Question

A wire of length ‘l’ and resistance 100 is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is ______.

Options

• 55 Ω

• 60 Ω

• 26 Ω

• 52 Ω

Answer 1

A wire of length ‘l’ and resistance 100 is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is 52 Ω.

Explanation:

Since resistance (R) is directly proportional to length, dividing a 100 Ω wire into 10 equal parts means each part has a resistance (r) of:

r = `100/10`

= 10 Ω

When n identical resistors are connected in series, the equivalent resistance (R_s) is n × r.

For the first 5 parts:

R_s = 5 × 10

= 50 Ω

When n identical resistors are connected in parallel, the equivalent resistance (R_p) is r/n.

For the first 5 parts:

R_p = `10/5`

= 2 Ω

The problem states that these two combinations (R_s and R_p) are connected in series. The total resistance (R_total) is the sum of their individual resistances:

`R_"total"` = R_s + R_p

= 50 + 2

= 52 Ω