Question

10. A cylindrical conductor of length I and uniform area of cross-
section A has resistance R. Another conductor of length 21 and
resistance R of the same material has area of cross section
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A
PLEASE EXPLAIN FULL ANSWER WITH STEPS.
12. If the Current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be
(a) 100%
(b) 200%
(c) 300%
(d) 400%
IN THE ABOVE QUESTION, PLEASE EXPLAIN WHY WE USED THE FORMULA P=I²R INSTEAD OF P=VI.​

Answer 1

ρ - Resistivity  - the factor in the resistance which takes into account the nature of the material is the resistivity

L - Length of the conductor

A - Area of a cross-section of the conductor.

From this relation, we observe that the length is directly proportional to the resistance and the area of cross-section is inversely proportional to the resistance.

In this case, the length of the conductor is doubled (2L) and so the resistance will be 2R. For the resistance to remain the same as R, the area of cross-section is also doubled as 2A.

Hence, the area of cross-section is 2A.

"Hope this will help you"

Answer 2

Answer:

10. A cylindrical conductor of length I and uniform area of cross-

section A has resistance R. Another conductor of length 21 and

resistance R of the same material has area of cross section

(a) A/2

(b) 3A/2

(c) 2A✔

(d) 3A

12. If the Current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

(a) 100%

(b) 200%

(c) 300%✔

(d) 400%

Explanation:

Hope this helps