Problem 94 - A Complete Resource Book in Physics for JEE Main

Question

In the arrangement shown, the magnitude of each resistance is \(2 \Omega\). The equivalent resistance between \(O\) and \(A\) is given by (A) \(\frac{14}{15} \Omega\) (B) \(\frac{7}{15} \Omega\) (C) \(\frac{4}{3} \Omega\) (D) \(\frac{5}{6} \Omega\)

Step-by-Step Solution

Answer

The equivalent resistance between points O and A can be found by first calculating the parallel combinations of resistances in the outer and inner networks. The outer network's combined resistance is \(1\Omega\), while the inner network's combined resistance is \(\frac{2}{3}\Omega\). These two resistances are connected in series, so their equivalent resistance is the sum of the individual resistances: \(R_{OA} = 1\Omega + \frac{2}{3}\Omega = \frac{5}{3}\Omega\). Therefore, the answer is \(\frac{5}{3}\Omega\), which corresponds to option (C).

Step 1: Identify resistances in series or parallel

First, we need to analyze the given circuit and find out which resistances are in series or parallel. We can see that the outer network's top resistor and middle resistor as well as the bottom resistor and middle resistor are in parallel with each other. The inner concentric network has three resistors (left, top-right, and bottom-right) also in parallel with each other.

Step 2: Calculate the parallel combinations of resistors

Next, we will calculate the parallel combinations of resistors in the outer and inner networks. The formula for calculating parallel resistors is: \[R_{parallel} = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}\] For the parallel combination of resistors in the outer network: \[R_{O} = \frac{1}{\frac{1}{2\Omega}+\frac{1}{2\Omega}} = \frac{1}{\frac{2}{2\Omega}} = 1\Omega\] For the parallel combination of resistors in the inner network: \[R_{I} = \frac{1}{\frac{1}{2\Omega}+\frac{1}{2\Omega}+\frac{1}{2\Omega}} = \frac{1}{\frac{3}{2\Omega}} =\frac{2}{3}\Omega\]

Step 3: Combine the resulting resistors in series

Now, we have simplified the network to two resistors connected in series, one obtained from the outer network (\(1\Omega\)) and another from the inner network (\(\frac{2}{3}\Omega\)). The equivalent resistance of resistors connected in series is the sum of the individual resistances. \[R_{OA} = R_O + R_I = 1\Omega + \frac{2}{3}\Omega = \frac{3}{3}\Omega + \frac{2}{3}\Omega = \frac{5}{3}\Omega\] Therefore, the equivalent resistance between points O and A is \(\frac{5}{3}\Omega\), which corresponds to option (C).