Question

Four resistances of $$15\Omega, \, 12\Omega, \, 4\Omega$$ and $$10 \Omega$$ respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of $$10\Omega$$ to balance the network is _______ $$\Omega$$.

Answer 1

Correct option is A. 10
The given work is: [Ref. image]

For a balanced wheatstonels bridge, we have the relation

$$\dfrac{15}{10 || R} = \dfrac{12}{4}$$,

where $$10||R$$ represents the equivalent Resistance of the parallel combination

$$10||R = \dfrac{10R}{10+R}$$

$$\therefore \dfrac{15}{10||R} = \dfrac{15(10 + R)}{10R} = \dfrac{12}{4} = 3$$

$$15(10 + R) = 30R, 150 + 15R = 30R$$.

$$150 = 15R, R = 10\Omega$$