The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in which the current takes different paths through a piece of material. Each of the rectangular pieces is made from a material whose resistivity is rho = 1.50 x 10-2 Ω*m, and the unit of length in the drawing is L0 = 7 cm. Each piece of material is connected to a 3.00-V battery. Find (a) the resistance and (b) the current in each case.

Resistance=0.5 ohms

Current=10 A

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