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Resonance in RLC Circuit

In the circui...

Question

In the circuit shown in fig. A and B are two cells of the same emf E and of internal resistances $r_{A}$ and $r_{B}$ respectively. L is an ideal inductor and C is an ideal capacitor. The key K is closed. When the current in the circuit becomes steady, what should be the value of R so that the potential difference across the terminals of cell A is zero.

A

R = r_{A} - r_{B}

if

r_{A}

>

r_{B}

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B

R = \sqrt{r_{A} r_{B}}

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C

R = \frac{1}{2} (r_{A} r_{B})

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D

For no value of R will the potential difference between the
terminals of cell A be equal to zero

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Solution

The correct option is A $R = r_{A} - r_{B}$ if $r_{A}$ > $r_{B}$
At steady state the capacitor behaves like infinite resistance and inductor behaves like short. Between x and y we have a balanced wheat stone network of resistance R. The circuit at steady state is

$i = \frac{2 E}{R + r_{A} + r_{B}}$
$V_{A} = 0$
But
$V_{A} = E - i r_{A}$
$\Rightarrow E = i r_{A}$
$E = \frac{2 E r_{A}}{R + r_{A} + r_{B}}$
$R + r_{A} + r_{B} = 2 r_{A}$
$R = r_{A} - r_{B} (r_{A}$ > $r_{B})$

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