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0 votes
78.3k views
in Physics by (36.3k points)

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

2 Answers

+1 vote
by (33.0k points)
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Best answer

Let the resistance to be connected is R.

For balanced wheatstone bridge,

15 × 4 = 12 × (10R/(10 + R))

R = 10 Ω

+1 vote
by (51.9k points)

(10R/(10 + R)) x 12 = 15 x 4

on solving

R = 10Ω

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