Question

A uniform metallic wire having resistance 4 $\Omega$ is bent to form a square loop (ABCD). A resistance of 2 $\Omega$ is connected between points B and D and a battery of 2 V is connected across points A and C as shown in the figure. Now the amount of current (I) is: ____.

• A. 4 A
• B. 8 A
• C. 4.5 A
• D. 2 A

Hint:

In a balanced Wheatstone bridge, the central resistor (the one connected between B and D in this case) can be completely ignored during calculation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This circuit forms a Wheatstone bridge. We must calculate the resistance of each arm of the square and determine if the bridge is balanced to find the total current $I$.

Step 2: Key Formula or Approach:
1. Total resistance of wire = 4 $\Omega$. Since it's a square, each side has resistance $r = 1\,\Omega$. 2. Analyze the network between points A and C.

Step 3: Detailed Explanation:
1. The four sides of the square are $AB = 1\,\Omega$, $BC = 1\,\Omega$, $CD = 1\,\Omega$, and $DA = 1\,\Omega$. 2. Resistance between B and D ($R_{BD}$) is 2 $\Omega$. 3. Points B and D are at the same potential because the arms $AB, BC, CD, DA$ are all equal ($1\,\Omega$). The bridge is balanced ($1/1 = 1/1$). 4. In a balanced bridge, no current flows through the central resistor ($2\,\Omega$). 5. The circuit simplifies to two parallel branches (ABC and ADC): - Branch ABC = $1 + 1 = 2\,\Omega$ - Branch ADC = $1 + 1 = 2\,\Omega$ 6. Equivalent Resistance ($R_{eq}$) = $2\,\Omega \parallel 2\,\Omega = 1\,\Omega$. 7. Total Current $I = V / R_{eq} = 2\text{ V} / 1\,\Omega = 2\text{ A}$.

Step 4: Final Answer:
The amount of current $I$ is 2 A.