Question

The Wheatstone bridge is balanced when \(R_3 = 144 \, \Omega\). If \(R_2\) and \(R_1\) are interchanged, the bridge balances for \(R_3 = 169 \, \Omega\). The value of \(R_4\) is: 

 

• A. \(136 \, \Omega\)
• B. \(152 \, \Omega\)
• C. \(146 \, \Omega\)
• D. \(156 \, \Omega\)

Hint:

In a Wheatstone bridge, the product of resistances across the diagonals must equal each other for the bridge to balance.

Answer 1

The Correct Option is D

For a Wheatstone bridge to be balanced, the following relation must hold:

\[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \]

We are given that the bridge is balanced when \( R_3 = 144 \, \Omega \), and when \( R_2 \) and \( R_1 \) are interchanged, the bridge balances for \( R_3 = 169 \, \Omega \). 
This implies:

\[ \frac{R_1}{R_2} = \frac{144}{R_4} \]

When \( R_1 \) and \( R_2 \) are interchanged, the relation becomes:

\[ \frac{R_2}{R_1} = \frac{169}{R_4} \]

Now, from the first equation, we can express \( R_1 \) in terms of \( R_2 \) and \( R_4 \):

\[ R_1 = \frac{R_2 \cdot 144}{R_4} \]

Substitute this into the second equation:

\[ \frac{R_2}{\frac{R_2 \cdot 144}{R_4}} = \frac{169}{R_4} \]

Simplifying:

\[ \frac{R_4}{144} = \frac{169}{R_4} \]

Now, solving for \( R_4 \):

\[ R_4^2 = 144 \times 169 \] \[ R_4 = \sqrt{144 \times 169} = \sqrt{24336} = 156 \, \Omega \]

Thus, the correct answer is Option (4), \( R_4 = 156 \, \Omega \).