In Wheatstone bridge, four resistors of resistances 15 Ω , 12 Ω , 4 Ω and 10 Ω respectively, are connected in cyclic order. The resistance (in Ω ) that is to be connected in parallel with the resistan

Question

In Wheatstone bridge, four resistors of resistances $15 Ω$ , $12 Ω$ , $4 Ω$ and $10 Ω$ respectively, are connected in cyclic order. The resistance (in $Ω$ ) that is to be connected in parallel with the resistance of $10 Ω$ to balance the network is

Answer:

10.00

Solution:

$\frac{10 R}{10 + R} \times 12 = 15 \times 4$

on solving $R = 10 Ω$

TestHub · Accepted Answer

10 R 10 + R × 12 = 15 × 4 on solving R = 10 Ω

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