A circuit consisting of five resistors each of resistance R, forming a wheatstone bridge. What is the equivalent resistance of the circuit?

Options

A. 2 R
B. R
C. 2R/3
D. R/2

Explanation

To find the equivalent resistance of a Wheatstone bridge circuit consisting of five resistors, each with resistance R, we can follow these steps:

### Step 1: Understand the Wheatstone Bridge Configuration
In a Wheatstone bridge, we typically have four resistors forming a diamond shape, with a fifth resistor (often a variable resistor) connected across the bridge. In this case, we have five resistors, each with resistance R.

### Step 2: Identify the Series and Parallel Combinations
In a standard Wheatstone bridge:
- Two resistors are in series on one side of the bridge.
- The other two resistors are in series on the opposite side.
- The fifth resistor is connected across the bridge.

### Step 3: Calculate the Resistance of the Series Combinations
1. The two resistors on one side (let's call them R1 and R2) are in series:
\[ R_{series1} = R + R = 2R \]
2. The two resistors on the other side (let's call them R3 and R4) are also in series:
\[ R_{series2} = R + R = 2R \]

### Step 4: Analyze the Fifth Resistor
The fifth resistor (let's call it R5) is connected across the bridge. In this case, we assume that R5 is also R.

### Step 5: Combine the Series Resistances with the Fifth Resistor
Now, we have:
- One side of the bridge has a total resistance of 2R.
- The other side also has a total resistance of 2R.
- The fifth resistor (R5) is connected across these two series combinations.

### Step 6: Calculate the Equivalent Resistance
The two series combinations (2R and 2R) are in parallel with the fifth resistor (R5). The equivalent resistance (R_eq) can be calculated using the formula for resistors in parallel:
\[ \frac{1}{R_{eq}} = \frac{1}{R_{series1}} + \frac{1}{R_{series2}} + \frac{1}{R5} \]
Substituting the values:
\[ \frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{2R} + \frac{1}{R} \]
Combining these fractions:
\[ \frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{2R} + \frac{2}{2R} = \frac{1 + 1 + 2}{2R} = \frac{4}{2R} = \frac{2}{R} \]
Thus, the equivalent resistance is:
\[ R_{eq} = \frac{R}{2} \]

### Final Answer
The equivalent resistance of the Wheatstone bridge circuit with five resistors, each of resistance R, is:
\[ R_{eq} = \frac{R}{2} \]