Question

A, B,C ad D are four resistances of 4,4,4 and $8\Omega$ respectively. They are used to form a wheatstone bridge.The resistance D is short circuited with a resistance R in order to get the bridge balanced. The value of R will be

• A. $4\Omega$
• B. $6\Omega$
• C. $8\Omega$
• D. $3\Omega$

Answer 1

The correct option is C $8\Omega$

$\frac{A}{B}=\frac{C}{\left(\frac{DR}{D+R}\right)}\frac{4}{4}=\frac{4}{\left(\frac{8R}{8+R}\right)}2R=8+RR=8\Omega$