Question

Four resistances of 3$15.18\Omega$, 3$81.15\Omega$, 3$51.18\Omega$ and 4$18.15\Omega$ respectively are used to form a Wheatstone bridge. The 4$r_{1}$ resistance is short circuited with a resistance R in order to get bridge balanced. The value of R will be

• A. 10 $r_{1}$
• B. 11$r_{2}$
• C. 12 $r_{1}$
• D. 13$r_{2}$

Answer 1

Answer: (C)

12 $r_{1}$

Answer 2

: The bridge will be balanced when the shunted resistance is of the value of $3\Omega$

$\therefore 3 = \frac{4 \times R}{4 + R}$

$12 + 3R = 4R$

$\therefore R = 12$