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(D) $1$. First,simplify the parallel combinations: The two $10 \ \Omega$ resistors in parallel give $R_1 = (10 	imes 10) / (10 + 10) = 5 \ \Omega$. T

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$3.75$

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(D) $1$. First,simplify the parallel combinations: The two $10 \ \Omega$ resistors in parallel give $R_1 = (10 	imes 10) / (10 + 10) = 5 \ \Omega$. The two $6 \ \Omega$ resistors in parallel give $R_2 = (6 	imes 6) / (6 + 6) = 3 \ \Omega$.$2$. The circuit now simplifies to a Wheatstone bridge structure. The arms are $5 \ \Omega$ and $5 \ \Omega$ on the left,and $3 \ \Omega$ and $3 \ \Omega$ on the right,with an $8 \ \Omega$ resistor in the middle.$3$. Check for the balanced Wheatstone bridge condition: $P/Q = 5/5 = 1$ and $R/S = 3/3 = 1$. Since $P/Q = R/S$,the bridge is balanced.$4$. In a balanced Wheatstone bridge,no current flows through the central $8 \ \Omega$ resistor. Thus,it can be removed.$5$. The equivalent resistance is now $(5 + 5) \ \Omega$ in parallel with $(3 + 3) \ \Omega$,which is $10 \ \Omega$ in parallel with $6 \ \Omega$.$6$. $R_{eq} = (10 	imes 6) / (10 + 6) = 60 / 16 = 3.75 \ \Omega$.

What is the equivalent resistance between $A$ and $B$ in the given circuit (in $\Omega$)?

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