Solveeit Logo

Question

Question: In an \(LCR\) circuit, capacitance is changed from \(C\) to \(2C\). For the resonant frequency to re...

In an LCRLCR circuit, capacitance is changed from CC to 2C2C. For the resonant frequency to remain unchanged, the inductance should be changed from LL to :
A. 4L4L
B. L2\dfrac{L}{2}
C. 2L2L
D. L4\dfrac{L}{4}

Explanation

Solution

a LCR circuit is an electronic circuit which is a combination of capacitor, inductor and resistor which are either connected in series or in parallel. For the resonance condition inductive reactance becomes equal to capacitive reactance.

Formula used:
XL=XCX_L = X_C
Where XLX_L= inductive reactance and XCX_C=capacitive reactance.

Complete step by step answer:
The resonance of a series LCR circuit occurs when the inductive and capacitive reactance are equal in magnitude but cancel each other because they are 180 degrees apart in phase. In resonance condition a circuit acts as a pure resistive circuit and has maximum current in it.
In LCR circuit, the impedance is given by:
Z2=R2+(XLXC)2{Z^2} = {R^2} + {\left( {X_L - X_C} \right)^2}
where RR is resistance, XLX_L=​ inductive reactance and XCX_C=capacitive reactance.

At Resonance in LCR circuit,
XL=XCX_L = X_C
ωL=1ωC\Rightarrow \omega L = \dfrac{1}{{\omega C}}
ω2=1LC\Rightarrow {\omega ^2} = \dfrac{1}{{LC}}
ω=1LC\Rightarrow \omega = \dfrac{1}{{\sqrt {LC} }}
In the resonance condition the frequency remains unchanged. Hence LC\sqrt {LC} =constant. It means that even LCLC is constant. Hence we can write that :
L1C1=L2C2{L_1}{C_1} = {L_2}{C_2}
Where L2L_2 is inductance initially and L2L_2 is final inductance.
We know that here C1=C{C_1} = C and C2=2C{C_2} = 2C let us now substitute these values, we get:
L1C=L22C{L_1}C = {L_2}2C
L2=L1C2C\Rightarrow {L_2} = \dfrac{{{L_1}C}}{{2C}}
L2=L12\therefore {L_2} = \dfrac{{{L_1}}}{2}

Hence the correct answer is option B.

Note: depending on the values of the reactance the behavior of the circuit changes, like: If XL>XCX_L > X_C the circuit is inductive in nature, if XL<XCX_L < X_C, the circuit behaves like a capacitive circuit and if XL=XCX_L = X_C. Also note that XL=ωLX_L = \omega L and not XL=1ωLX_L = \dfrac{1}{{\omega L}} whereas XC=1ωCX_C = \dfrac{1}{{\omega C}} and not XC=ωCX_C = \omega C.