Question

The value of current \(I\) in the electrical circuit as given below, when potential at \(A\) is equal to the potential at \(B\), will be _____ A. graphics[width=0.5\linewidth]{Q5.png}

• A. \(2A\)
• B. \(3A\)
• C. \(4A\)
• D. \(5A\)

Hint:

For a balanced Wheatstone bridge: \[ \frac{P}{Q}=\frac{R}{S} \] No current flows through the middle branch.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The circuit contains a bridge network with a central resistor.
The condition "potential at A is equal to the potential at B" (\(V_A = V_B\)) is the defining characteristic of a balanced Wheatstone bridge.
In a balanced state, there is no potential difference across the central branch (\(30 \Omega\)), and therefore, no current flows through it.
This simplifies the circuit by allowing us to effectively remove the central resistor from our calculations.
Step 2: Key Formula or Approach:
For a balanced bridge with resistors \(P, Q, R, S\) in the arms:
\[ \frac{P}{Q} = \frac{R}{S} \implies PS = QR \]
Total current \(I\) from the battery is found using Ohm's Law: \(I = V / R_{eq}\).
Step 3: Detailed Explanation:
Let's define the bridge resistors from the diagram:
Top-left arm \(P = 10 \Omega\).
Bottom-left arm \(R = R\) (unknown).
Top-right arm \(Q = 20 \Omega\).
Bottom-right arm \(S = 40 \Omega\).
Central arm = \(30 \Omega\) (can be ignored since \(V_A = V_B\)).
Applying the balance condition to find \(R\):
\[ \frac{10}{R} = \frac{20}{40} \implies \frac{10}{R} = \frac{1}{2} \implies R = 20 \Omega. \]
Now, let's find the equivalent resistance (\(R_{eq}\)) of the whole circuit.
The top branch consists of \(10 \Omega\) and \(20 \Omega\) in series: \(R_{top} = 10 + 20 = 30 \Omega\).
The bottom branch consists of \(20 \Omega\) and \(40 \Omega\) in series: \(R_{bot} = 20 + 40 = 60 \Omega\).
These two branches are in parallel across the \(40 V\) battery:
\[ \frac{1}{R_{eq}} = \frac{1}{30} + \frac{1}{60} \]
\[ \frac{1}{R_{eq}} = \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20} \implies R_{eq} = 20 \Omega. \]
Using Ohm's Law for the total current:
\[ I = \frac{V}{R_{eq}} = \frac{40 V}{20 \Omega} = 2 A. \]
Step 4: Final Answer:
The total current \(I\) in the circuit is 2 A.