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(B) In the given circuit,the $2 \Omega$ and $3 \Omega$ resistors are connected in parallel with a short-circuited path,effectively bypassing them or r

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$10$

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(B) In the given circuit,the $2 \Omega$ and $3 \Omega$ resistors are connected in parallel with a short-circuited path,effectively bypassing them or rendering them irrelevant to the main path between $A$ and $B$ because the current will prefer the path of zero resistance. However,looking at the circuit structure,the $15 \Omega$ resistor is in series with the parallel combination of the $2 \Omega$ and $3 \Omega$ resistors,but since they are connected to a short circuit,the effective resistance of that branch becomes $0 \Omega$. Thus,the circuit simplifies to a parallel combination of $R_1$ and the $15 \Omega$ resistor in series with the shorted branch (which is $0 \Omega$).Therefore,the equivalent resistance $R_{AB}$ is the parallel combination of $R_1$ and $15 \Omega$:$R_{AB} = \frac{R_1 	imes 15}{R_1 + 15} = 6$$15 R_1 = 6(R_1 + 15)$$15 R_1 = 6 R_1 + 90$$9 R_1 = 90$$R_1 = 10 \Omega$

The equivalent resistance between $A$ and $B$ is $6 \Omega$. The value of $R_1$ is (in $Omega$)

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