Answer: (a). Resistance = 0.4286ohms and Current (I) = 7A
(b). Resistance (R) = 0.027 ohms and Current (I) = 111.1A
(c). Resistance (R) = 0.1071 ohms and Current (I) = 28A
Explanation:
From the question, given that;
ρ = 1.5*10-2ῼ.m
Lo = 7cm = 0.07m
V = 3V
From the formula R = ρL/A, where A is the area of cross section, L is the length of material and ρ is the resistivity.
(A)
L = 4Lo and A = 2Lo*Lo
R = ρL/A
R = ρ4Lo/(2Lo*Lo)
R = 2ρ /Lo = 2*1.5*10-2/0.07
R = 0.4286 ῼ
From this the current becomes;
I = V/R = 3/ 0.4286 = 6.99 = 7A
(B)
L = Lo and A = 4Lo * 2Lo
R = ρL/A
R = ρLo/ (4Lo*2Lo) after eliminating Lo from both sides we get,
R = ρ/8Lo = 1.5*10-2 / 8*0.07
R = 0.027
Current (I) = V/R = 3/0.027 = 111.1A
(C)
L = 2Lo and A = Lo * 4Lo
R = ρL/A
R = ρ2Lo/ (Lo*4Lo) eliminating Lo from both sides we get,
R = ρ/2Lo = 1.5*10-2 / 2*0.07 = 0.1071
The current becomes;
I = V/R = 3/0.1071 = 28A
