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In a Wheatstone bridge circuit, `P=5 Omega, Q=6 Omega, R=10 Omega` and `S= 5 Omega`. Find the additional resistance to be used in series with S, so that the bridge is balanced. |
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Answer» Let the bridge be balanced when additional resitance x is put in series with S. Then, `(S +x)= Q/P R` or `x=Q/P R - S = 6/5 xx 10 - 5= 7 Omega`. |
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