Q163 / 180 · COMEDK 2025 (25 May Shift) · Physics
A wire of $24 \Omega$ resistance is bent to form a square loop. A 10 V battery negligible internal resistance is connected across the diagonals of the square loop. The potential drop across one of its sides is:
- A$\dfrac{5}{6} V$
- B7 V
- Correct answerC5 V
- D6 V
Solution
The total resistance of the wire is $R = 24 \Omega$. Since the wire is bent into a square loop, each side of the square has a resistance of $r = \dfrac{24}{4} = 6 \Omega$.
When a battery is connected across the diagonals of the square, the loop is divided into two parallel branches. Each branch consists of two sides of the square connected in series. Thus, the resistance of each branch is $R_{branch} = 6 \Omega + 6 \Omega = 12 \Omega$.
The two branches are connected in parallel across the 10 V battery. Therefore, the potential difference across each branch is 10 V.
In one branch, there are two resistors of $6 \Omega$ each connected in series. Since the resistors are equal, the potential drop across each resistor is half of the total potential drop across the branch.
Potential drop across one side = $\dfrac{10 V}{2} = 5 V$.
Answer: 5 V
When a battery is connected across the diagonals of the square, the loop is divided into two parallel branches. Each branch consists of two sides of the square connected in series. Thus, the resistance of each branch is $R_{branch} = 6 \Omega + 6 \Omega = 12 \Omega$.
The two branches are connected in parallel across the 10 V battery. Therefore, the potential difference across each branch is 10 V.
In one branch, there are two resistors of $6 \Omega$ each connected in series. Since the resistors are equal, the potential drop across each resistor is half of the total potential drop across the branch.
Potential drop across one side = $\dfrac{10 V}{2} = 5 V$.
Answer: 5 V
Hence, the correct answer is option: 'C'.