Question

In the circuit shown in figure, the equivalent resistance between A and B is _______ $\Omega$.

• A. 40 $\Omega$
• B. 20 $\Omega$
• C. 10 $\Omega$
• D. 5 $\Omega$

Answer 1

Answer: (B)

20 $\Omega$

Answer 2

1. Identify the common nodes in the circuit. The leftmost vertical wire connects the top and bottom points, forming a single node (let's call it X). Similarly, the rightmost vertical wire forms another single node (Y). Points A and B are connected by a wire, making them a single node (let's call it K). Let the node between the 25 $\Omega$ and 40 $\Omega$ resistors on the right side be C.

2. Simplify the parallel combinations:

   • The two 80 $\Omega$ resistors connected between X and K are in parallel. Their equivalent resistance is $R_{XK} = \frac{80 \times 80}{80 + 80} = 40 \, \Omega$.
   • The 25 $\Omega$ and 40 $\Omega$ resistors connected between C and Y are in parallel. Their equivalent resistance is $R_{CY} = \frac{25 \times 40}{25 + 40} = \frac{1000}{65} = \frac{200}{13} \, \Omega$.

3. Now, consider the path from X to Y through K and C. This path consists of $R_{XK}$, the 40 $\Omega$ resistor between K and C, and $R_{CY}$ in series. The equivalent resistance of this series path is $R_{series} = R_{XK} + 40 \, \Omega + R_{CY} = 40 \, \Omega + 40 \, \Omega + \frac{200}{13} \, \Omega = 80 + \frac{200}{13} = \frac{80 \times 13 + 200}{13} = \frac{1040 + 200}{13} = \frac{1240}{13} \, \Omega$.

4. Finally, the 35 $\Omega$ resistor is connected directly between X and Y, making it parallel to the $R_{series}$ path. The total equivalent resistance between X and Y is $R_{eq} = \frac{35 \times R_{series}}{35 + R_{series}} = \frac{35 \times \frac{1240}{13}}{35 + \frac{1240}{13}}$. $R_{eq} = \frac{35 \times 1240}{35 \times 13 + 1240} = \frac{43400}{455 + 1240} = \frac{43400}{1695} \, \Omega$.

5. Calculating the numerical value: $R_{eq} \approx 25.60 \, \Omega$.

Since the calculated value (25.60 $\Omega$) is not among the given options, and assuming there might be a slight approximation or a rounding intended in the problem's design, the closest option to 25.60 $\Omega$ is 20 $\Omega$. However, based on direct calculation, none of the options are exact. If this were a precise problem, none of the options would be correct. Given the context of multiple-choice questions, the closest option is sometimes the intended answer.