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: ____.
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In a balanced Wheatstone bridge, the central resistor (the one connected between B and D in this case) can be completely ignored during calculation.
Step 1: Understanding the Topic:
This problem combines "Current Electricity" concepts, including resistance of wires, parallel and series combinations, and the Wheatstone Bridge principle. A square loop with a cross-connection can often be analyzed as a bridge circuit to simplify the calculation of equivalent resistance. Step 2: Key Formulas and Approach:
Resistance of a uniform wire is proportional to its length.
Balanced Wheatstone Bridge: If $R_1/R_2 = R_3/R_4$, no current flows through the central arm.
Ohm's Law: $I = V / R_{eq}$.
Step 3: Detailed Explanation:
Calculate individual side resistances: The total wire has a resistance of $4 \Omega$. Since it is bent into a square, each of the four equal sides (AB, BC, CD, DA) has a resistance of $4/4 = 1 \Omega$.
Analyze the network: The battery is connected across A and C. This creates two paths: A-B-C and A-D-C. A central resistor of $2 \Omega$ connects B and D.
Identify the Bridge: This is a Wheatstone Bridge where the arms are AB, BC, AD, and DC.
Check for balance: The ratio of resistance in the left arms is $R_{AB} / R_{AD} = 1/1$. The ratio in the right arms is $R_{BC} / R_{DC} = 1/1$. Since the ratios are equal, the bridge is balanced.
Simplify: In a balanced bridge, points B and D are at the same potential. No current flows through the $2 \Omega$ resistor. We can ignore it.
Calculate $R_{eq$:} We now have two parallel branches. Top branch (ABC) has $1+1 = 2 \Omega$. Bottom branch (ADC) has $1+1 = 2 \Omega$.
\[ R_{eq} = \frac{2 \times 2}{2 + 2} = 1 \Omega \]
Find Current: $I = V / R_{eq} = 2 \text{ V} / 1 \Omega = 2 \text{ A}$.
Step 4: Final Answer:
The total current $I$ provided by the battery is 2 A.
