The resistances in the four arms of a Wheatstone network in cyclic order are `5 Omega, 2 Omega, 6 Omega` and `15 Omega` . If a current of 2.8 A enters the junction of `5 Omega` and `15 Omega`, then the current through `2 Omega` resistor is-

Options

A. 1.5 A
B. 2.8 A
C. 0.7 A
D. 2.1 A

Explanation

To solve the problem, we need to analyze the Wheatstone bridge circuit and apply the current divider rule. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the resistances in the Wheatstone bridge
The resistances in the four arms of the Wheatstone network are:
- R1 = 5 Ω
- R2 = 2 Ω
- R3 = 6 Ω
- R4 = 15 Ω

### Step 2: Calculate the equivalent resistances
The resistors R1 (5 Ω) and R2 (2 Ω) are in series, so their equivalent resistance (R_eq1) is:
\[ R_{eq1} = R1 + R2 = 5 \, \Omega + 2 \, \Omega = 7 \, \Omega \]

The resistors R3 (6 Ω) and R4 (15 Ω) are also in series, so their equivalent resistance (R_eq2) is:
\[ R_{eq2} = R3 + R4 = 6 \, \Omega + 15 \, \Omega = 21 \, \Omega \]

### Step 3: Determine how the equivalent resistances are connected
The two equivalent resistances R_eq1 (7 Ω) and R_eq2 (21 Ω) are connected in parallel.

### Step 4: Apply the current divider rule
We know that a current of 2.8 A enters the junction of the 5 Ω and 15 Ω resistors. We need to find the current through the 2 Ω resistor (which is in series with the 5 Ω resistor).

Using the current divider rule, the current through the 2 Ω resistor (I_2) can be calculated as follows:
\[ I_2 = I \times \frac{R_{eq2}}{R_{eq1} + R_{eq2}} \]
Where:
- I = 2.8 A (total current entering the junction)
- R_eq1 = 7 Ω
- R_eq2 = 21 Ω

### Step 5: Substitute the values
Substituting the values into the equation:
\[ I_2 = 2.8 \, A \times \frac{21 \, \Omega}{7 \, \Omega + 21 \, \Omega} \]
\[ I_2 = 2.8 \, A \times \frac{21 \, \Omega}{28 \, \Omega} \]

### Step 6: Simplify the equation
\[ I_2 = 2.8 \, A \times \frac{21}{28} \]
\[ I_2 = 2.8 \, A \times \frac{3}{4} \]
\[ I_2 = 2.8 \, A \times 0.75 \]
\[ I_2 = 2.1 \, A \]

### Final Answer
The current through the 2 Ω resistor is **2.1 A**.

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