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Question: In a \(R - L - C\) series circuit, the potential difference across each element is \(20{\text{V}}\)....

In a R−L−CR - L - C series circuit, the potential difference across each element is 20V20{\text{V}}. Now the value of resistance alone is doubled, then P.D. across RR, LL and CC is respectively
(A) 20V, 10V, 10V20{\text{V, 10V, 10V}}
(B) 20V, 20V, 20V20{\text{V, 20V, 20V}}
(C) 20V, 40V, 40V20{\text{V, 40V, 40V}}
(D) 10V, 20V, 20V10{\text{V, 20V, 20V}}

Explanation

Solution

To solve this question, we need to obtain the emf and the frequency of the source. Then putting them in the formulae of the voltages across the three elements, we will get the required values.

Complete step-by-step solution
Let VR′{V_R}', VR′{V_R}', and VR′{V_R}' be the respective required voltages on the resistor, inductor and the capacitor.
Let ii be the current in the circuit.
We know that the net emf of the source in a series R−L−CR - L - Ccircuit is given by
V=VR2+(VL−VC)2V = \sqrt {{V_R}^2 + {{\left( {{V_L} - {V_C}} \right)}^2}}
According to the question, VR=VL=VC=20V{V_R} = {V_L} = {V_C} = 20{\text{V}}
So, V=202+(20−20)2V = \sqrt {{{20}^2} + {{\left( {20 - 20} \right)}^2}}
V=20VV = 20{\text{V}}
Therefore, the source emf is of 20V20{\text{V}}
Now, as VL=VC{V_L} = {V_C}
Writing the voltages in terms of the impedances, we have
iXL=iXCi{X_L} = i{X_C}
Cancelling ii from both the sides
XL=XC{X_L} = {X_C}
As we know, XL=ωL{X_L} = \omega L and XC=1ωC{X_C} = \dfrac{1}{{\omega C}}
So, ωL=1ωC\omega L = \dfrac{1}{{\omega C}}
Or ω2=1LC{\omega ^2} = \dfrac{1}{{LC}}
Taking square root, we get
ω=1LC\omega = \dfrac{1}{{\sqrt {LC} }}
Therefore, the source has a frequency equal to the resonant frequency.
We know that in the resonance condition, the entire source voltage appears on the resistance.
So, the voltage on the resistor is always equal to the net emf of the source and is independent of the value of the resistance.
Therefore, doubling the value of the resistance does not change its voltage.
Hence, VR′=20V{V_R}' = 20V
But, according to the ohm’s law, we have
VR′=i′R′{V_R}' = i'R'
As VR′=VR{V_R}' = {V_R}, and R′=2RR' = 2R, we have
VR=i′(2R){V_R} = i'(2R)
Substituting VR=iR{V_R} = iR
iR=2i′RiR = 2i'R
i′=i2i' = \dfrac{i}{2}
So, the current is reduced to half.
Now, we have
VL′=i′XL{V_L}' = i'{X_L}
⇒VL′=i2XL\Rightarrow {V_L}' = \dfrac{i}{2}{X_L}
Substituting VL=iXL{V_L} = i{X_L}, we get
⇒VL′=VL2\Rightarrow {V_L}' = \dfrac{{{V_L}}}{2}
⇒VL′=202=10V\Rightarrow {V_L}' = \dfrac{{20}}{2} = 10{\text{V}}
For resonance, VC′=VL′=10V{V_C}' = {V_L}' = 10{\text{V}}
Thus, the P.D. across RR, LL and CC are respectively 20V, 10V, 10V20{\text{V, 10V, 10V}}

Hence, the correct answer is option A.

Note: Do not try to obtain the value of net source emf by the algebraic addition of the voltages. Always remember that the voltages in a series R−L−CR - L - C circuit are actually phasors which are treated as vectors, so the net emf is obtained as a vector addition of the three voltages given.