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Asked in GUJCET 2025 · Wheatstone bridge
Given: from the figure the upper arms are 15 Ω and 10 Ω; the lower left arm is r, and the lower right arm is r in parallel with r/n. The galvanometer joins the top and bottom nodes, the cell the left and right nodes.
Idea: at balance no current crosses the galvanometer, so the two arms on the left of it are in the same ratio as the two arms on the right.
Lower right arm: 1/S=1/r+n/r=(1+n)/r, so S=r/(n+1).
Balance: (15)/(10)=r/S=n+1.
n+1=3/2, so n=1/2.
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