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Question

(A) From the circuit diagram,we can simplify the branches connected in parallel between points $A$ and $B$:$1$. The top branch has two resistors of $1

Vedclass · Accepted Answer

$\frac{2}{3} \Omega$

Vedclass · Answer

(A) From the circuit diagram,we can simplify the branches connected in parallel between points $A$ and $B$:$1$. The top branch has two resistors of $1.5 \, \Omega$ and $0.5 \, \Omega$ in series,so $R_1 = 1.5 + 0.5 = 2 \, \Omega$.$2$. The second branch is a single resistor of $12 \, \Omega$.$3$. The third branch has $8 \, \Omega$ and $4 \, \Omega$ in parallel,so $R_3 = \frac{8 	imes 4}{8 + 4} = \frac{32}{12} = \frac{8}{3} \, \Omega$. Wait,looking at the diagram,the middle section is a bridge. Let's re-evaluate the parallel branches based on the simplified circuit image provided: The simplified circuit shows five parallel branches with resistances: $2 \, \Omega$,$12 \, \Omega$,$(1.6 + 2.4) = 4 \, \Omega$,$6 \, \Omega$,and $2 \, \Omega$.Now,calculate the equivalent resistance $R_{eq}$ for these parallel branches:$\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{12} + \frac{1}{4} + \frac{1}{6} + \frac{1}{2}$$\frac{1}{R_{eq}} = \frac{6 + 1 + 3 + 2 + 6}{12} = \frac{18}{12} = \frac{3}{2} \, \Omega^{-1}$Therefore,$R_{eq} = \frac{2}{3} \, \Omega$.

The equivalent resistance between $A$ and $B$ is $......$

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