The current through the 1 Omega resistance in | vedclass

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(B) Let the total current $I = 2.1 	ext{ A}$ enter at point $P$ and leave at point $R$. Let the current through the branch $PQ$ be $i$. Then the curr

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$1.5$

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(B) Let the total current $I = 2.1 	ext{ A}$ enter at point $P$ and leave at point $R$. Let the current through the branch $PQ$ be $i$. Then the current through the branch $QR$ is also $i$ (assuming the galvanometer branch $QS$ has zero current,which is true if the bridge is balanced,but here we must check the potentials). Applying Kirchhoff's Voltage Law $(KVL)$ to the loops or using the potential divider principle:The potential difference between $P$ and $R$ is the same for both parallel paths $PQR$ and $PSR$.For path $PQR$,the total resistance is $R_1 = 5 \Omega + 1 \Omega = 6 \Omega$.For path $PSR$,the total resistance is $R_2 = 12.5 \Omega + 2.5 \Omega = 15 \Omega$.Let $i_1$ be the current through path $PQR$ and $i_2$ be the current through path $PSR$.Since $i_1 + i_2 = 2.1 	ext{ A}$ and $i_1 R_1 = i_2 R_2$:$i_1 (6) = i_2 (15) \Rightarrow i_2 = \frac{6}{15} i_1 = 0.4 i_1$.Substituting into the total current equation:$i_1 + 0.4 i_1 = 2.1 	ext{ A} \Rightarrow 1.4 i_1 = 2.1 	ext{ A} \Rightarrow i_1 = \frac{2.1}{1.4} = 1.5 	ext{ A}$.However,looking at the provided diagram,the current $i$ is flowing through the $1 \Omega$ resistor. The current $i_1$ calculated above is the current through the entire branch $PQR$,which includes the $1 \Omega$ resistor. 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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