Question

The figure shows a wire of resistance 40 Ω bent to form a circle and included in an electric circuit by connecting it from the opposite ends of a diameter of the circle. The current in the circuit is:

• A. 0.125 A
• B. 0.5 A
• C. 0.25 A
• D. 0.1 A

Answer 1

Answer:

B. 0.5 A

• When a wire of uniform resistance is bent into a complete circle and connected to a circuit from the opposite ends of its diameter, it gets divided into two equal halves.

• Total Resistance of the wire = $40\ \Omega$

• Since the connection points are at the opposite ends of a diameter, the wire is split into two semicircular paths.

• Resistance of each half (R1 &R2)=40/2=20

• Because the current enters at one end of the diameter and exits from the other, these two halves are connected in parallel with each other.

• 1/R=1/R₁+1/R₂=1/20+1/20=1/10

• Using Ohm's Law I=V/R

• I =V/10

• The current passing through each individual branch of the circle, the total current divides equally because both paths have the exact same resistance ($20\ \Omega$). Therefore, the current in each half would be exactly half of the total current ($\frac{I}{2}$).

• total current ($\frac{I}{2}$).=0.5A