Question

A Wheatstone network is balanced with respective resisto$ 5Ω. l0Ω. 20 Ω and 40 Ω in tbe P. 0. R and S arms. If a 40 Ω) resisror is connecred across S arm. Then the bridge is again balanced by connecting

• A. 20Ω across R
• B. l0Ω across P
• C. 10Ω across R
• D. 20Ω across P
• E. 10Ω across Q

Answer 1

The Correct Option is A

Wheatstone Bridge Balancing 

A Wheatstone bridge is initially balanced with \(P = 5\Omega\), \(Q = 10\Omega\), \(R = 20\Omega\), and \(S = 40\Omega\). A \(40\Omega\) resistor is connected in parallel across \(S\). We need to find what resistor to add in parallel across which arm to re-balance the bridge.

Step 1: Initial Balanced Condition

For a balanced Wheatstone bridge, the following condition holds:

\(\frac{P}{Q} = \frac{R}{S}\)

In our case: \(\frac{5}{10} = \frac{20}{40} = \frac{1}{2}\), so the bridge is initially balanced.

Step 2: Calculate the New Resistance \(S'\)

A \(40\Omega\) resistor is connected in parallel with the original \(40\Omega\) resistor \(S\). The equivalent resistance \(S'\) is:

\(\frac{1}{S'} = \frac{1}{S} + \frac{1}{40} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20}\)

\(S' = 20\Omega\)

Step 3: New Balance Condition

To re-balance the bridge, we need to find a resistor \(R_x\) to connect in parallel with \(R\) such that:

\(\frac{P}{Q} = \frac{R'}{S'}\)

Where \(R'\) is the parallel combination of the original R and \(R_x\). Since we want \(\frac{P}{Q} = \frac{5}{10} = \frac{1}{2}\) and we know \(S'=20\), we have:

\(\frac{1}{2} = \frac{R'}{20}\)

\(R'=10\)

Step 4: Calculate Resistance \(R_x\) to add in parallel with R

We need to find \(R_x\) such that the parallel combination of \(R = 20\Omega\) and \(R_x\) is equal to \(10\Omega\):

\(\frac{1}{R'} = \frac{1}{R} + \frac{1}{R_x}\)

\(\frac{1}{10} = \frac{1}{20} + \frac{1}{R_x}\)

\(\frac{1}{R_x} = \frac{1}{10} - \frac{1}{20} = \frac{2}{20} - \frac{1}{20} = \frac{1}{20}\)

\(R_x = 20\Omega\)

Step 5: Check each potential answer

The problem states that the answer should be 10Ω. It turns out that all the potential solutions are incorrect. With the correct answer (20Ω in parallel to R), the calculation is:

\(\frac{R'}{20}\) = \(\frac{\frac{20*20}{20+20}}{20}\) = \(\frac{\frac{400}{40}}{20}\) = \(\frac{10}{20}\) = \(\frac{1}{2}\)

The ratio of P to Q is \(\frac{5}{10} = \frac{1}{2}\)

Conclusion

The problem is flawed as the real answer is 20Ω across R. The bridge is balanced again by connecting 20Ω across R

Answer 2

A Wheatstone bridge is balanced when: \[ \frac{P}{Q} = \frac{R}{S} \] Given initially: \[ P = 5\ \Omega,\ Q = 10\ \Omega,\ R = 20\ \Omega,\ S = 40\ \Omega \] Check initial balance: \[ \frac{5}{10} = \frac{1}{2}, \quad \frac{20}{40} = \frac{1}{2} \Rightarrow \text{Bridge is balanced} \] Now, a 40 Ω resistor is connected in parallel across the S arm: New \( S' \) becomes: \[ \frac{1}{S'} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} \Rightarrow S' = 20\ \Omega \] To rebalance the bridge: New \( R' \) must satisfy: \[ \frac{P}{Q} = \frac{R'}{S'} \Rightarrow \frac{5}{10} = \frac{R'}{20} \Rightarrow R' = 10\ \Omega \] So, we must connect a resistor in parallel with R (which was 20 Ω) such that the equivalent becomes 10 Ω. Let the resistance to be connected be \( x \): \[ \frac{1}{R'} = \frac{1}{20} + \frac{1}{x} = \frac{1}{10} \Rightarrow \frac{1}{x} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20} \Rightarrow x = 20\ \Omega \] So, a 20 Ω resistor must be connected across R to make its equivalent resistance 10 Ω. But the option says "10 Ω across R", which implies adding a **10 Ω resistor directly across R**. Let's recheck that: Actually, that would give: \[ \frac{1}{R'} = \frac{1}{20} + \frac{1}{10} = \frac{3}{20} \Rightarrow R' = \frac{20}{3} \neq 10\ \Omega \] 
Correction: The correct resistor to be added across R should be 20 Ω, so that the total becomes 10 Ω. But the answer key shows (A) 10 Ω across R, suggesting a mismatch. Based on calculations:
Required: make R = 10 Ω using parallel resistance
Original R = 20 Ω
Required parallel resistance \( x = 20 \) Ω So the correct resistor to connect across R is 20 Ω, not 10 Ω. Thus, Answer should be: Not (A), but option matching this condition.Final Note However, as per the marked answer, (A) 10 Ω across R is considered correct, possibly assuming new total R value adjustment from a different scenario.