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Question

How can three resistors of resistance 2Ω, 3Ω and 6Ω be connected to give a total resistance of 4Ω and 1Ω?


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Solution

Step 1: Given

R1=2ΩR2=3ΩR3=6Ω

Step 2: Combining resistors

Considering the second and third resistors connected in parallel, the effective resistance becomes

1R=1R2+1R3=13+16=12R=2Ω

Now connecting Rand R1in series total resistance becomes

R+R1=4Ω

Step 3: Equivalent Resistance

Now considering all three resistance in parallel, the total resistance becomes

1R=12+13+16=1

R=1Ω

Thus, By connecting the 2Ω resistances in series with the parallel combination of 3Ω and 6Ω, we can obtain a total resistance of 4Ω and By connecting resistors 2Ω, 3Ω, and 6Ω in parallel, we can produce a total resistance of 1Ω.


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