Question

Four resistors A, B, C and D form a Wheatstone bridges. 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. $100\Omega$
• C. $110\Omega$
• D. $120\Omega$

Answer 1

The correct option is C $110\Omega$

For balanced Wheatstone bridge, $A/B=D/C=D/100$

When A and B are interchanged, $B/A=D/C^{\prime }=D/121$

Thus, $D/100=121/D$

or $D^2=100\times 121$ or $D=110\Omega$