Question

Five equal resistances each \( 2R \) are connected as shown in figure. A battery of \( V \) volts connected between A and B. Then current through FC is: 

 

• A. \( \frac{V}{4R} \)
• B. \( \frac{V}{8R} \)
• C. \( \frac{V}{R} \)
• D. \( \frac{V}{2R} \)

Hint:

For complex resistor networks, first simplify the circuit by combining resistors in series and parallel. Use equivalent resistances to reduce the circuit step by step.

Answer 1

The Correct Option is A

We are given a circuit with five equal resistances of \( 2R \) each. The resistances are connected in a configuration as shown in the image. A battery of voltage \( V \) is connected between points A and B, and we are tasked to find the current through the resistance \( FC \). 
Step 1: First, we simplify the given circuit by recognizing the series and parallel combinations of resistors. The resistors connected between A and B can be grouped into simpler series and parallel combinations. 
Step 2: The resistors between points A and B, starting from \( A \) to \( C \), can be simplified as parallel and series resistances. Let's break it into parts: - The two resistances \( 2R \) connected in parallel at \( B \) will give an equivalent resistance of \( R \). - Now, combine this \( R \) in series with the remaining \( 2R \) resistor from point A to point B. This gives an equivalent resistance of \( 3R \). 
Step 3: Now, between points B and C, the remaining resistances combine to give an equivalent resistance of \( 2R \). 
Step 4: Now, the total equivalent resistance between points A and B is the sum of the resistances \( 3R \) and \( 2R \), which is \( 5R \). The current flowing from the battery is \( I = \frac{V}{5R} \). 
Step 5: Now, the current through FC is the current flowing through the equivalent resistance between points A and B, which is \( \frac{V}{4R} \).
Thus, the current through FC is \( \frac{V}{4R} \).