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

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Last updated date: 16th Jun 2024
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Answer
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Hint: 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 ${V_R}'$, ${V_R}'$, and ${V_R}'$ be the respective required voltages on the resistor, inductor and the capacitor.
Let $i$ be the current in the circuit.
We know that the net emf of the source in a series $R - L - C$circuit is given by
$V = \sqrt {{V_R}^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} $
According to the question, ${V_R} = {V_L} = {V_C} = 20{\text{V}}$
So, $V = \sqrt {{{20}^2} + {{\left( {20 - 20} \right)}^2}} $
$V = 20{\text{V}}$
Therefore, the source emf is of $20{\text{V}}$
Now, as ${V_L} = {V_C}$
Writing the voltages in terms of the impedances, we have
$i{X_L} = i{X_C}$
Cancelling $i$ from both the sides
${X_L} = {X_C}$
As we know, ${X_L} = \omega L$ and ${X_C} = \dfrac{1}{{\omega C}}$
So, $\omega L = \dfrac{1}{{\omega C}}$
Or ${\omega ^2} = \dfrac{1}{{LC}}$
Taking square root, we get
$\omega = \dfrac{1}{{\sqrt {LC} }}$
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, ${V_R}' = 20V$
But, according to the ohm’s law, we have
${V_R}' = i'R'$
As ${V_R}' = {V_R}$, and $R' = 2R$, we have
${V_R} = i'(2R)$
Substituting ${V_R} = iR$
$iR = 2i'R$
$i' = \dfrac{i}{2}$
So, the current is reduced to half.
Now, we have
${V_L}' = i'{X_L}$
$\Rightarrow {V_L}' = \dfrac{i}{2}{X_L}$
Substituting ${V_L} = i{X_L}$, we get
$\Rightarrow {V_L}' = \dfrac{{{V_L}}}{2}$
$\Rightarrow {V_L}' = \dfrac{{20}}{2} = 10{\text{V}}$
For resonance, \[{V_C}' = {V_L}' = 10{\text{V}}\]
Thus, the P.D. across $R$, $L$ and $C$ are respectively $20{\text{V, 10V, 10V}}$

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 $R - L - C$ circuit are actually phasors which are treated as vectors, so the net emf is obtained as a vector addition of the three voltages given.