JEE MAIN - Physics (2020 - 8th January Morning Slot - No. 12)
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
10
Explanation
_8th_January_Morning_Slot_en_12_1.png)
Wheatstone bridge balance condition
$${{{R_1}} \over {{R_2}}} = {{{R_3}} \over {{R_4}}}$$
$$ \Rightarrow $$ $${{15} \over 4} = {{15} \over {{{10R} \over {10 + R}}}}$$
$$ \Rightarrow $$ 2R = 10 + R
$$ \Rightarrow $$ R = 10 $$\Omega $$
Comments (0)
