A Wheatstone bridge has the resistances 10 Om | vedclass

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(D) In a balanced Wheatstone bridge,the ratio of resistances in adjacent arms must be equal,i.e.,$\frac{P}{Q} = \frac{R}{S}$.Let the four arms be $P =

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$15$

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(D) In a balanced Wheatstone bridge,the ratio of resistances in adjacent arms must be equal,i.e.,$\frac{P}{Q} = \frac{R}{S}$.Let the four arms be $P = 10 \Omega$,$Q = 10 \Omega$,$R = 10 \Omega$,and $S = 30 \Omega$.To balance the bridge,the effective resistance $S'$ in the fourth arm must satisfy $\frac{10}{10} = \frac{10}{S'}$,which gives $S' = 10 \Omega$.Let a resistance $x$ be connected in parallel to the $30 \Omega$ resistor to make the equivalent resistance $10 \Omega$.The formula for parallel resistance is $\frac{1}{S'} = \frac{1}{30} + \frac{1}{x}$.Substituting the values: $\frac{1}{10} = \frac{1}{30} + \frac{1}{x}$.$\frac{1}{x} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}$.Therefore,$x = 15 \Omega$.

$A$ Wheatstone bridge has the resistances $10 \Omega, 10 \Omega, 10 \Omega$ and $30 \Omega$ in its four arms. What resistance joined in parallel to the $30 \Omega$ resistance will bring it to the balanced condition (in $Omega$)?

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