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(B) Let the total current entering at point $P$ be $I = 2.1 	ext{ A}$.Let $i$ be the current flowing through the upper branch ($PQ$ and $QR$).Then,th

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$1.5$

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(B) Let the total current entering at point $P$ be $I = 2.1 	ext{ A}$.Let $i$ be the current flowing through the upper branch ($PQ$ and $QR$).Then,the current flowing through the lower branch ($PS$ and $SR$) is $(I - i) = (2.1 - i) 	ext{ A}$.Since the potential at $Q$ and $S$ is the same if the bridge were balanced,but here it is not,we use Kirchhoff's laws or potential division.The potential difference between $P$ and $R$ is the same along both paths:$V_{PR} = i(R_{PQ} + R_{QR}) = (I - i)(R_{PS} + R_{SR})$$V_{PR} = i(5 + 1) = (2.1 - i)(12.5 + 2.5)$$6i = (2.1 - i)(15)$$6i = 31.5 - 15i$$21i = 31.5$$i = \frac{31.5}{21} = 1.5 	ext{ A}$.Thus,the current through the $1 \Omega$ resistor is $1.5 	ext{ A}$.

The current through the $1 \Omega$ resistance in the following circuit is (in $A$)

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