Q131 / 160 · AP EAMCET 2012 · Physics

Four resistors $A, B, C$ and $D$ form a Wheatstones bridge. The bridge is balanced, when $C=100 \Omega$. If $A$ and $B$ are interchanged, the bridge balances for $C=121 \Omega$. The value of $D$ is

  • A
    $10 \Omega$
  • B
    $100 \Omega$
  • Correct answer
    C
    $110 \Omega$
  • D
    $120 \Omega$

Solution

Balancing condition of bridge balance $\begin{aligned} \frac{A}{B} & =\frac{100}{x} \\ \frac{B}{A} & =\frac{121}{x} \\ \therefore \quad \frac{100}{x} & =\frac{x}{121} \\ x^2 & =12100 \\ x & =110 \Omega\end{aligned}$

Hence, the correct answer is option: 'C'.

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