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Question

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


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Solution

Step 1. Given data,

Four resistance, 15Ω,12Ω,4Ω, and10Ω,

From the diagram,

Let, Resistance connected on AB be, RAB=15Ω

Resistance connected on BC be, RBC=12Ω

Resistance connected on CD be, RCD=4Ω

Resistance connected on DA be, RDA=Req

Step 2. Finding the resistance, R that is to be connected with 10Ω resistance,

To calculate the Resistances connected on DA we have to use the formula of equivalent resistance connected in parallel,

By using the formula of equivalent resistance connected in parallel, Req

1Req=1R+110

1Req=10+R10R

Req=10R10+R

By balancing the Wheatstone bridge,

RABRDA=RBCRCD

15Req=124

1510R10+R=124

1510+R10R=124

120R=600+60R

120R-60R=600

60R=600

R=10Ω

Therefore, the resistance of 10Ω is to be connected to balance the network.


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