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Q.

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

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Detailed Solution

Balancing Wheatstone bridge, ​1510R10+R = 124⇒10R10+R×12=15×4   ​⇒120R=600+60R⇒60R=600​⇒ R= 10Ω
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