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

To determine the current through FC, we need to analyze the given circuit configuration. The circuit consists of five equal resistances, each of \(2R\), connected in a specific arrangement.

Assume the points in the circuit are labeled as follows: A, D, B, C, and F.

First, identify the equivalent resistance:

• The resistances between A and D, and D and B are in series, each having a value of \(2R\):

\[ R_{AD} = R_{DB} = 2R + 2R = 4R \]

• These two equivalent resistances \(R_{AD}\) and \(R_{DB}\) are in parallel with a resistance between D and B. Considering all points:

\[ R_{ADB} = \frac{R_{AD} \cdot R_{DB}}{R_{AD} + R_{DB}} = \frac{4R \times 4R}{4R + 4R} = \frac{16R^2}{8R} = 2R \]

The resistance between C and D and C and F, being in series, is:

\[ R_{CD} = R_{CF} = 2R \]

Now, determine the total circuit resistance:

• The circuit can then be simplified to having two parallel resistances \(R_{ADB}\) and \(R_{CD}\) connected between the points A and B:

\[ R_{total} = \frac{R_{ADB} \cdot R_{CD}}{R_{ADB} + R_{CD}} = \frac{2R \times 2R}{2R + 2R} = \frac{4R^2}{4R} = R \]

Using Ohm’s Law, the total current \(I\) through the circuit is:

\[ I = \frac{V}{R_{total}} = \frac{V}{R} \]

The current divides equally, so the current through each series branch between A and D, and C and F will be halved:

\[ I_{AD} = I_{CF} = \frac{I}{2} = \frac{V}{2R} \]

Finally, the current through FC is half of the current through the branch A to D, which gives:

\[ I_{FC} = \frac{I_{AD}}{2} = \frac{V}{4R} \]

Thus, the current through FC is \(\frac{V}{4R}\).

Answer 2

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 \( \frac{V}{4R} \).