Step 1: Understanding the Question:
The given circuit is a Wheatstone bridge with five identical resistors \( R \).
The voltage source is \( 16 \text{ V} \).
We need to find the current flowing specifically through the branch ACB. Step 2: Key Formula or Approach:
For a balanced Wheatstone bridge, the ratio of resistances in adjacent arms is equal:
\[ \frac{R_{AC}}{R_{AD}} = \frac{R_{CB}}{R_{DB}} \]
Given all \( R = 4 \Omega \), the bridge is balanced because \( \frac{4}{4} = \frac{4}{4} = 1 \).
In a balanced bridge, no current flows through the central branch CD. Step 3: Detailed Explanation:
Since no current flows through branch CD, the circuit effectively consists of two parallel branches:
1. Upper branch ACB: Total resistance \( R_{ACB} = R + R = 4 + 4 = 8 \Omega \).
2. Lower branch ADB: Total resistance \( R_{ADB} = R + R = 4 + 4 = 8 \Omega \).
The potential difference across the branch ACB is equal to the battery voltage, \( V = 16 \text{ V} \).
The current through branch ACB (\( I_{ACB} \)) is calculated using Ohm's Law:
\[ I_{ACB} = \frac{V}{R_{ACB}} \]
\[ I_{ACB} = \frac{16 \text{ V}}{8 \Omega} = 2 \text{ A} \]
Step 4: Final Answer:
The current through the branch ACB is \( 2 \text{ A} \).
