Question

Choose the correct option:

A circular loop has a resistance of 40 Ω. Two points P and Q of the loop, which are one-quarter of the circumference apart are connected to a 24 V battery, having an internal resistance of 0.5 Ω. What is the current flowing through the battery?

Options

• 0.5A

• 1A

• 2A

• 3A

Answer 1

3A

Explanation:

For a wire of uniform cross-sectional area, the resistance of a stretch of wire is directly proportional to its length.

Since a quarter of the circular wire is contained between P and Q, the resistance of that portion of the wire is 14^th of the whole loop.

Denote the total resistance of the wire as R.

Therefore, the resistance R_eq between point P and Q can be written as:

`R_(eq) = 1/4 R`

The resistance R_r of the rest of the wire is then parallel to R_eq

Therefore the equivalent resistance R_eq between points, P and Q can be written as:

R_eq = `(R_(pq) R_r)/(R_r + R_(pq))`

Substitute 4R for R_pq 3R/4 for R_r

R_eq = `((R/4) * ((3R)/4))/(R/4 + (3R)/4)`

R_eq = `(3R)/16`

Substitute the value of R as 40 Ω.

R_eq = `((3)*(40Ω))/16`  .....(1)

Since the resistance and the internal resistance of the battery are in series, the current flowing through the battery with voltage can be determined as follows:

`I = V/(R_(eq) + r)`  .....(2)

Substitute the value of R_eq from equation (1), 0.5 Ω for r, and 24V for V in equation (2),

`I = (24V)/(7.5Ω + 0.5Ω)`

I = 3A