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The four arms of a Wheatstone bridge have the resistances R₁ = 100 Ω, R₂ = 10 Ω, R₃ = 500 Ω and R₄. What will be the value of R₄ if the Wheatstone bridge is balanced?

Asked in GSEB Board August 2020 · Wheatstone bridge

• (1) 5 kΩ
• (2) 2 Ω
• (3) 2 kΩ
• (4) 50 Ω

Step-by-step solution

Why the other options are wrong

• (1) Pairs the arms the wrong way: R₄ = (R₁R₃)/(R₂) = (100×500)/(10) = 5000 Ω. At balance the products of opposite arms are equal, R₁R₄ = R₂R₃.
• (2) Comes from R₄ = (R₁R₂)/(R₃) = (100×10)/(500) = 2 Ω, which swaps the roles of R₃ and R₄. Balance needs (R₁)/(R₂) = (R₃)/(R₄).
• (3) No single slip produces 2 kΩ. Sanity check: (R₁)/(R₂) = 10, so R₄ must be one tenth of R₃ = 500 Ω, which is 50 Ω.

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