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(B) First,simplify the circuit. The two resistors on the left ($6 \, \Omega$ and $3 \, \Omega$) are in parallel,and the two resistors on the right ($3

Vedclass · Accepted Answer

$i_3 = \frac{5}{3} \, A$

Vedclass · Answer

(B) First,simplify the circuit. The two resistors on the left ($6 \, \Omega$ and $3 \, \Omega$) are in parallel,and the two resistors on the right ($3 \, \Omega$ and $6 \, \Omega$) are in parallel. These two parallel combinations are in series with each other.Equivalent resistance of the left part: $R_L = \frac{6 	imes 3}{6 + 3} = \frac{18}{9} = 2 \, \Omega$.Equivalent resistance of the right part: $R_R = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2 \, \Omega$.Total equivalent resistance: $R_{eq} = R_L + R_R = 2 + 2 = 4 \, \Omega$.Total current $i_4 = \frac{V}{R_{eq}} = \frac{20}{4} = 5 \, A$.Using the current divider rule for the left side:$i_1 = i_4 	imes \frac{3}{6 + 3} = 5 	imes \frac{3}{9} = \frac{5}{3} \, A$.$i_2 = i_4 	imes \frac{6}{6 + 3} = 5 	imes \frac{6}{9} = \frac{10}{3} \, A$.For the right side,the current $i_4$ splits into $i_3$ and another branch current. By symmetry,$i_3 = i_4 	imes \frac{6}{3 + 6} = 5 	imes \frac{6}{9} = \frac{10}{3} \, A$.Comparing with the options,statement $B$ $(i_3 = \frac{5}{3} \, A)$ is incorrect.

The diagram shows a circuit. Choose the incorrect statement.

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