JEE Physics 2023 Question with Solution

Given: The equivalent resistance is to be found between $a$ and $b$ for the given resistor network.

Find: The value of $R_{ab}$.

For a balanced Wheatstone Bridge, the equivalent resistance between points $a$ and $b$ is given by:

$$\frac{1}{R_{ab}} = \frac{1}{16} + \frac{1}{8} + \frac{1}{8}$$

Combine the terms by finding a common denominator:

$$\frac{1}{R_{ab}} = \frac{1}{16} + \frac{2}{16} + \frac{2}{16}$$ $$\frac{1}{R_{ab}} = \frac{5}{16}$$

Take the reciprocal to calculate $R_{ab}$:

$$R_{ab} = \frac{16}{5} = 3.2 \, \Omega$$

Therefore, the equivalent resistance between points $a$ and $b$ is $3.2 \, \Omega$. The solution working gives the numerical result matching option D, although the solution incorrectly labels the correct option as C.

Given: A resistor network between $a$ and $b$ that is treated as a balanced Wheatstone bridge in the provided solution.

Find: The equivalent resistance between $a$ and $b$.

The provided solution identifies the circuit as a balanced bridge, so the network reduces to three effective parallel branches contributing:

$$\frac{1}{R_{ab}} = \frac{1}{16} + \frac{1}{8} + \frac{1}{8}$$

Now add the parallel terms:

$$\frac{1}{R_{ab}} = \frac{1 + 2 + 2}{16} = \frac{5}{16}$$

Hence,

$$R_{ab} = \frac{16}{5}$$ $$R_{ab} = 3.2 \, \Omega$$

Thus, the equivalent resistance is $3.2 \, \Omega$, so the defensible correct option is D.