Question

Five resistors, each of resistance \(R\), are connected between points \(A\) and \(B\) as shown in the figure. The equivalent resistance between \(A\) and \(B\) is:

• A. \(\dfrac{3R}{8}\)
• B. \(5R\)
• C. \(\dfrac{8R}{3}\)
• D. \(2R\)

Hint:

For two equal resistors \(R\) in parallel: \[ R\parallel R=\frac{R}{2} \] A useful strategy for complex resistor networks is: \[ \boxed{\text{Simplify small parallel groups first, then combine series and parallel branches.}} \]

Answer 1

The Correct Option is A

Step 1: Simplify the upper branch. In the upper branch, one resistor \(R\) is in series with two resistors \(R\) connected in parallel. The parallel combination is: \[ R_p = \frac{R\cdot R}{R+R} = \frac{R}{2} \] Therefore, the equivalent resistance of the upper branch is: \[ R_{\text{upper}} = R+\frac{R}{2} = \frac{3R}{2} \]

Step 2: Simplify the lower branch. The lower branch consists of two resistors \(R\) connected in parallel. Thus, \[ R_{\text{lower}} = \frac{R\cdot R}{R+R} = \frac{R}{2} \]

Step 3: Combine the two branches. The upper and lower branches are connected in parallel between \(A\) and \(B\). Hence, \[ R_{\text{eq}} = \frac{R_{\text{upper}}\,R_{\text{lower}}} {R_{\text{upper}}+R_{\text{lower}}} \] \[ = \frac{\left(\frac{3R}{2}\right)\left(\frac{R}{2}\right)} {\frac{3R}{2}+\frac{R}{2}} \] \[ = \frac{\frac{3R^2}{4}} {2R} \] \[ = \frac{3R}{8} \] Therefore, \[ \boxed{R_{\text{eq}}=\frac{3R}{8}} \] Hence, the correct answer is: \[ \boxed{\text{(A)}} \]