In the given Wheatstone Bridge no current flows through \( 15\,\Omega \) resistance. The value of the unknown resistance 'R' is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 16\,\Omega \)
No current through the middle \( 15\,\Omega \) resistor means the Wheatstone bridge is balanced: the ratio of the two resistances in the upper arms equals the ratio in the lower arms.
From the figure:
\( R_2 = 4\,\Omega \text{ (A to top)} \)
\( R_1 = 2\,\Omega \text{ (A to bottom)} \)
\( R_3 = 4\,\Omega \text{ (bottom to B)} \)
The arm from the top to B is \( 16\,\Omega \) in parallel with \( R \); call it \( R_4 \).
\( \displaystyle \frac{R_2}{R_4} = \frac{R_1}{R_3} \;\Rightarrow\; \frac{4}{R_4} = \frac{2}{4} \;\Rightarrow\; R_4 = 8\,\Omega \)
For the parallel pair:
\( \displaystyle \frac{16\,R}{16 + R} = 8 \)
\( 16R = 128 + 8R \)
\( 8R = 128 \)
\( R = 16\,\Omega \)