (1), the resonant frequency in the RLC circuit is inversely proportional to the inductance in a series RLC circuit and also inversely proportional to the capacitance in a series RLC circuit. Resonant frequency increases as both the inductance and capacitance decreases.
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