Question

Four resistances $5\Omega ,10\Omega ,15\Omega$ and $20\Omega$ are connected in a cyclic order so as to form an unbalanced Wheatstone's bridge. The bridge can be balanced by
(i) connecting $25\Omega$ in series with $15\Omega$
(ii) connecting $12\Omega$ in parallel with $20\Omega$

• A. Only (i) is correct
• B. Only (ii) is correct
• C. Both (i) and (ii) are correct
• D. Both (i) and (ii) are incorrect

Answer 1

Answer: (C)

Given situation is shown in the figure.
(i) Connecting $25\Omega$ in series with $15\Omega$
Net resistance in the arm $=25+15=40\Omega$
$\frac{5}{10}=\frac{20}{40}$ which is the required condition for balanced Wheat stone bridge.
(ii) Connecting $12\Omega$ in parallel with $20\Omega$
Net resistance ,$=\frac{20\times 12}{20+12}=\frac{15}{2}\Omega$
Hence, $\frac{5}{10}=\frac{7.5}{15}$ which is the required condition for balanced Wheat-stone bridge.